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Class 11_Chapter 3_Lecture_5.pdf

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Pranav Sharma

The document presents various trigonometric identities and functions as taught in a mathematics lecture by Dr. Pranav Sharma at the Maths Learning Centre, Jalandhar. It includes formulas for double and triple angles for sine, cosine, and tangent, along with examples and detailed calculations for specific angle values. Additional comparisons and proofs regarding the relationships between trigonometric functions are also provided.

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“The author makes no claim to the content's originality” Class XI Mathematics Chapter- 3 Trigonometric Functions Lecture - 5 Dr. Pranav Sharma Maths Learning Centre. Jalandhar. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures TRIGONOMETRIC FUNCTIONS OF MULTIPLES OF ANGLES (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 (𝒊𝒊) 𝐜𝐨𝐬 𝟐𝒙 = (𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙) = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙) (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝒙 (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝐬𝐢𝐧 (𝒙 + 𝒙) = 𝐬𝒊𝐧𝒙 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝒙𝐬𝒊𝐧𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙. (ii) 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬 (𝒙 + 𝒙) = 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒙 − 𝐬𝐦 ̇ 𝒙𝐬𝒊𝐧𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙. Now, 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝒊𝒏𝟐 𝒙 = (𝟏 − 𝐬𝐢𝐧𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙) = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙) Also, 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − (𝟏 − 𝐜𝐨𝐬𝟐 𝒙) = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) . 𝐜𝐨𝐬 𝟐𝒙 = (𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙) = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙) = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐭𝐚𝐧 (𝒙 + 𝒙) = 𝐭𝐚𝐧 𝒙+ 𝐭𝐚𝐧 𝒙 𝟏−( 𝐭𝐚𝐧 𝒙× 𝐭𝐚𝐧 𝒙) = 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝒙 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures The above results may be expressed as (i) (𝟏 − 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐 𝒙 (ii) (𝟏 + 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐜𝐨𝐬𝟐 𝒙 (iii) (𝟏− 𝐜𝐨𝐬 𝟐𝒙) (𝟏+ 𝐜𝐨𝐬 𝟐𝒙) = 𝐭𝐚𝐧𝟐 𝒙 If 𝒙 is not an odd multiple of 𝝅 𝟐 then prove that (𝒊) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 (𝒊𝒊) 𝐜𝐨𝐬 𝟐𝒙 = 𝟏−𝐭𝐚𝐧𝟐𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 Since 𝒙 is not an odd multiple of 𝝅 𝟐 , we have 𝐜𝐨𝐬 𝒙 ≠ 𝟎. (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬𝟐𝒙+𝐬𝐢𝐧𝟐𝒙 = ( 𝟐𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙 ) 𝟏+𝐭𝐚𝐧𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙 = 𝐜𝐨𝐬𝟐𝒙−𝐬𝐢𝐧𝟐𝒙 𝐜𝐨𝐬𝟐𝒙+𝐬𝐢𝐧𝟐𝒙 = 𝟏− 𝐬𝒊𝒏𝟐𝒙 𝐜𝐨𝐬𝟐𝒙 𝟏+ 𝐬𝐢𝐧𝟐𝒙 𝐜𝐨𝐬𝟐𝒙 = 𝟏−𝐭𝐚𝐧𝟐𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures (i) 𝐬𝐢𝐧 𝟑𝒙 = 𝟑 𝐬𝐢𝐧 𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙 (ii) 𝐜𝐨𝐬 𝟑𝒙 = 𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙 (iii) 𝐭𝐚𝐧 𝟑𝒙 = 𝟑 𝐭𝐚𝐧 𝒙−𝐭𝐚𝐧𝟑𝒙 𝟏−𝟑𝐭𝐚𝐧𝟐𝒙 (i) 𝐬𝐢𝐧 𝟑𝒙 = 𝐬𝐢𝐧 (𝟐𝒙 + 𝒙) = 𝐬𝐢𝐧 𝟐𝒙 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝟐𝒙 𝐬𝐢𝐧 𝒙 = 𝟐 𝐬𝐢𝐧 𝒙𝐜𝐨𝐬𝟐 𝒙 + 𝐜𝐨𝐬 𝟐𝒙 𝐬𝐢𝐧 𝒙 = 𝟐 𝐬𝐢𝐧 𝒙(𝟏 − 𝐬𝐢𝐧𝟐 𝒙) + (𝟏 − 𝟐𝐬𝒊𝒏𝟐 𝒙)𝐬𝒊𝐧𝒙 = 𝟑 𝐬𝐢𝐧 𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙. (ii) 𝐜𝐨𝐬 𝟑𝒙 = 𝐜𝐨𝐬 (𝟐𝒙 + 𝒙) = 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬 𝒙 − 𝐬𝐢𝐧 𝟐𝒙 𝐬𝐢𝐧 𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) 𝐜𝐨𝐬 𝒙 − 𝟐𝐬𝐢𝐧𝟐 𝒙 𝐜𝐨𝐬 𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) 𝐜𝐨𝐬 𝒙 + 𝟐(𝐜𝐨𝐬𝟐 𝒙 − 𝟏) 𝐜𝐨𝐬 𝒙 = 𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙. (iii) 𝐭𝐚𝐧 𝟑𝒙 = 𝐭𝐚𝐧 (𝟐𝒙 + 𝒙) = 𝐭𝐚𝐧 𝟐𝒙+ 𝐭𝐚𝐧 𝒙 𝟏− 𝐭𝐚𝐧 𝟐𝒙 𝐭𝐚𝐧 𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 (𝟏−𝐭𝐚𝐧𝟐𝒙) +𝐭.𝐚𝐧𝒙 𝟏−( 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝒙 ) 𝐭𝐚𝐧 𝒙 = 𝟑 𝐭𝐚𝐧 𝒙−𝐭𝐚𝐧𝟑𝒙 𝟏−𝟑𝐭𝐚𝐧𝟐𝒙 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures If 𝐬𝐢𝐧 𝒙 = − 𝟏 𝟐 and 𝝅 < 𝒙 < 𝟑𝝅 𝟐 , find the values of (i) 𝐬𝐢𝐧 𝟐𝒙, (ii) 𝐜𝐨𝐬 𝟐𝒙, (iii) 𝐭𝐚𝐧 𝟐𝒙. Since 𝒙 lies in Quadrant III, we have 𝐬𝐢𝐧 𝒙 < 𝟎, 𝐜𝐨𝐬 𝒙 < 𝟎 and 𝐭𝐚𝐧 𝒙 > 𝟎. Now, 𝐬𝐢𝐧 𝒙 = − 𝟏 𝟐 (given). 𝐜𝐨𝐬 𝒙 = −√𝟏 − 𝐬𝐢𝐧𝟐𝒙 = −√(𝟏 − 𝟏 𝟒 ) = −√ 𝟑 𝟒 = −√𝟑 𝟐 and 𝐭𝐚𝐧 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = (− 𝟏 𝟐 × 𝟐 −√𝟑 ) = 𝟏 √𝟑 . (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = 𝟐 × (− 𝟏 𝟐 ) × (− √𝟑 𝟐 ) = √𝟑 𝟐 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) = (𝟐 × 𝟑 𝟒 ) − 𝟏 = ( 𝟑 𝟐 − 𝟏) = 𝟏 𝟐 . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐬𝐢𝐧 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = ( √𝟑 𝟐 × 𝟐 𝟏 ) = √𝟑. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures If 𝐬𝐞𝐜 𝒙 = − 𝟏𝟑 𝟏𝟐 and 𝝅 𝟐 < 𝒙 < 𝝅,find the values of (i) 𝐬𝐢𝐧 𝟐𝒙, (ii) 𝐜𝐨𝐬 𝟐𝒙, (iii) 𝐭𝐚𝐧 𝟐𝒙. Since 𝒙 lies in Quadrant II, we have 𝐬𝒊𝐧𝒙 > 𝟎, 𝐜𝐨𝐬 𝒙 < 𝟎 and 𝐭𝐚𝐧 𝒙 < 𝟎. Now, 𝐬𝐞𝐜 𝒙 = −𝟏𝟑 𝟏𝟐 ⇒ 𝐜𝐨𝐬 𝒙 = 𝟏 𝐬𝐞𝐜 𝒙 = −𝟏𝟐 𝟏𝟑 . 𝐬𝐢𝐧 𝒙 = +√𝟏 − 𝐜𝐨𝐬𝟐𝒙 = +√𝟏 − 𝟏𝟒𝟒 𝟏𝟔𝟗 = +√ 𝟐𝟓 𝟏𝟔𝟗 = 𝟓 𝟏𝟑 . And, 𝐭𝐚𝐧 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = 𝟓 𝟏𝟑 × ( −𝟏𝟑 𝟏𝟐 ) = −𝟓 𝟏𝟐 . (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = {𝟐 × 𝟓 𝟏𝟑 × (−𝟏𝟐) 𝟏𝟑 } = −𝟏𝟐𝟎 𝟏𝟔𝟗 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = 𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏 = (𝟐 × 𝟏𝟒𝟒 𝟏𝟔𝟗 − 𝟏) = 𝟏𝟏𝟗 𝟏𝟔𝟗 . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐬𝐦 ̇ 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = ( −𝟏𝟐𝟎 𝟏𝟔𝟗 × 𝟏𝟔𝟗 𝟏𝟏𝟗 ) = −𝟏𝟐𝟎 𝟏𝟏𝟗 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures If 𝐭𝐚𝐧 𝒙 = −𝟑 𝟒 and 𝟑𝝅 𝟐 < 𝒙 < 𝟐𝝅, find the values of (i) 𝐬𝐢𝐧 𝟐𝒙, (ii) 𝐜𝐨𝐬 𝟐𝒙, (iii) 𝐭𝐚𝐧 𝟐𝒙. Since 𝒙 lies in Quadrant IV, we have 𝒄𝒐𝒔 𝒙 > 𝟎, 𝒔𝒊𝒏 𝒙 < 𝟎 and 𝒕𝒂𝒏 𝒙 < 𝟎. So, 𝒔𝒆𝒄 𝒙 = √𝟏 + 𝒕𝒂𝒏𝟐𝒙 = √𝟏 + 𝟗 𝟏𝟔 = 𝟓 𝟒 ⇒ 𝒄𝒐𝒔 𝒙 = 𝟒/𝟓 And, 𝐬𝐢𝐧 𝒙 = −√𝟏 − 𝐜𝐨𝐬𝟐𝒙 = −√(𝟏 − 𝟏𝟔 𝟐𝟓 ) = −√ 𝟗 𝟐𝟓 = −𝟑 𝟓 . 𝐭𝐚𝐧 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = ( −𝟑 𝟓 × 𝟓 𝟒 ) = −𝟑 𝟒 . (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = {𝟐 × (−𝟑) 𝟓 × 𝟒 𝟓 } = −𝟐𝟒 𝟐𝟓 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) = (𝟐 × 𝟏𝟔 𝟐𝟓 − 𝟏) = 𝟕 𝟐𝟓 . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐬𝒊𝒏 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = ( −𝟐𝟒 𝟐𝟓 × 𝟐𝟓 𝟕 ) = −𝟐𝟒 𝟕 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures (i) If 𝐬𝐢𝐧 𝒙 = 𝟏 𝟑 ,find the value of 𝐬𝐢𝐧 𝟑𝒙. (ii) If 𝐜𝐨𝐬 𝒙 = 𝟏 𝟐 ,find the value of 𝐜𝐨𝐬 𝟑𝒙. (i) 𝐬𝐢𝐧 𝟑𝒙 =(𝟑𝐬i𝐧𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙) = { (𝟑 × 𝟏 𝟑 ) − 𝟒 × ( 𝟏 𝟑 ) 𝟑 } = (𝟏 − 𝟒 𝟐𝟕 ) = 𝟐𝟑 𝟐𝟕 . (ii) 𝐜𝐨𝐬 𝟑𝒙 = (𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙) = { (𝟒 × 𝟏 𝟖 ) − (𝟑 ×𝟏 𝟐 )} = ( 𝟏 𝟐 − 𝟑 𝟐 ) = −𝟏. If 𝐜𝐨𝐬 𝒙 = 𝟒 𝟓 and 𝒙 is acute, find the value of 𝐭𝐚𝐧 𝟐𝒙. 𝐜𝐨𝐬 𝒙 = 𝟒 𝟓 ⇒ 𝒔𝐞𝒄𝒙 = 𝟓 𝟒 ⇒ 𝐭𝐚𝐧 𝒙 = √𝐬𝐞𝐜𝟐𝒙 − 𝟏 = √ 𝟐𝟓 𝟏𝟔 − 𝟏 = √ 𝟗 𝟏𝟔 = 𝟑 𝟒 . 𝐭𝐚𝐧 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏 − 𝐭𝐚𝐧𝟐𝒙 = 𝟐 × 𝟑 𝟒 𝟏 − 𝟗 𝟏𝟔 = ( 𝟑 𝟐 × 𝟏𝟔 𝟕 ) = 𝟐𝟒 𝟕 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures If 𝐭𝐚𝐧 𝒙 = 𝟏 𝟕 and 𝐭𝐚𝐧 𝒚 = 𝟏 𝟑 , show that 𝐜𝐨𝐬 𝟐𝒙 = 𝐬𝐢𝐧 𝟒𝒚. We have 𝐜𝐨𝐬 𝟐𝒙 = (𝟏−𝐭𝐚𝐧𝟐𝒙) (𝟏+𝐭𝐚𝐧𝟐𝒙) = (𝟏− 𝟏 𝟒𝟗 ) (𝟏+ 𝟏 𝟒𝟗 ) = ( 𝟒𝟖 𝟒𝟗 × 𝟒𝟗 𝟓𝟎 ) = 𝟐𝟒 𝟐𝟓 . 𝐬𝐢𝐧 𝟒𝒚 = 𝟐 𝐬𝐢𝐧 𝟐𝒚 𝐜𝐨𝐬 𝟐𝒚 = 𝟐 × (𝟐 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧𝟐𝒚) × (𝟏 − 𝐭𝐚𝐧𝟐 𝒚) (𝟏 + 𝐭𝐚𝐧𝟐𝒚) = 𝟐 × (𝟐 × 𝟏 𝟑 ) (𝟏 + 𝟏 𝟗 ) × ( 𝟏 − 𝟏 𝟗 𝟏 + 𝟏 𝟗 ) = ( 𝟒 𝟑 × 𝟗 𝟏𝟎 × 𝟖 𝟗 × 𝟗 𝟏𝟎 ) = 𝟐𝟒 𝟐𝟓 . So, 𝐜𝐨𝐬 𝟐𝒙 = 𝐬𝐢𝐧 𝟒𝒚. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that (i) 𝐬𝐢𝐧 𝝅 𝟔 𝐜𝐨𝐬 𝝅 𝟔 = √𝟑 𝟒 (ii) 𝐜𝐨𝐬𝟐 𝝅 𝟏𝟐 —𝒔𝒊𝒏𝟐 𝝅 𝟏𝟐 = √𝟑 𝟐 (i) 𝐬𝐢𝐧 𝝅 𝟔 𝐜𝐨𝐬 𝝅 𝟔 = 𝟏 𝟐 (𝟐 𝐬𝐢𝐧 𝝅 𝟔 𝐜𝐨𝐬 𝝅 𝟔 ) = 𝟏 𝟐 × 𝐬𝐢𝐧 (𝟐 × 𝝅 𝟔 ) [ 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = 𝐬𝐢𝐧 𝟐𝒙] = 𝟏 𝟐 𝐬𝐢𝐧 𝝅 𝟑 = ( 𝟏 𝟐 × √𝟑 𝟐 ) = √𝟑 𝟒 . (ii) 𝐜𝐨𝐬𝟐 𝝅 𝟏𝟐 − 𝐬𝐢𝐧𝟐 𝝅 𝟏𝟐 = 𝐜𝐨𝐬 (𝟐 × 𝝅 𝟏𝟐 ) [ 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙 = 𝐜𝐨𝐬 𝟐𝒙] = 𝐜𝐨𝐬 𝝅 𝟔 = 𝟑 𝟐 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that (𝒊) 𝐬𝐢𝐧 𝟐𝒙 𝟏− 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐭 𝒙 (𝒊𝒊) 𝟏− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐜𝐨𝐬 𝟐𝒙 = 𝐭𝐚𝐧𝟐 𝒙 (i) LHS = 𝐬𝐢𝐧 𝟐𝒙 𝟏− 𝐜𝐨𝐬 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝟐𝐬𝒊𝒏𝟐𝒙 [ 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙, (𝟏 − 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐 𝒙] = 𝐜𝐨𝐭 𝒙 = 𝐑𝐇𝐒. Hence, 𝐬𝐢𝐧 𝟐𝒙 𝟏− 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐭 𝒙. (ii) 𝐋𝐇𝐒 = 𝟏− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐜𝐨𝐬 𝟐𝒙 = 𝟐𝐬𝐢𝐧𝟐𝒙 𝟐𝐜𝐨𝐬𝟐𝒙 = 𝐭𝐚𝐧𝟐 𝒙 = 𝐑𝐇𝐒. Hence, 𝟏− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐜𝐨𝐬 𝟐𝒙 = 𝐭𝐚𝐧𝟐 𝒙. { 𝟏 − 𝐜𝐨𝐬𝟐𝒙 = 𝟐𝐬𝒊𝒏𝟐 𝒙 𝟏 + 𝐜𝐨𝐬𝟐𝒙 = 𝟐𝐜𝐨𝐬𝟐 𝒙 } Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟒𝒙 = 𝟏 − 𝟖𝐬𝐢𝐧𝟐 𝒙𝐜𝐨𝐬𝟐 𝒙. We have, 𝐋𝐇𝐒 = 𝐜𝐨𝐬 𝟒𝒙 = 𝐜𝐨𝐬 𝟐(𝟐𝒙) = 𝐜𝐨𝐬 𝟐𝜽, where 𝟐𝒙 = 𝜽 = 𝐜𝐨𝐬𝟐 𝜽 − 𝐬𝐢𝐧𝟐 𝜽 [ 𝐜𝐨𝐬 𝟐𝜽 = 𝐜𝐨𝐬𝟐 𝜽 − 𝐬𝐢𝐧𝟐 𝜽] = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝜽) = {𝟏 − 𝟐( 𝐬𝐢𝐧 𝟐𝒙)𝟐} = 𝟏 − 𝟐(𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙)𝟐 [ 𝐬i𝐧𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙] = 𝟏 − 𝟖𝐬𝐢𝐧𝟐 𝒙𝐜𝐨𝐬𝟐 𝒙 = 𝐑𝐇𝐒. Prove that 𝟏+ 𝐬𝐢𝐧 𝟐𝒙− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙+ 𝐜𝐨𝐬 𝟐𝒙 = 𝐭𝐚𝐧 𝒙. LHS = 𝟏+ 𝐬𝐢𝐧 𝟐𝒙− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙+ 𝐜𝐨𝐬 𝟐𝒙 = (𝟏− 𝐜𝐨𝐬 𝟐𝒙)+ 𝐬𝐢𝐧 𝟐𝒙 (𝟏+ 𝐜𝐨𝐬 𝟐𝒙)+ 𝐬𝐢𝐧 𝟐𝒙 = 𝟐𝐬𝒊𝒏𝟐𝒙+𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝟐𝐜𝐨𝐬𝟐𝒙+𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 [ (𝐥 − 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐 𝒙, (𝟏 + 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐜𝐨𝐬𝟐 𝒙; 𝐬𝐢𝐧𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙] = 𝟐𝐬𝐢𝐧 𝒙( 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙) 𝟐 𝐜𝐨𝐬 𝒙( 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙) = 𝐭𝐚𝐧 𝒙 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝟏− 𝐬𝐢𝐧 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙 = 𝐭𝐚𝐧𝟐 ( 𝝅 𝟒 − 𝒙) . We have LHS = 𝟏− 𝐬𝐢𝐧 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙 = 𝟏− 𝐜𝐨𝐬 ( 𝝅 𝟐 −𝟐𝒙) 𝟏+ 𝐜𝐨𝐬 ( 𝝅 𝟐 −𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐( 𝝅 𝟒 −𝒙) 𝟐𝐜𝐨𝐬𝟐( 𝝅 𝟒 −𝒙) = 𝐭𝐚𝐧𝟐 ( 𝝅 𝟒 − 𝒙) = 𝐑𝐇𝐒. Prove that 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 − 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 = 𝟐 𝐭𝐚𝐧 𝟐𝒙. We have LHS = 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 − 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 = ( 𝐜𝐨𝐬 𝒙+ 𝐬𝐢𝐧 𝒙)𝟐−( 𝐜𝐨𝐬 𝒙− 𝐬𝐢𝐧 𝒙)𝟐 𝐜𝐨𝐬𝟐𝒙−𝐬𝐢𝐧𝟐𝒙 = 𝟒 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬𝟐𝒙−𝐬𝐢𝐧𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝟐𝒙 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐭𝐚𝐧𝟒𝜽 = 𝟒 𝐭𝐚𝐧 𝜽(𝟏−𝐭𝐚𝐧𝟐𝜽) 𝟏−𝟔𝐭𝐚𝐧𝟐𝜽+𝐭𝐚𝐧𝟒𝜽 . We have, 𝐋𝐇𝐒 = 𝐭𝐚𝐧 𝟒𝜽 = 𝐭𝐚𝐧 𝟐(𝟐𝜽) = 𝐭𝐚𝐧 𝟐𝒙, where 𝒙 = 𝟐𝜽 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝝌 = 𝟐 𝐭𝐚𝐧 𝟐𝜽 𝟏−𝐭𝐚𝐧𝟐𝟐𝜽 = 𝟐× 𝟐 𝐭𝐚𝐧 𝜽 (𝟏−𝐭𝐚𝐧𝟐𝜽) 𝟏−( 𝟐 𝐭𝐚𝐧 𝜽 𝟏−𝐭𝐚𝐧𝟐𝜽 ) 𝟐 = 𝟒 𝐭𝐚𝐧 𝜽(𝟏−𝐭𝐚𝐧𝟐𝜽) (𝟏−𝐭𝐚𝐧𝟐𝜽)𝟐−𝟒𝐭𝐚𝐧𝟐𝜽 = 𝟒 𝐭𝐚𝐧 𝜽(𝟏−𝐭𝐚𝐧𝟐𝜽) 𝟏−𝟔𝐭𝐚𝐧𝟐𝜽+𝐭𝐚𝐧𝟒𝜽 = 𝐑𝐇𝐒. Prove that 𝐭𝐚𝐧 𝟓𝜽+ 𝐭𝐚𝐧 𝟑𝜽 𝐭𝐚𝐧 𝟓𝜽− 𝐭𝐚𝐧 𝟑𝜽 = 𝟒 𝐜𝐨𝐬 𝟐𝜽 𝐜𝐨𝐬 𝟒𝜽. We have, LHS = 𝐭𝐚𝐧 𝟓𝜽+ 𝐭𝐚𝐧 𝟑𝜽 𝐭𝐚𝐧 𝟓𝜽− 𝐭𝐚𝐧 𝟑𝜽 = ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟓𝜽 + 𝐬𝐢𝐧 𝟑𝜽 𝐜𝐨𝐬 𝟑𝜽 ) ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟓𝜽 − 𝐬𝐢𝐧 𝟑𝜽 𝐜𝐨𝐬 𝟑𝜽 ) = ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟑𝜽+ 𝐜𝐨𝐬 𝟓𝜽 𝐬𝐢𝐧 𝟑𝜽) ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟑𝜽− 𝐜𝐨𝐬 𝟓𝜽 𝐬𝐢𝐧 𝟑𝜽) = 𝐬𝐢𝐧 (𝟓𝜽+𝟑𝜽) 𝐬𝐢𝐧 (𝟓𝜽−𝟑𝜽) = 𝐬𝐢𝐧 𝟖𝜽 𝐬𝐢𝐧 𝟐𝜽 = 𝟐 𝐬𝐢𝐧 𝟒𝜽 𝐜𝐨𝐬 𝟒𝜽 𝐬𝐢𝐧 𝟐𝜽 = 𝟒 𝐬𝐢𝐧 𝟐𝜽 𝐜𝐨𝐬 𝟐𝜽 𝐜𝐨𝐬 𝟒𝜽 𝐬𝐢𝐧 𝟐𝜽 = 𝟒 𝐜𝐨𝐬 𝟐𝜽 𝐜𝐨𝐬 𝟒𝜽 Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐬𝐞𝐜 𝟖𝜽−𝟏 𝐬𝐞𝐜 𝟒𝜽−𝟏 = 𝐭𝐚𝐧 𝟖𝜽 𝐭𝐚𝐧 𝟐𝜽 . We have, LHS = 𝐬𝐞𝐜 𝟖𝜽−𝟏 𝐬𝐞𝐜 𝟒𝜽−𝟏 = ( 𝟏 𝐜𝐨𝐬 𝟖𝜽 −𝟏) ( 𝟏 𝐜𝐨𝐬 𝟒𝜽 −𝟏) = (𝟏− 𝐜𝐨𝐬 𝟖𝜽) (𝟏− 𝐜𝐨𝐬 𝟒𝜽) ⋅ 𝐜𝐨𝐬 𝟒𝜽 𝐜𝐨𝐬 𝟖𝜽 = (𝟐𝐬𝐢𝐧𝟐𝟒𝜽)( 𝐜𝐨𝐬 𝟒𝜽) (𝟐𝐬𝐢𝐧𝟐𝟐𝜽)( 𝐜𝐨𝐬 𝟖𝜽) = (𝟐 𝐬𝐢𝐧 𝟒𝜽 𝐜𝐨𝐬 𝟒𝜽)( 𝐬𝐢𝐧 𝟒𝜽) (𝟐𝐬𝐢𝐧𝟐𝟐𝜽)( 𝐜𝐨𝐬 𝟖𝜽) = ( 𝐬𝐢𝐧 𝟖𝜽)(𝟐 𝐬𝐢𝐧 𝟐𝜽 𝐜𝐨𝐬 𝟐𝜽) ( 𝐜𝐨𝐬 𝟖𝜽)(𝟐𝐬𝐢𝐧𝟐𝟐𝜽) = 𝐭𝐚𝐧 𝟖𝜽 𝐜𝐨𝐭 𝟐𝜽 = 𝐭𝐚𝐧 𝟖𝜽 𝐭𝐚𝐧 𝟐𝜽 = 𝐑𝐇𝐒. Show that √𝟐 + √𝟐 + 𝟐 𝐜𝐨𝐬 𝟒𝜽 = 𝟐 𝐜𝐨𝐬 𝜽. LHS = √𝟐 + √𝟐 + 𝟐 𝐜𝐨𝐬 𝟒𝜽 = √𝟐 + √𝟐(𝟏 + 𝐜𝐨𝐬 𝟒𝜽) = √𝟐 + √𝟒𝐜𝐨𝐬𝟐𝟐𝜽 = √𝟐 + 𝟐 𝐜𝐨𝐬 𝟐𝜽 = √𝟐(𝟏 + 𝐜𝐨𝐬 𝟐𝜽) = √𝟒𝐜𝐨𝐬𝟐𝜽 = 𝟐 𝐜𝐨𝐬 𝜽 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟓𝒙 = 𝟏𝟔𝐜𝐨𝐬𝟓 𝒙 − 𝟐𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟓 𝐜𝐨𝐬 𝒙. We have 𝐜𝐨𝐬 𝟓𝒙 = 𝐜𝐨𝐬 (𝟑𝒙 + 𝟐𝒙) = 𝐜𝐨𝐬 𝟑𝒙 𝐜𝐨𝐬 𝟐𝒙 − 𝐬𝐢𝐧 𝟑𝒙 𝐬𝐢𝐧 𝟐𝒙 = (𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙)(𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) − (𝟑 𝐬𝐢𝐧 𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙)(𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙) = (𝟖𝐜𝐨𝐬𝟓 𝒙 − 𝟏𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟑 𝐜𝐨𝐬 𝒙) − 𝟔𝐬𝐢𝐧𝟐 𝒙 𝐜𝐨𝐬 𝒙 + 𝟖𝐬𝐢𝐧𝟒 𝒙 𝐜𝐨𝐬 𝒙 = (𝟖𝐜𝐨𝐬𝟓 𝒙 − 𝟏𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟑 𝐜𝐨𝐬 𝒙) − 𝟔(𝟏 − 𝐜𝐨𝐬𝟐 𝒙) 𝐜𝐨𝐬 𝒙 +𝟖(𝟏 − 𝐜𝐨𝐬𝟐 𝒙) 𝟐 ⋅ 𝐜𝐨𝐬 𝒙 = 𝟏𝟔𝐜𝐨𝐬𝟓 𝒙 − 𝟐𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟓 𝐜𝐨𝐬 𝒙. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟔𝒙 = 𝟑𝟐𝐜𝐨𝐬𝟔 𝒙 − 𝟒𝟖𝐜𝐨𝐬𝟒 𝒙 + 𝟏𝟖𝐜𝐨𝐬𝟐 𝒙 − 𝟏. We have 𝐜𝐨𝐬 𝟔𝒙 = 𝐜𝐨𝐬 𝟐(𝟑𝒙) = 𝐜𝐨𝐬 𝟐𝜽, where 𝟑𝒙 = 𝜽 = 𝟐𝐜𝐨𝐬𝟐 𝜽 − 𝟏 = 𝟐𝐜𝐨𝐬𝟐 𝟑𝒙 − 𝟏 = 𝟐(𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙) 𝟐 − 𝟏 [ 𝐜𝐨𝐬 𝟑𝒙 = (𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙)] = 𝟐(𝟗𝐜𝐨𝐬𝟐 𝒙 + 𝟏𝟔𝐜𝐨𝐬𝟔 𝒙 − 𝟐𝟒𝐜𝐨𝐬𝟒 𝒙) − 𝟏 = 𝟑𝟐𝐜𝐨𝐬𝟔 𝒙 − 𝟒𝟖𝐜𝐨𝐬𝟒 𝒙 + 𝟏𝟖𝐜𝐨𝐬𝟐 𝒙 − 𝟏. Hence, 𝐜𝐨𝐬 𝟔𝒙 = 𝟑𝟐𝐜𝐨𝐬𝟔 𝒙 − 𝟒𝟖𝐜𝐨𝐬𝟒 𝒙 + 𝟏𝟖𝐜𝐨𝐬𝟐 𝒙 − 𝟏. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝟏. We have, 𝐜𝐨𝐭 𝟑𝒙 = 𝐜𝐨𝐭 (𝟐𝒙 + 𝒙) ⇔ 𝐜𝐨𝐭 𝟑𝒙 = 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙−𝟏 𝐜𝐨𝐭 𝟐𝒙+ 𝐜𝐨𝐭 𝒙 ⇔ 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 + 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝟏 ⇔ 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝟏. Hence, 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝟏. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Find the value of (i) 𝐬𝐢𝐧 𝟏𝟖∘ (ii) 𝐜𝐨𝐬 𝟏𝟖∘ (iii) 𝐜𝐨𝐬 𝟑𝟔∘ (iv) 𝐬𝐢𝐧 𝟑𝟔∘ (v) 𝐬𝐢𝐧 𝟕𝟐∘ (vi) 𝐜𝐨𝐬 𝟕𝟐∘ (vii) 𝐬𝐢𝐧 𝟓𝟒∘ (viii) 𝐜𝐨𝐬 𝟓𝟒∘ (i) Let 𝜽 = 𝟏𝟖∘ . Then, 𝜽 = 𝟏𝟖∘ ⇒ 𝟓𝜽 = 𝟗𝟎∘ ⇒ 𝟐𝜽 = (𝟗𝟎∘ − 𝟑𝜽) ⇒ 𝐬𝐢𝐧 𝟐𝜽 = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟑𝜽) = 𝐜𝐨𝐬 𝟑𝜽 ⇒ 𝟐 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 = 𝟒𝐜𝐨𝐬𝟑 𝜽 − 𝟑 𝐜𝐨𝐬 𝜽 ⇒ 𝟐 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 − 𝟒𝐜𝐨𝐬𝟑 𝜽 + 𝟑 𝐜𝐨𝐬 𝜽 = 𝟎 ⇒ 𝐜𝐨𝐬 𝜽(𝟐 𝐬𝐢𝐧 𝜽 − 𝟒𝐜𝐨𝐬𝟐 𝜽 + 𝟑) = 𝟎 ⇒ 𝟐 𝐬𝐢𝐧 𝜽 − 𝟒𝐜𝐨𝐬𝟐 𝜽 + 𝟑 = 𝟎 [ 𝐜𝐨𝐬 𝜽 = 𝐜𝐨𝐬 𝟏𝟖∘ ≠ 𝟎] ⇒ 𝟐 𝐬𝐢𝐧 𝜽 − 𝟒(𝟏 − 𝐬𝐢𝐧𝟐 𝜽) + 𝟑 = 𝟎 ⇒ 𝟒𝐬𝐢𝐧𝟐 𝜽 + 𝟐 𝐬𝐢𝐧 𝜽 − 𝟏 = 𝟎 ⇒ 𝐬𝐢𝐧 𝜽 = −𝟐±√𝟒+𝟏𝟔 𝟖 = (−𝟏±√𝟓) 𝟒 ⇒ 𝐬𝐢𝐧 𝜽 = (√𝟓−𝟏) 𝟒 [ 𝐬𝐢𝐧 𝜽 = 𝐬𝐢𝐧 𝟏𝟖∘ > 𝟎]. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures (ii) 𝐜𝐨𝐬𝟐 𝟏𝟖∘ = (𝟏 − 𝐬𝒊𝒏𝟐 𝟏𝟖∘ ) = {𝟏 − (√𝟓−𝟏) 𝟐 𝟏𝟔 } = {𝟏 − (𝟔−𝟐√𝟓) 𝟏𝟔 } = 𝟏𝟎+𝟐√𝟓 𝟏𝟔 ⇔ 𝐜𝐨𝐬 𝟏𝟖∘ = √𝟏𝟎 + 𝟐√𝟓 𝟒 [ 𝐜𝐨𝐬 𝟏𝟖∘ > 𝟎]. (iii) 𝐜𝐨𝐬 𝟑𝟔∘ = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝟏𝟖∘ ) = {𝟏 − 𝟐 ⋅ (√𝟓−𝟏) 𝟐 𝟏𝟔 } = {𝟏 − (𝟔−𝟐√𝟓) 𝟖 } = √𝟓+𝟏 𝟒 . (iv) 𝐬𝐢𝐧 𝟑𝟔∘ = √𝟏 − 𝐜𝐨𝐬𝟐𝟑𝟔∘ = {𝟏 − (√𝟓+𝟏) 𝟐 𝟏𝟔 } 𝟏 𝟐 = { 𝟏𝟎−𝟐√𝟓) 𝟏𝟔 } 𝟏 𝟐 = √𝟏𝟎−𝟐√𝟓 𝟒 . (v) 𝐬𝐢𝐧 𝟕𝟐∘ = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐜𝐨𝐬 𝟏𝟖∘ = √𝟏𝟎+𝟐√𝟓 𝟒 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures (vi) 𝐜𝐨𝐬 𝟕𝟐∘ = 𝐜𝐨𝐬 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐬𝐢𝐧 𝟏𝟖∘ = (√𝟓−𝟏) 𝟒 . (vii) 𝐬𝐢𝐧 𝟓𝟒∘ = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟑𝟔∘) = 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 . (viii) 𝐜𝐨𝐬 𝟓𝟒∘ = 𝐜𝐨𝐬 (𝟗𝟎∘ − 𝟑𝟔∘) = 𝐬𝐢𝐧 𝟑𝟔∘ = √𝟏𝟎−𝟐√𝟓 𝟒 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐬𝒊𝐧 𝝅 𝟏𝟎 + 𝐬𝐢𝐧 𝟏𝟑𝝅 𝟏𝟎 = −𝟏 𝟐 . We have, LHS = 𝐬𝒊𝐧 𝝅 𝟏𝟎 + 𝐬𝐢𝐧 𝟏𝟑𝝅 𝟏𝟎 = 𝐬𝐢𝐧 𝝅 𝟏𝟎 + 𝐬𝐦 ̇ (𝝅 + 𝟑𝝅 𝟏𝟎 ) = 𝐬𝐢𝐧 𝝅 𝟏𝟎 − 𝐬𝐢𝐧 𝟑𝝅 𝟏𝟎 [ 𝐬𝐢𝐧 (𝝅 + 𝜽) = − 𝐬𝐢𝐧 𝜽] = 𝐬𝐢𝐧 𝟏𝟖∘ − 𝐬𝐢𝐧 𝟓𝟒∘ = 𝐬𝐢𝐧𝟏𝟖∘ − 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟑𝟔∘) = ( 𝐬𝐢𝐧 𝟏𝟖∘ − 𝐜𝐨𝐬 𝟑𝟔∘) = { (√𝟓−𝟏) 𝟒 − (√𝟓+𝟏) 𝟒 } = −𝟏 𝟐 [⋅.⋅ 𝐬𝐢𝐧 𝟏𝟖∘ = (√𝟓−𝟏) 𝟒 and 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that (𝐜𝐨𝐬𝟐 𝟒𝟖∘ − 𝐬𝒊𝒏𝟐 𝟏𝟐∘ ) = (√𝟓+𝟏) 𝟖 . We have, LHS = (𝐜𝐨𝐬𝟐 𝟒𝟖∘ − 𝐬𝐢𝐧𝟐 𝟏𝟐∘ ) = 𝟏 𝟐 (𝟐𝐜𝐨𝐬𝟐 𝟒𝟖∘ − 𝟐𝐬𝐢𝐧𝟐 𝟏𝟐∘ ) = 𝟏 𝟐 {(𝟏 + 𝐜𝐨𝐬 𝟗𝟔∘) − (𝟏 − 𝐜𝐨𝐬 𝟐𝟒∘)} [ 𝟐𝐜𝐨𝐬𝟐 𝜽 = (𝟏 + 𝐜𝐨𝐬 𝟐𝜽) and 𝟐𝐬𝐢𝐧𝟐 𝜽 = (𝟏 − 𝐜𝐨𝐬 𝟐𝜽)] = 𝟏 𝟐 ( 𝐜𝐨𝐬 𝟗𝟔∘ + 𝐜𝐨𝐬 𝟐𝟒∘) = 𝟏 𝟐 [𝟐 𝐜𝐨𝐬 ( 𝟗𝟔∘ + 𝟐𝟒∘ 𝟐 ) 𝐜𝐨𝐬 ( 𝟗𝟔∘ − 𝟐𝟒∘ 𝟐 )] = 𝟏 𝟐 × 𝟐 𝐜𝐨𝐬 𝟔𝟎∘ 𝐜𝐨𝐬 𝟑𝟔∘ = 𝟏 𝟐 × (√𝟓+𝟏) 𝟒 [⋅ . 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] = (√𝟓 + 𝟏) 𝟖 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that (𝐬𝐢𝐧𝟐 𝟕𝟐∘ − 𝐬𝐢𝐧𝟐 𝟔𝟎∘ ) = (√𝟓−𝟏) 𝟖 . We have, 𝐋𝐇𝐒 = (𝐬𝐢𝐧𝟐 𝟕𝟐∘ − 𝐬𝒊𝒏𝟐 𝟔𝟎∘ ) = 𝟏 𝟐 (𝟐𝐬𝒊𝒏𝟐 𝟕𝟐∘ − 𝟐𝐬𝐢𝐧𝟐 𝟔𝟎∘ ) = 𝟏 𝟐 {(𝟏 − 𝐜𝐨𝐬 𝟏𝟒𝟒∘) − (𝟏 − 𝐜𝐨𝐬 𝟏𝟐𝟎∘)} [ 𝟐𝐬𝐢𝐧𝟐 𝜽 = (𝟏 − 𝐜𝐨𝐬 𝟐𝜽)] = 𝟏 𝟐 ( 𝐜𝐨𝐬 𝟏𝟐𝟎∘ − 𝐜𝐨𝐬 𝟏𝟒𝟒∘) = − 𝟏 𝟒 − 𝟏 𝟐 𝐜𝐨𝐬 𝟏𝟒𝟒∘ = − 𝟏 𝟒 − 𝟏 𝟐 𝐜𝐨𝐬 (𝟏𝟖𝟎∘ − 𝟑𝟔∘) [⋅.⋅ 𝐜𝐨𝐬 𝟏𝟐𝟎∘ = − 𝟏 𝟐 ] = − 𝟏 𝟒 + 𝟏 𝟐 𝐜𝐨𝐬 𝟑𝟔∘ = − 𝟏 𝟒 + (√𝟓+𝟏) 𝟖 [⋅.⋅ 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] = (√𝟓 − 𝟏) 𝟖 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟔∘ 𝐜𝐨𝐬 𝟒𝟐∘ 𝐜𝐨𝐬 𝟔𝟔∘ 𝐜𝐨𝐬 𝟕𝟖∘ = 𝟏 𝟏𝟔 . We have, 𝐋𝐇𝐒 = 𝐜𝐨𝐬 𝟔∘ 𝐜𝐨𝐬 𝟒𝟐∘ 𝐜𝐨𝐬 𝟔𝟔∘ 𝐜𝐨𝐬 𝟕𝟖∘ = 𝟏 𝟒 (𝟐 𝐜𝐨𝐬 𝟔𝟔∘ 𝐜𝐨𝐬 𝟔∘)(𝟐 𝐜𝐨𝐬 𝟕𝟖∘ 𝐜𝐨𝐬 𝟒𝟐∘) = 𝟏 𝟒 [ 𝐜𝐨𝐬 (𝟔𝟔∘ + 𝟔∘) + 𝐜𝐨𝐬 (𝟔𝟔∘ − 𝟔∘)] × [ 𝐜𝐨𝐬 (𝟕𝟖∘ + 𝟒𝟐∘) + 𝐜𝐨𝐬 (𝟕𝟖∘ − 𝟒𝟐∘)] = 𝟏 𝟒 ( 𝐜𝐨𝐬 𝟕𝟐∘ + 𝐜𝐨𝐬 𝟔𝟎∘)( 𝐜𝐨𝐬 𝟏𝟐𝟎∘ + 𝐜𝐨𝐬 𝟑𝟔∘) = 𝟏 𝟒 ( 𝐬𝐢𝐧 𝟏𝟖∘ + 𝟏 𝟐 ) (− 𝟏 𝟐 + 𝐜𝐨𝐬 𝟑𝟔∘) [ 𝐜𝐨𝐬 𝟕𝟐∘ = 𝐜𝐨𝐬 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐬𝐢𝐧 𝟏𝟖∘] = 𝟏 𝟒 [ (√𝟓−𝟏) 𝟒 + 𝟏 𝟐 ] [− 𝟏 𝟐 + (√𝟓+𝟏) 𝟒 ] [⋅.⋅ 𝐬𝐢𝐧 𝟏𝟖∘ = (√𝟓−𝟏) 𝟒 and 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] = 𝟏 𝟒 ⋅ (√𝟓+𝟏) 𝟒 ⋅ (√𝟓−𝟏) 𝟒 = (𝟓−𝟏) 𝟔𝟒 = 𝟒 𝟔𝟒 = 𝟏 𝟏𝟔 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Prove that 𝐬𝒊𝐧 𝝅 𝟓 𝐬𝐢𝐧 𝟐𝝅 𝟓 𝐬𝒊𝐧 𝟑𝝅 𝟓 𝐬𝐢𝐧 𝟒𝝅 𝟓 = 𝟓 𝟏𝟔 . We have, 𝐋𝐇𝐒 = 𝐬𝐢𝐧 𝝅 𝟓 𝐬𝐢𝐧 𝟐𝝅 𝟓 𝐬𝐢𝐧 𝟑𝝅 𝟓 𝐬𝐢𝐧 𝟒𝝅 𝟓 = 𝐬𝐢𝐧 𝝅 𝟓 𝐬𝐢𝐧 𝟐𝝅 𝟓 𝐬𝐢𝐧 (𝝅 − 𝟐𝝅 𝟓 ) 𝐬𝐢𝐧 (𝝅 − 𝝅 𝟓 ) = 𝐬𝐢𝐧𝟐 𝝅 𝟓 𝐬𝐢𝐧𝟐 𝟐𝝅 𝟓 [ 𝐬𝐢𝐧 (𝝅 − 𝜽) = 𝐬𝒊𝒏𝜽] = ( 𝐬𝐢𝐧 𝟑𝟔∘)𝟐 × (𝐬𝒊𝒏 𝟕𝟐∘)𝟐 = ( 𝐬𝐢𝐧 𝟑𝟔∘)𝟐 × ( 𝐜𝐨𝐬 𝟏𝟖∘)𝟐 [ 𝐬𝒊𝐧𝟕𝟐∘ = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐜𝐨𝐬 𝟏𝟖∘ ] = (𝟏𝟎−𝟐√𝟓) 𝟏𝟔 × (𝟏𝟎+𝟐√𝟓) 𝟏𝟔 = (𝟏𝟎𝟎−𝟐𝟎) (𝟏𝟔×𝟏𝟔) [⋅.⋅ 𝐬𝐢𝐧 𝟑𝟔∘ = √𝟏𝟎−𝟐√𝟓 𝟒 and 𝐜𝐨𝐬 𝟏𝟖∘ = √𝟏𝟎+𝟐√𝟓 𝟒 ] = 𝟖𝟎 (𝟏𝟔×𝟏𝟔) = 𝟓 𝟏𝟔 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
youtube.com/@MathematicsOnlineLectures Find the value of (i) 𝐬𝐢𝐧 𝟐𝟐∘ 𝟑𝟎′ (ii) 𝐜𝐨𝐬 𝟐𝟐∘ 𝟑𝟎′ (iii) 𝐭𝐚𝐧 𝟐𝟐∘ 𝟑𝟎′ (i) 𝐬𝐢𝐧𝟐 𝜽 = (𝟏− 𝐜𝐨𝐬 𝟐𝜽) 𝟐 ⇒ 𝐬𝐢𝐧𝟐(𝟐𝟐∘ 𝟑𝟎’) = (𝟏− 𝐜𝐨𝐬 𝟒𝟓∘) 𝟐 = (𝟏− 𝟏 √𝟐 ) 𝟐 = √𝟐−𝟏 𝟐√𝟐 𝐬𝐢𝐧 𝟐𝟐∘ 𝟑𝟎′ = √ √𝟐 − 𝟏 𝟐√𝟐 (ii) 𝐜𝐨𝐬𝟐 𝜽 = (𝟏+ 𝐜𝐨𝐬 𝟐𝜽) 𝟐 ⇒ 𝐜𝐨𝐬𝟐(𝟐𝟐∘ 𝟑𝟎′) = (𝟏+ 𝐜𝐨𝐬 𝟒𝟓∘) 𝟐 = (𝟏+ 𝟏 √𝟐 ) 𝟐 = (√𝟐+𝟏) 𝟐√𝟐 𝐜𝐨𝐬 𝟐𝟐∘ 𝟑𝟎′ = √ (√𝟐 + 𝟏) 𝟐√𝟐 (iii) 𝐭𝐚𝐧𝟐(𝟐𝟐∘ 𝟑𝟎′) = 𝐬𝐢𝐧𝟐(𝟐𝟐∘𝟑𝟎′) 𝐜𝐨𝐬𝟐(𝟐𝟐∘𝟑𝟎’) = (√𝟐−𝟏) (𝟐√𝟐) × (𝟐√𝟐) (√𝟐+𝟏) = √𝟐−𝟏 √𝟐+𝟏 = (√𝟐−𝟏) (√𝟐+𝟏) × (√𝟐−𝟏) (√𝟐−𝟏) = (√𝟐 − 𝟏) 𝟐 𝐭𝐚𝐧 𝟐𝟐∘ 𝟑𝟎′ = (√𝟐 − 𝟏) Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar

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Class 11_Chapter 3_Lecture_5.pdf

  • 1. “The author makes no claim to the content's originality” Class XI Mathematics Chapter- 3 Trigonometric Functions Lecture - 5 Dr. Pranav Sharma Maths Learning Centre. Jalandhar. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 2. youtube.com/@MathematicsOnlineLectures TRIGONOMETRIC FUNCTIONS OF MULTIPLES OF ANGLES (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 (𝒊𝒊) 𝐜𝐨𝐬 𝟐𝒙 = (𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙) = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙) (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝒙 (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝐬𝐢𝐧 (𝒙 + 𝒙) = 𝐬𝒊𝐧𝒙 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝒙𝐬𝒊𝐧𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙. (ii) 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬 (𝒙 + 𝒙) = 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒙 − 𝐬𝐦 ̇ 𝒙𝐬𝒊𝐧𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙. Now, 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝒊𝒏𝟐 𝒙 = (𝟏 − 𝐬𝐢𝐧𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙) = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙) Also, 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − (𝟏 − 𝐜𝐨𝐬𝟐 𝒙) = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) . 𝐜𝐨𝐬 𝟐𝒙 = (𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙) = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙) = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐭𝐚𝐧 (𝒙 + 𝒙) = 𝐭𝐚𝐧 𝒙+ 𝐭𝐚𝐧 𝒙 𝟏−( 𝐭𝐚𝐧 𝒙× 𝐭𝐚𝐧 𝒙) = 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝒙 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 3. youtube.com/@MathematicsOnlineLectures The above results may be expressed as (i) (𝟏 − 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐 𝒙 (ii) (𝟏 + 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐜𝐨𝐬𝟐 𝒙 (iii) (𝟏− 𝐜𝐨𝐬 𝟐𝒙) (𝟏+ 𝐜𝐨𝐬 𝟐𝒙) = 𝐭𝐚𝐧𝟐 𝒙 If 𝒙 is not an odd multiple of 𝝅 𝟐 then prove that (𝒊) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 (𝒊𝒊) 𝐜𝐨𝐬 𝟐𝒙 = 𝟏−𝐭𝐚𝐧𝟐𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 Since 𝒙 is not an odd multiple of 𝝅 𝟐 , we have 𝐜𝐨𝐬 𝒙 ≠ 𝟎. (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬𝟐𝒙+𝐬𝐢𝐧𝟐𝒙 = ( 𝟐𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙 ) 𝟏+𝐭𝐚𝐧𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙 = 𝐜𝐨𝐬𝟐𝒙−𝐬𝐢𝐧𝟐𝒙 𝐜𝐨𝐬𝟐𝒙+𝐬𝐢𝐧𝟐𝒙 = 𝟏− 𝐬𝒊𝒏𝟐𝒙 𝐜𝐨𝐬𝟐𝒙 𝟏+ 𝐬𝐢𝐧𝟐𝒙 𝐜𝐨𝐬𝟐𝒙 = 𝟏−𝐭𝐚𝐧𝟐𝒙 𝟏+𝐭𝐚𝐧𝟐𝒙 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 4. youtube.com/@MathematicsOnlineLectures (i) 𝐬𝐢𝐧 𝟑𝒙 = 𝟑 𝐬𝐢𝐧 𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙 (ii) 𝐜𝐨𝐬 𝟑𝒙 = 𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙 (iii) 𝐭𝐚𝐧 𝟑𝒙 = 𝟑 𝐭𝐚𝐧 𝒙−𝐭𝐚𝐧𝟑𝒙 𝟏−𝟑𝐭𝐚𝐧𝟐𝒙 (i) 𝐬𝐢𝐧 𝟑𝒙 = 𝐬𝐢𝐧 (𝟐𝒙 + 𝒙) = 𝐬𝐢𝐧 𝟐𝒙 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝟐𝒙 𝐬𝐢𝐧 𝒙 = 𝟐 𝐬𝐢𝐧 𝒙𝐜𝐨𝐬𝟐 𝒙 + 𝐜𝐨𝐬 𝟐𝒙 𝐬𝐢𝐧 𝒙 = 𝟐 𝐬𝐢𝐧 𝒙(𝟏 − 𝐬𝐢𝐧𝟐 𝒙) + (𝟏 − 𝟐𝐬𝒊𝒏𝟐 𝒙)𝐬𝒊𝐧𝒙 = 𝟑 𝐬𝐢𝐧 𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙. (ii) 𝐜𝐨𝐬 𝟑𝒙 = 𝐜𝐨𝐬 (𝟐𝒙 + 𝒙) = 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬 𝒙 − 𝐬𝐢𝐧 𝟐𝒙 𝐬𝐢𝐧 𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) 𝐜𝐨𝐬 𝒙 − 𝟐𝐬𝐢𝐧𝟐 𝒙 𝐜𝐨𝐬 𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) 𝐜𝐨𝐬 𝒙 + 𝟐(𝐜𝐨𝐬𝟐 𝒙 − 𝟏) 𝐜𝐨𝐬 𝒙 = 𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙. (iii) 𝐭𝐚𝐧 𝟑𝒙 = 𝐭𝐚𝐧 (𝟐𝒙 + 𝒙) = 𝐭𝐚𝐧 𝟐𝒙+ 𝐭𝐚𝐧 𝒙 𝟏− 𝐭𝐚𝐧 𝟐𝒙 𝐭𝐚𝐧 𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 (𝟏−𝐭𝐚𝐧𝟐𝒙) +𝐭.𝐚𝐧𝒙 𝟏−( 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝒙 ) 𝐭𝐚𝐧 𝒙 = 𝟑 𝐭𝐚𝐧 𝒙−𝐭𝐚𝐧𝟑𝒙 𝟏−𝟑𝐭𝐚𝐧𝟐𝒙 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 5. youtube.com/@MathematicsOnlineLectures If 𝐬𝐢𝐧 𝒙 = − 𝟏 𝟐 and 𝝅 < 𝒙 < 𝟑𝝅 𝟐 , find the values of (i) 𝐬𝐢𝐧 𝟐𝒙, (ii) 𝐜𝐨𝐬 𝟐𝒙, (iii) 𝐭𝐚𝐧 𝟐𝒙. Since 𝒙 lies in Quadrant III, we have 𝐬𝐢𝐧 𝒙 < 𝟎, 𝐜𝐨𝐬 𝒙 < 𝟎 and 𝐭𝐚𝐧 𝒙 > 𝟎. Now, 𝐬𝐢𝐧 𝒙 = − 𝟏 𝟐 (given). 𝐜𝐨𝐬 𝒙 = −√𝟏 − 𝐬𝐢𝐧𝟐𝒙 = −√(𝟏 − 𝟏 𝟒 ) = −√ 𝟑 𝟒 = −√𝟑 𝟐 and 𝐭𝐚𝐧 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = (− 𝟏 𝟐 × 𝟐 −√𝟑 ) = 𝟏 √𝟑 . (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = 𝟐 × (− 𝟏 𝟐 ) × (− √𝟑 𝟐 ) = √𝟑 𝟐 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) = (𝟐 × 𝟑 𝟒 ) − 𝟏 = ( 𝟑 𝟐 − 𝟏) = 𝟏 𝟐 . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐬𝐢𝐧 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = ( √𝟑 𝟐 × 𝟐 𝟏 ) = √𝟑. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 6. youtube.com/@MathematicsOnlineLectures If 𝐬𝐞𝐜 𝒙 = − 𝟏𝟑 𝟏𝟐 and 𝝅 𝟐 < 𝒙 < 𝝅,find the values of (i) 𝐬𝐢𝐧 𝟐𝒙, (ii) 𝐜𝐨𝐬 𝟐𝒙, (iii) 𝐭𝐚𝐧 𝟐𝒙. Since 𝒙 lies in Quadrant II, we have 𝐬𝒊𝐧𝒙 > 𝟎, 𝐜𝐨𝐬 𝒙 < 𝟎 and 𝐭𝐚𝐧 𝒙 < 𝟎. Now, 𝐬𝐞𝐜 𝒙 = −𝟏𝟑 𝟏𝟐 ⇒ 𝐜𝐨𝐬 𝒙 = 𝟏 𝐬𝐞𝐜 𝒙 = −𝟏𝟐 𝟏𝟑 . 𝐬𝐢𝐧 𝒙 = +√𝟏 − 𝐜𝐨𝐬𝟐𝒙 = +√𝟏 − 𝟏𝟒𝟒 𝟏𝟔𝟗 = +√ 𝟐𝟓 𝟏𝟔𝟗 = 𝟓 𝟏𝟑 . And, 𝐭𝐚𝐧 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = 𝟓 𝟏𝟑 × ( −𝟏𝟑 𝟏𝟐 ) = −𝟓 𝟏𝟐 . (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = {𝟐 × 𝟓 𝟏𝟑 × (−𝟏𝟐) 𝟏𝟑 } = −𝟏𝟐𝟎 𝟏𝟔𝟗 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = 𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏 = (𝟐 × 𝟏𝟒𝟒 𝟏𝟔𝟗 − 𝟏) = 𝟏𝟏𝟗 𝟏𝟔𝟗 . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐬𝐦 ̇ 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = ( −𝟏𝟐𝟎 𝟏𝟔𝟗 × 𝟏𝟔𝟗 𝟏𝟏𝟗 ) = −𝟏𝟐𝟎 𝟏𝟏𝟗 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 7. youtube.com/@MathematicsOnlineLectures If 𝐭𝐚𝐧 𝒙 = −𝟑 𝟒 and 𝟑𝝅 𝟐 < 𝒙 < 𝟐𝝅, find the values of (i) 𝐬𝐢𝐧 𝟐𝒙, (ii) 𝐜𝐨𝐬 𝟐𝒙, (iii) 𝐭𝐚𝐧 𝟐𝒙. Since 𝒙 lies in Quadrant IV, we have 𝒄𝒐𝒔 𝒙 > 𝟎, 𝒔𝒊𝒏 𝒙 < 𝟎 and 𝒕𝒂𝒏 𝒙 < 𝟎. So, 𝒔𝒆𝒄 𝒙 = √𝟏 + 𝒕𝒂𝒏𝟐𝒙 = √𝟏 + 𝟗 𝟏𝟔 = 𝟓 𝟒 ⇒ 𝒄𝒐𝒔 𝒙 = 𝟒/𝟓 And, 𝐬𝐢𝐧 𝒙 = −√𝟏 − 𝐜𝐨𝐬𝟐𝒙 = −√(𝟏 − 𝟏𝟔 𝟐𝟓 ) = −√ 𝟗 𝟐𝟓 = −𝟑 𝟓 . 𝐭𝐚𝐧 𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = ( −𝟑 𝟓 × 𝟓 𝟒 ) = −𝟑 𝟒 . (i) 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = {𝟐 × (−𝟑) 𝟓 × 𝟒 𝟓 } = −𝟐𝟒 𝟐𝟓 . (ii) 𝐜𝐨𝐬 𝟐𝒙 = (𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) = (𝟐 × 𝟏𝟔 𝟐𝟓 − 𝟏) = 𝟕 𝟐𝟓 . (iii) 𝐭𝐚𝐧 𝟐𝒙 = 𝐬𝒊𝒏 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = ( −𝟐𝟒 𝟐𝟓 × 𝟐𝟓 𝟕 ) = −𝟐𝟒 𝟕 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 8. youtube.com/@MathematicsOnlineLectures (i) If 𝐬𝐢𝐧 𝒙 = 𝟏 𝟑 ,find the value of 𝐬𝐢𝐧 𝟑𝒙. (ii) If 𝐜𝐨𝐬 𝒙 = 𝟏 𝟐 ,find the value of 𝐜𝐨𝐬 𝟑𝒙. (i) 𝐬𝐢𝐧 𝟑𝒙 =(𝟑𝐬i𝐧𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙) = { (𝟑 × 𝟏 𝟑 ) − 𝟒 × ( 𝟏 𝟑 ) 𝟑 } = (𝟏 − 𝟒 𝟐𝟕 ) = 𝟐𝟑 𝟐𝟕 . (ii) 𝐜𝐨𝐬 𝟑𝒙 = (𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙) = { (𝟒 × 𝟏 𝟖 ) − (𝟑 ×𝟏 𝟐 )} = ( 𝟏 𝟐 − 𝟑 𝟐 ) = −𝟏. If 𝐜𝐨𝐬 𝒙 = 𝟒 𝟓 and 𝒙 is acute, find the value of 𝐭𝐚𝐧 𝟐𝒙. 𝐜𝐨𝐬 𝒙 = 𝟒 𝟓 ⇒ 𝒔𝐞𝒄𝒙 = 𝟓 𝟒 ⇒ 𝐭𝐚𝐧 𝒙 = √𝐬𝐞𝐜𝟐𝒙 − 𝟏 = √ 𝟐𝟓 𝟏𝟔 − 𝟏 = √ 𝟗 𝟏𝟔 = 𝟑 𝟒 . 𝐭𝐚𝐧 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏 − 𝐭𝐚𝐧𝟐𝒙 = 𝟐 × 𝟑 𝟒 𝟏 − 𝟗 𝟏𝟔 = ( 𝟑 𝟐 × 𝟏𝟔 𝟕 ) = 𝟐𝟒 𝟕 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 9. youtube.com/@MathematicsOnlineLectures If 𝐭𝐚𝐧 𝒙 = 𝟏 𝟕 and 𝐭𝐚𝐧 𝒚 = 𝟏 𝟑 , show that 𝐜𝐨𝐬 𝟐𝒙 = 𝐬𝐢𝐧 𝟒𝒚. We have 𝐜𝐨𝐬 𝟐𝒙 = (𝟏−𝐭𝐚𝐧𝟐𝒙) (𝟏+𝐭𝐚𝐧𝟐𝒙) = (𝟏− 𝟏 𝟒𝟗 ) (𝟏+ 𝟏 𝟒𝟗 ) = ( 𝟒𝟖 𝟒𝟗 × 𝟒𝟗 𝟓𝟎 ) = 𝟐𝟒 𝟐𝟓 . 𝐬𝐢𝐧 𝟒𝒚 = 𝟐 𝐬𝐢𝐧 𝟐𝒚 𝐜𝐨𝐬 𝟐𝒚 = 𝟐 × (𝟐 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧𝟐𝒚) × (𝟏 − 𝐭𝐚𝐧𝟐 𝒚) (𝟏 + 𝐭𝐚𝐧𝟐𝒚) = 𝟐 × (𝟐 × 𝟏 𝟑 ) (𝟏 + 𝟏 𝟗 ) × ( 𝟏 − 𝟏 𝟗 𝟏 + 𝟏 𝟗 ) = ( 𝟒 𝟑 × 𝟗 𝟏𝟎 × 𝟖 𝟗 × 𝟗 𝟏𝟎 ) = 𝟐𝟒 𝟐𝟓 . So, 𝐜𝐨𝐬 𝟐𝒙 = 𝐬𝐢𝐧 𝟒𝒚. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 10. youtube.com/@MathematicsOnlineLectures Prove that (i) 𝐬𝐢𝐧 𝝅 𝟔 𝐜𝐨𝐬 𝝅 𝟔 = √𝟑 𝟒 (ii) 𝐜𝐨𝐬𝟐 𝝅 𝟏𝟐 —𝒔𝒊𝒏𝟐 𝝅 𝟏𝟐 = √𝟑 𝟐 (i) 𝐬𝐢𝐧 𝝅 𝟔 𝐜𝐨𝐬 𝝅 𝟔 = 𝟏 𝟐 (𝟐 𝐬𝐢𝐧 𝝅 𝟔 𝐜𝐨𝐬 𝝅 𝟔 ) = 𝟏 𝟐 × 𝐬𝐢𝐧 (𝟐 × 𝝅 𝟔 ) [ 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 = 𝐬𝐢𝐧 𝟐𝒙] = 𝟏 𝟐 𝐬𝐢𝐧 𝝅 𝟑 = ( 𝟏 𝟐 × √𝟑 𝟐 ) = √𝟑 𝟒 . (ii) 𝐜𝐨𝐬𝟐 𝝅 𝟏𝟐 − 𝐬𝐢𝐧𝟐 𝝅 𝟏𝟐 = 𝐜𝐨𝐬 (𝟐 × 𝝅 𝟏𝟐 ) [ 𝐜𝐨𝐬𝟐 𝒙 − 𝐬𝐢𝐧𝟐 𝒙 = 𝐜𝐨𝐬 𝟐𝒙] = 𝐜𝐨𝐬 𝝅 𝟔 = 𝟑 𝟐 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 11. youtube.com/@MathematicsOnlineLectures Prove that (𝒊) 𝐬𝐢𝐧 𝟐𝒙 𝟏− 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐭 𝒙 (𝒊𝒊) 𝟏− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐜𝐨𝐬 𝟐𝒙 = 𝐭𝐚𝐧𝟐 𝒙 (i) LHS = 𝐬𝐢𝐧 𝟐𝒙 𝟏− 𝐜𝐨𝐬 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝟐𝐬𝒊𝒏𝟐𝒙 [ 𝐬𝐢𝐧 𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙, (𝟏 − 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐 𝒙] = 𝐜𝐨𝐭 𝒙 = 𝐑𝐇𝐒. Hence, 𝐬𝐢𝐧 𝟐𝒙 𝟏− 𝐜𝐨𝐬 𝟐𝒙 = 𝐜𝐨𝐭 𝒙. (ii) 𝐋𝐇𝐒 = 𝟏− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐜𝐨𝐬 𝟐𝒙 = 𝟐𝐬𝐢𝐧𝟐𝒙 𝟐𝐜𝐨𝐬𝟐𝒙 = 𝐭𝐚𝐧𝟐 𝒙 = 𝐑𝐇𝐒. Hence, 𝟏− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐜𝐨𝐬 𝟐𝒙 = 𝐭𝐚𝐧𝟐 𝒙. { 𝟏 − 𝐜𝐨𝐬𝟐𝒙 = 𝟐𝐬𝒊𝒏𝟐 𝒙 𝟏 + 𝐜𝐨𝐬𝟐𝒙 = 𝟐𝐜𝐨𝐬𝟐 𝒙 } Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 12. youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟒𝒙 = 𝟏 − 𝟖𝐬𝐢𝐧𝟐 𝒙𝐜𝐨𝐬𝟐 𝒙. We have, 𝐋𝐇𝐒 = 𝐜𝐨𝐬 𝟒𝒙 = 𝐜𝐨𝐬 𝟐(𝟐𝒙) = 𝐜𝐨𝐬 𝟐𝜽, where 𝟐𝒙 = 𝜽 = 𝐜𝐨𝐬𝟐 𝜽 − 𝐬𝐢𝐧𝟐 𝜽 [ 𝐜𝐨𝐬 𝟐𝜽 = 𝐜𝐨𝐬𝟐 𝜽 − 𝐬𝐢𝐧𝟐 𝜽] = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝜽) = {𝟏 − 𝟐( 𝐬𝐢𝐧 𝟐𝒙)𝟐} = 𝟏 − 𝟐(𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙)𝟐 [ 𝐬i𝐧𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙] = 𝟏 − 𝟖𝐬𝐢𝐧𝟐 𝒙𝐜𝐨𝐬𝟐 𝒙 = 𝐑𝐇𝐒. Prove that 𝟏+ 𝐬𝐢𝐧 𝟐𝒙− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙+ 𝐜𝐨𝐬 𝟐𝒙 = 𝐭𝐚𝐧 𝒙. LHS = 𝟏+ 𝐬𝐢𝐧 𝟐𝒙− 𝐜𝐨𝐬 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙+ 𝐜𝐨𝐬 𝟐𝒙 = (𝟏− 𝐜𝐨𝐬 𝟐𝒙)+ 𝐬𝐢𝐧 𝟐𝒙 (𝟏+ 𝐜𝐨𝐬 𝟐𝒙)+ 𝐬𝐢𝐧 𝟐𝒙 = 𝟐𝐬𝒊𝒏𝟐𝒙+𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝟐𝐜𝐨𝐬𝟐𝒙+𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 [ (𝐥 − 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐 𝒙, (𝟏 + 𝐜𝐨𝐬 𝟐𝒙) = 𝟐𝐜𝐨𝐬𝟐 𝒙; 𝐬𝐢𝐧𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙] = 𝟐𝐬𝐢𝐧 𝒙( 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙) 𝟐 𝐜𝐨𝐬 𝒙( 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙) = 𝐭𝐚𝐧 𝒙 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 13. youtube.com/@MathematicsOnlineLectures Prove that 𝟏− 𝐬𝐢𝐧 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙 = 𝐭𝐚𝐧𝟐 ( 𝝅 𝟒 − 𝒙) . We have LHS = 𝟏− 𝐬𝐢𝐧 𝟐𝒙 𝟏+ 𝐬𝐢𝐧 𝟐𝒙 = 𝟏− 𝐜𝐨𝐬 ( 𝝅 𝟐 −𝟐𝒙) 𝟏+ 𝐜𝐨𝐬 ( 𝝅 𝟐 −𝟐𝒙) = 𝟐𝐬𝐢𝐧𝟐( 𝝅 𝟒 −𝒙) 𝟐𝐜𝐨𝐬𝟐( 𝝅 𝟒 −𝒙) = 𝐭𝐚𝐧𝟐 ( 𝝅 𝟒 − 𝒙) = 𝐑𝐇𝐒. Prove that 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 − 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 = 𝟐 𝐭𝐚𝐧 𝟐𝒙. We have LHS = 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 − 𝐜𝐨𝐬 𝒙−𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝒙+𝐬𝒊𝒏 𝒙 = ( 𝐜𝐨𝐬 𝒙+ 𝐬𝐢𝐧 𝒙)𝟐−( 𝐜𝐨𝐬 𝒙− 𝐬𝐢𝐧 𝒙)𝟐 𝐜𝐨𝐬𝟐𝒙−𝐬𝐢𝐧𝟐𝒙 = 𝟒 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬𝟐𝒙−𝐬𝐢𝐧𝟐𝒙 = 𝟐 𝐬𝐢𝐧 𝟐𝒙 𝐜𝐨𝐬 𝟐𝒙 = 𝟐 𝐭𝐚𝐧 𝟐𝒙 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 14. youtube.com/@MathematicsOnlineLectures Prove that 𝐭𝐚𝐧𝟒𝜽 = 𝟒 𝐭𝐚𝐧 𝜽(𝟏−𝐭𝐚𝐧𝟐𝜽) 𝟏−𝟔𝐭𝐚𝐧𝟐𝜽+𝐭𝐚𝐧𝟒𝜽 . We have, 𝐋𝐇𝐒 = 𝐭𝐚𝐧 𝟒𝜽 = 𝐭𝐚𝐧 𝟐(𝟐𝜽) = 𝐭𝐚𝐧 𝟐𝒙, where 𝒙 = 𝟐𝜽 = 𝟐 𝐭𝐚𝐧 𝒙 𝟏−𝐭𝐚𝐧𝟐𝝌 = 𝟐 𝐭𝐚𝐧 𝟐𝜽 𝟏−𝐭𝐚𝐧𝟐𝟐𝜽 = 𝟐× 𝟐 𝐭𝐚𝐧 𝜽 (𝟏−𝐭𝐚𝐧𝟐𝜽) 𝟏−( 𝟐 𝐭𝐚𝐧 𝜽 𝟏−𝐭𝐚𝐧𝟐𝜽 ) 𝟐 = 𝟒 𝐭𝐚𝐧 𝜽(𝟏−𝐭𝐚𝐧𝟐𝜽) (𝟏−𝐭𝐚𝐧𝟐𝜽)𝟐−𝟒𝐭𝐚𝐧𝟐𝜽 = 𝟒 𝐭𝐚𝐧 𝜽(𝟏−𝐭𝐚𝐧𝟐𝜽) 𝟏−𝟔𝐭𝐚𝐧𝟐𝜽+𝐭𝐚𝐧𝟒𝜽 = 𝐑𝐇𝐒. Prove that 𝐭𝐚𝐧 𝟓𝜽+ 𝐭𝐚𝐧 𝟑𝜽 𝐭𝐚𝐧 𝟓𝜽− 𝐭𝐚𝐧 𝟑𝜽 = 𝟒 𝐜𝐨𝐬 𝟐𝜽 𝐜𝐨𝐬 𝟒𝜽. We have, LHS = 𝐭𝐚𝐧 𝟓𝜽+ 𝐭𝐚𝐧 𝟑𝜽 𝐭𝐚𝐧 𝟓𝜽− 𝐭𝐚𝐧 𝟑𝜽 = ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟓𝜽 + 𝐬𝐢𝐧 𝟑𝜽 𝐜𝐨𝐬 𝟑𝜽 ) ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟓𝜽 − 𝐬𝐢𝐧 𝟑𝜽 𝐜𝐨𝐬 𝟑𝜽 ) = ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟑𝜽+ 𝐜𝐨𝐬 𝟓𝜽 𝐬𝐢𝐧 𝟑𝜽) ( 𝐬𝐢𝐧 𝟓𝜽 𝐜𝐨𝐬 𝟑𝜽− 𝐜𝐨𝐬 𝟓𝜽 𝐬𝐢𝐧 𝟑𝜽) = 𝐬𝐢𝐧 (𝟓𝜽+𝟑𝜽) 𝐬𝐢𝐧 (𝟓𝜽−𝟑𝜽) = 𝐬𝐢𝐧 𝟖𝜽 𝐬𝐢𝐧 𝟐𝜽 = 𝟐 𝐬𝐢𝐧 𝟒𝜽 𝐜𝐨𝐬 𝟒𝜽 𝐬𝐢𝐧 𝟐𝜽 = 𝟒 𝐬𝐢𝐧 𝟐𝜽 𝐜𝐨𝐬 𝟐𝜽 𝐜𝐨𝐬 𝟒𝜽 𝐬𝐢𝐧 𝟐𝜽 = 𝟒 𝐜𝐨𝐬 𝟐𝜽 𝐜𝐨𝐬 𝟒𝜽 Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 15. youtube.com/@MathematicsOnlineLectures Prove that 𝐬𝐞𝐜 𝟖𝜽−𝟏 𝐬𝐞𝐜 𝟒𝜽−𝟏 = 𝐭𝐚𝐧 𝟖𝜽 𝐭𝐚𝐧 𝟐𝜽 . We have, LHS = 𝐬𝐞𝐜 𝟖𝜽−𝟏 𝐬𝐞𝐜 𝟒𝜽−𝟏 = ( 𝟏 𝐜𝐨𝐬 𝟖𝜽 −𝟏) ( 𝟏 𝐜𝐨𝐬 𝟒𝜽 −𝟏) = (𝟏− 𝐜𝐨𝐬 𝟖𝜽) (𝟏− 𝐜𝐨𝐬 𝟒𝜽) ⋅ 𝐜𝐨𝐬 𝟒𝜽 𝐜𝐨𝐬 𝟖𝜽 = (𝟐𝐬𝐢𝐧𝟐𝟒𝜽)( 𝐜𝐨𝐬 𝟒𝜽) (𝟐𝐬𝐢𝐧𝟐𝟐𝜽)( 𝐜𝐨𝐬 𝟖𝜽) = (𝟐 𝐬𝐢𝐧 𝟒𝜽 𝐜𝐨𝐬 𝟒𝜽)( 𝐬𝐢𝐧 𝟒𝜽) (𝟐𝐬𝐢𝐧𝟐𝟐𝜽)( 𝐜𝐨𝐬 𝟖𝜽) = ( 𝐬𝐢𝐧 𝟖𝜽)(𝟐 𝐬𝐢𝐧 𝟐𝜽 𝐜𝐨𝐬 𝟐𝜽) ( 𝐜𝐨𝐬 𝟖𝜽)(𝟐𝐬𝐢𝐧𝟐𝟐𝜽) = 𝐭𝐚𝐧 𝟖𝜽 𝐜𝐨𝐭 𝟐𝜽 = 𝐭𝐚𝐧 𝟖𝜽 𝐭𝐚𝐧 𝟐𝜽 = 𝐑𝐇𝐒. Show that √𝟐 + √𝟐 + 𝟐 𝐜𝐨𝐬 𝟒𝜽 = 𝟐 𝐜𝐨𝐬 𝜽. LHS = √𝟐 + √𝟐 + 𝟐 𝐜𝐨𝐬 𝟒𝜽 = √𝟐 + √𝟐(𝟏 + 𝐜𝐨𝐬 𝟒𝜽) = √𝟐 + √𝟒𝐜𝐨𝐬𝟐𝟐𝜽 = √𝟐 + 𝟐 𝐜𝐨𝐬 𝟐𝜽 = √𝟐(𝟏 + 𝐜𝐨𝐬 𝟐𝜽) = √𝟒𝐜𝐨𝐬𝟐𝜽 = 𝟐 𝐜𝐨𝐬 𝜽 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 16. youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟓𝒙 = 𝟏𝟔𝐜𝐨𝐬𝟓 𝒙 − 𝟐𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟓 𝐜𝐨𝐬 𝒙. We have 𝐜𝐨𝐬 𝟓𝒙 = 𝐜𝐨𝐬 (𝟑𝒙 + 𝟐𝒙) = 𝐜𝐨𝐬 𝟑𝒙 𝐜𝐨𝐬 𝟐𝒙 − 𝐬𝐢𝐧 𝟑𝒙 𝐬𝐢𝐧 𝟐𝒙 = (𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙)(𝟐𝐜𝐨𝐬𝟐 𝒙 − 𝟏) − (𝟑 𝐬𝐢𝐧 𝒙 − 𝟒𝐬𝐢𝐧𝟑 𝒙)(𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙) = (𝟖𝐜𝐨𝐬𝟓 𝒙 − 𝟏𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟑 𝐜𝐨𝐬 𝒙) − 𝟔𝐬𝐢𝐧𝟐 𝒙 𝐜𝐨𝐬 𝒙 + 𝟖𝐬𝐢𝐧𝟒 𝒙 𝐜𝐨𝐬 𝒙 = (𝟖𝐜𝐨𝐬𝟓 𝒙 − 𝟏𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟑 𝐜𝐨𝐬 𝒙) − 𝟔(𝟏 − 𝐜𝐨𝐬𝟐 𝒙) 𝐜𝐨𝐬 𝒙 +𝟖(𝟏 − 𝐜𝐨𝐬𝟐 𝒙) 𝟐 ⋅ 𝐜𝐨𝐬 𝒙 = 𝟏𝟔𝐜𝐨𝐬𝟓 𝒙 − 𝟐𝟎𝐜𝐨𝐬𝟑 𝒙 + 𝟓 𝐜𝐨𝐬 𝒙. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 17. youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟔𝒙 = 𝟑𝟐𝐜𝐨𝐬𝟔 𝒙 − 𝟒𝟖𝐜𝐨𝐬𝟒 𝒙 + 𝟏𝟖𝐜𝐨𝐬𝟐 𝒙 − 𝟏. We have 𝐜𝐨𝐬 𝟔𝒙 = 𝐜𝐨𝐬 𝟐(𝟑𝒙) = 𝐜𝐨𝐬 𝟐𝜽, where 𝟑𝒙 = 𝜽 = 𝟐𝐜𝐨𝐬𝟐 𝜽 − 𝟏 = 𝟐𝐜𝐨𝐬𝟐 𝟑𝒙 − 𝟏 = 𝟐(𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙) 𝟐 − 𝟏 [ 𝐜𝐨𝐬 𝟑𝒙 = (𝟒𝐜𝐨𝐬𝟑 𝒙 − 𝟑 𝐜𝐨𝐬 𝒙)] = 𝟐(𝟗𝐜𝐨𝐬𝟐 𝒙 + 𝟏𝟔𝐜𝐨𝐬𝟔 𝒙 − 𝟐𝟒𝐜𝐨𝐬𝟒 𝒙) − 𝟏 = 𝟑𝟐𝐜𝐨𝐬𝟔 𝒙 − 𝟒𝟖𝐜𝐨𝐬𝟒 𝒙 + 𝟏𝟖𝐜𝐨𝐬𝟐 𝒙 − 𝟏. Hence, 𝐜𝐨𝐬 𝟔𝒙 = 𝟑𝟐𝐜𝐨𝐬𝟔 𝒙 − 𝟒𝟖𝐜𝐨𝐬𝟒 𝒙 + 𝟏𝟖𝐜𝐨𝐬𝟐 𝒙 − 𝟏. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 18. youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝟏. We have, 𝐜𝐨𝐭 𝟑𝒙 = 𝐜𝐨𝐭 (𝟐𝒙 + 𝒙) ⇔ 𝐜𝐨𝐭 𝟑𝒙 = 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙−𝟏 𝐜𝐨𝐭 𝟐𝒙+ 𝐜𝐨𝐭 𝒙 ⇔ 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 + 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝟏 ⇔ 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝟏. Hence, 𝐜𝐨𝐭 𝟐𝒙 𝐜𝐨𝐭 𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝟐𝒙 − 𝐜𝐨𝐭 𝟑𝒙 𝐜𝐨𝐭 𝒙 = 𝟏. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 19. youtube.com/@MathematicsOnlineLectures Find the value of (i) 𝐬𝐢𝐧 𝟏𝟖∘ (ii) 𝐜𝐨𝐬 𝟏𝟖∘ (iii) 𝐜𝐨𝐬 𝟑𝟔∘ (iv) 𝐬𝐢𝐧 𝟑𝟔∘ (v) 𝐬𝐢𝐧 𝟕𝟐∘ (vi) 𝐜𝐨𝐬 𝟕𝟐∘ (vii) 𝐬𝐢𝐧 𝟓𝟒∘ (viii) 𝐜𝐨𝐬 𝟓𝟒∘ (i) Let 𝜽 = 𝟏𝟖∘ . Then, 𝜽 = 𝟏𝟖∘ ⇒ 𝟓𝜽 = 𝟗𝟎∘ ⇒ 𝟐𝜽 = (𝟗𝟎∘ − 𝟑𝜽) ⇒ 𝐬𝐢𝐧 𝟐𝜽 = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟑𝜽) = 𝐜𝐨𝐬 𝟑𝜽 ⇒ 𝟐 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 = 𝟒𝐜𝐨𝐬𝟑 𝜽 − 𝟑 𝐜𝐨𝐬 𝜽 ⇒ 𝟐 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 − 𝟒𝐜𝐨𝐬𝟑 𝜽 + 𝟑 𝐜𝐨𝐬 𝜽 = 𝟎 ⇒ 𝐜𝐨𝐬 𝜽(𝟐 𝐬𝐢𝐧 𝜽 − 𝟒𝐜𝐨𝐬𝟐 𝜽 + 𝟑) = 𝟎 ⇒ 𝟐 𝐬𝐢𝐧 𝜽 − 𝟒𝐜𝐨𝐬𝟐 𝜽 + 𝟑 = 𝟎 [ 𝐜𝐨𝐬 𝜽 = 𝐜𝐨𝐬 𝟏𝟖∘ ≠ 𝟎] ⇒ 𝟐 𝐬𝐢𝐧 𝜽 − 𝟒(𝟏 − 𝐬𝐢𝐧𝟐 𝜽) + 𝟑 = 𝟎 ⇒ 𝟒𝐬𝐢𝐧𝟐 𝜽 + 𝟐 𝐬𝐢𝐧 𝜽 − 𝟏 = 𝟎 ⇒ 𝐬𝐢𝐧 𝜽 = −𝟐±√𝟒+𝟏𝟔 𝟖 = (−𝟏±√𝟓) 𝟒 ⇒ 𝐬𝐢𝐧 𝜽 = (√𝟓−𝟏) 𝟒 [ 𝐬𝐢𝐧 𝜽 = 𝐬𝐢𝐧 𝟏𝟖∘ > 𝟎]. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 20. youtube.com/@MathematicsOnlineLectures (ii) 𝐜𝐨𝐬𝟐 𝟏𝟖∘ = (𝟏 − 𝐬𝒊𝒏𝟐 𝟏𝟖∘ ) = {𝟏 − (√𝟓−𝟏) 𝟐 𝟏𝟔 } = {𝟏 − (𝟔−𝟐√𝟓) 𝟏𝟔 } = 𝟏𝟎+𝟐√𝟓 𝟏𝟔 ⇔ 𝐜𝐨𝐬 𝟏𝟖∘ = √𝟏𝟎 + 𝟐√𝟓 𝟒 [ 𝐜𝐨𝐬 𝟏𝟖∘ > 𝟎]. (iii) 𝐜𝐨𝐬 𝟑𝟔∘ = (𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝟏𝟖∘ ) = {𝟏 − 𝟐 ⋅ (√𝟓−𝟏) 𝟐 𝟏𝟔 } = {𝟏 − (𝟔−𝟐√𝟓) 𝟖 } = √𝟓+𝟏 𝟒 . (iv) 𝐬𝐢𝐧 𝟑𝟔∘ = √𝟏 − 𝐜𝐨𝐬𝟐𝟑𝟔∘ = {𝟏 − (√𝟓+𝟏) 𝟐 𝟏𝟔 } 𝟏 𝟐 = { 𝟏𝟎−𝟐√𝟓) 𝟏𝟔 } 𝟏 𝟐 = √𝟏𝟎−𝟐√𝟓 𝟒 . (v) 𝐬𝐢𝐧 𝟕𝟐∘ = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐜𝐨𝐬 𝟏𝟖∘ = √𝟏𝟎+𝟐√𝟓 𝟒 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 21. youtube.com/@MathematicsOnlineLectures (vi) 𝐜𝐨𝐬 𝟕𝟐∘ = 𝐜𝐨𝐬 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐬𝐢𝐧 𝟏𝟖∘ = (√𝟓−𝟏) 𝟒 . (vii) 𝐬𝐢𝐧 𝟓𝟒∘ = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟑𝟔∘) = 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 . (viii) 𝐜𝐨𝐬 𝟓𝟒∘ = 𝐜𝐨𝐬 (𝟗𝟎∘ − 𝟑𝟔∘) = 𝐬𝐢𝐧 𝟑𝟔∘ = √𝟏𝟎−𝟐√𝟓 𝟒 . Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 22. youtube.com/@MathematicsOnlineLectures Prove that 𝐬𝒊𝐧 𝝅 𝟏𝟎 + 𝐬𝐢𝐧 𝟏𝟑𝝅 𝟏𝟎 = −𝟏 𝟐 . We have, LHS = 𝐬𝒊𝐧 𝝅 𝟏𝟎 + 𝐬𝐢𝐧 𝟏𝟑𝝅 𝟏𝟎 = 𝐬𝐢𝐧 𝝅 𝟏𝟎 + 𝐬𝐦 ̇ (𝝅 + 𝟑𝝅 𝟏𝟎 ) = 𝐬𝐢𝐧 𝝅 𝟏𝟎 − 𝐬𝐢𝐧 𝟑𝝅 𝟏𝟎 [ 𝐬𝐢𝐧 (𝝅 + 𝜽) = − 𝐬𝐢𝐧 𝜽] = 𝐬𝐢𝐧 𝟏𝟖∘ − 𝐬𝐢𝐧 𝟓𝟒∘ = 𝐬𝐢𝐧𝟏𝟖∘ − 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟑𝟔∘) = ( 𝐬𝐢𝐧 𝟏𝟖∘ − 𝐜𝐨𝐬 𝟑𝟔∘) = { (√𝟓−𝟏) 𝟒 − (√𝟓+𝟏) 𝟒 } = −𝟏 𝟐 [⋅.⋅ 𝐬𝐢𝐧 𝟏𝟖∘ = (√𝟓−𝟏) 𝟒 and 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 23. youtube.com/@MathematicsOnlineLectures Prove that (𝐜𝐨𝐬𝟐 𝟒𝟖∘ − 𝐬𝒊𝒏𝟐 𝟏𝟐∘ ) = (√𝟓+𝟏) 𝟖 . We have, LHS = (𝐜𝐨𝐬𝟐 𝟒𝟖∘ − 𝐬𝐢𝐧𝟐 𝟏𝟐∘ ) = 𝟏 𝟐 (𝟐𝐜𝐨𝐬𝟐 𝟒𝟖∘ − 𝟐𝐬𝐢𝐧𝟐 𝟏𝟐∘ ) = 𝟏 𝟐 {(𝟏 + 𝐜𝐨𝐬 𝟗𝟔∘) − (𝟏 − 𝐜𝐨𝐬 𝟐𝟒∘)} [ 𝟐𝐜𝐨𝐬𝟐 𝜽 = (𝟏 + 𝐜𝐨𝐬 𝟐𝜽) and 𝟐𝐬𝐢𝐧𝟐 𝜽 = (𝟏 − 𝐜𝐨𝐬 𝟐𝜽)] = 𝟏 𝟐 ( 𝐜𝐨𝐬 𝟗𝟔∘ + 𝐜𝐨𝐬 𝟐𝟒∘) = 𝟏 𝟐 [𝟐 𝐜𝐨𝐬 ( 𝟗𝟔∘ + 𝟐𝟒∘ 𝟐 ) 𝐜𝐨𝐬 ( 𝟗𝟔∘ − 𝟐𝟒∘ 𝟐 )] = 𝟏 𝟐 × 𝟐 𝐜𝐨𝐬 𝟔𝟎∘ 𝐜𝐨𝐬 𝟑𝟔∘ = 𝟏 𝟐 × (√𝟓+𝟏) 𝟒 [⋅ . 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] = (√𝟓 + 𝟏) 𝟖 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 24. youtube.com/@MathematicsOnlineLectures Prove that (𝐬𝐢𝐧𝟐 𝟕𝟐∘ − 𝐬𝐢𝐧𝟐 𝟔𝟎∘ ) = (√𝟓−𝟏) 𝟖 . We have, 𝐋𝐇𝐒 = (𝐬𝐢𝐧𝟐 𝟕𝟐∘ − 𝐬𝒊𝒏𝟐 𝟔𝟎∘ ) = 𝟏 𝟐 (𝟐𝐬𝒊𝒏𝟐 𝟕𝟐∘ − 𝟐𝐬𝐢𝐧𝟐 𝟔𝟎∘ ) = 𝟏 𝟐 {(𝟏 − 𝐜𝐨𝐬 𝟏𝟒𝟒∘) − (𝟏 − 𝐜𝐨𝐬 𝟏𝟐𝟎∘)} [ 𝟐𝐬𝐢𝐧𝟐 𝜽 = (𝟏 − 𝐜𝐨𝐬 𝟐𝜽)] = 𝟏 𝟐 ( 𝐜𝐨𝐬 𝟏𝟐𝟎∘ − 𝐜𝐨𝐬 𝟏𝟒𝟒∘) = − 𝟏 𝟒 − 𝟏 𝟐 𝐜𝐨𝐬 𝟏𝟒𝟒∘ = − 𝟏 𝟒 − 𝟏 𝟐 𝐜𝐨𝐬 (𝟏𝟖𝟎∘ − 𝟑𝟔∘) [⋅.⋅ 𝐜𝐨𝐬 𝟏𝟐𝟎∘ = − 𝟏 𝟐 ] = − 𝟏 𝟒 + 𝟏 𝟐 𝐜𝐨𝐬 𝟑𝟔∘ = − 𝟏 𝟒 + (√𝟓+𝟏) 𝟖 [⋅.⋅ 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] = (√𝟓 − 𝟏) 𝟖 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 25. youtube.com/@MathematicsOnlineLectures Prove that 𝐜𝐨𝐬 𝟔∘ 𝐜𝐨𝐬 𝟒𝟐∘ 𝐜𝐨𝐬 𝟔𝟔∘ 𝐜𝐨𝐬 𝟕𝟖∘ = 𝟏 𝟏𝟔 . We have, 𝐋𝐇𝐒 = 𝐜𝐨𝐬 𝟔∘ 𝐜𝐨𝐬 𝟒𝟐∘ 𝐜𝐨𝐬 𝟔𝟔∘ 𝐜𝐨𝐬 𝟕𝟖∘ = 𝟏 𝟒 (𝟐 𝐜𝐨𝐬 𝟔𝟔∘ 𝐜𝐨𝐬 𝟔∘)(𝟐 𝐜𝐨𝐬 𝟕𝟖∘ 𝐜𝐨𝐬 𝟒𝟐∘) = 𝟏 𝟒 [ 𝐜𝐨𝐬 (𝟔𝟔∘ + 𝟔∘) + 𝐜𝐨𝐬 (𝟔𝟔∘ − 𝟔∘)] × [ 𝐜𝐨𝐬 (𝟕𝟖∘ + 𝟒𝟐∘) + 𝐜𝐨𝐬 (𝟕𝟖∘ − 𝟒𝟐∘)] = 𝟏 𝟒 ( 𝐜𝐨𝐬 𝟕𝟐∘ + 𝐜𝐨𝐬 𝟔𝟎∘)( 𝐜𝐨𝐬 𝟏𝟐𝟎∘ + 𝐜𝐨𝐬 𝟑𝟔∘) = 𝟏 𝟒 ( 𝐬𝐢𝐧 𝟏𝟖∘ + 𝟏 𝟐 ) (− 𝟏 𝟐 + 𝐜𝐨𝐬 𝟑𝟔∘) [ 𝐜𝐨𝐬 𝟕𝟐∘ = 𝐜𝐨𝐬 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐬𝐢𝐧 𝟏𝟖∘] = 𝟏 𝟒 [ (√𝟓−𝟏) 𝟒 + 𝟏 𝟐 ] [− 𝟏 𝟐 + (√𝟓+𝟏) 𝟒 ] [⋅.⋅ 𝐬𝐢𝐧 𝟏𝟖∘ = (√𝟓−𝟏) 𝟒 and 𝐜𝐨𝐬 𝟑𝟔∘ = (√𝟓+𝟏) 𝟒 ] = 𝟏 𝟒 ⋅ (√𝟓+𝟏) 𝟒 ⋅ (√𝟓−𝟏) 𝟒 = (𝟓−𝟏) 𝟔𝟒 = 𝟒 𝟔𝟒 = 𝟏 𝟏𝟔 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 26. youtube.com/@MathematicsOnlineLectures Prove that 𝐬𝒊𝐧 𝝅 𝟓 𝐬𝐢𝐧 𝟐𝝅 𝟓 𝐬𝒊𝐧 𝟑𝝅 𝟓 𝐬𝐢𝐧 𝟒𝝅 𝟓 = 𝟓 𝟏𝟔 . We have, 𝐋𝐇𝐒 = 𝐬𝐢𝐧 𝝅 𝟓 𝐬𝐢𝐧 𝟐𝝅 𝟓 𝐬𝐢𝐧 𝟑𝝅 𝟓 𝐬𝐢𝐧 𝟒𝝅 𝟓 = 𝐬𝐢𝐧 𝝅 𝟓 𝐬𝐢𝐧 𝟐𝝅 𝟓 𝐬𝐢𝐧 (𝝅 − 𝟐𝝅 𝟓 ) 𝐬𝐢𝐧 (𝝅 − 𝝅 𝟓 ) = 𝐬𝐢𝐧𝟐 𝝅 𝟓 𝐬𝐢𝐧𝟐 𝟐𝝅 𝟓 [ 𝐬𝐢𝐧 (𝝅 − 𝜽) = 𝐬𝒊𝒏𝜽] = ( 𝐬𝐢𝐧 𝟑𝟔∘)𝟐 × (𝐬𝒊𝒏 𝟕𝟐∘)𝟐 = ( 𝐬𝐢𝐧 𝟑𝟔∘)𝟐 × ( 𝐜𝐨𝐬 𝟏𝟖∘)𝟐 [ 𝐬𝒊𝐧𝟕𝟐∘ = 𝐬𝐢𝐧 (𝟗𝟎∘ − 𝟏𝟖∘) = 𝐜𝐨𝐬 𝟏𝟖∘ ] = (𝟏𝟎−𝟐√𝟓) 𝟏𝟔 × (𝟏𝟎+𝟐√𝟓) 𝟏𝟔 = (𝟏𝟎𝟎−𝟐𝟎) (𝟏𝟔×𝟏𝟔) [⋅.⋅ 𝐬𝐢𝐧 𝟑𝟔∘ = √𝟏𝟎−𝟐√𝟓 𝟒 and 𝐜𝐨𝐬 𝟏𝟖∘ = √𝟏𝟎+𝟐√𝟓 𝟒 ] = 𝟖𝟎 (𝟏𝟔×𝟏𝟔) = 𝟓 𝟏𝟔 = 𝐑𝐇𝐒. Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar
  • 27. youtube.com/@MathematicsOnlineLectures Find the value of (i) 𝐬𝐢𝐧 𝟐𝟐∘ 𝟑𝟎′ (ii) 𝐜𝐨𝐬 𝟐𝟐∘ 𝟑𝟎′ (iii) 𝐭𝐚𝐧 𝟐𝟐∘ 𝟑𝟎′ (i) 𝐬𝐢𝐧𝟐 𝜽 = (𝟏− 𝐜𝐨𝐬 𝟐𝜽) 𝟐 ⇒ 𝐬𝐢𝐧𝟐(𝟐𝟐∘ 𝟑𝟎’) = (𝟏− 𝐜𝐨𝐬 𝟒𝟓∘) 𝟐 = (𝟏− 𝟏 √𝟐 ) 𝟐 = √𝟐−𝟏 𝟐√𝟐 𝐬𝐢𝐧 𝟐𝟐∘ 𝟑𝟎′ = √ √𝟐 − 𝟏 𝟐√𝟐 (ii) 𝐜𝐨𝐬𝟐 𝜽 = (𝟏+ 𝐜𝐨𝐬 𝟐𝜽) 𝟐 ⇒ 𝐜𝐨𝐬𝟐(𝟐𝟐∘ 𝟑𝟎′) = (𝟏+ 𝐜𝐨𝐬 𝟒𝟓∘) 𝟐 = (𝟏+ 𝟏 √𝟐 ) 𝟐 = (√𝟐+𝟏) 𝟐√𝟐 𝐜𝐨𝐬 𝟐𝟐∘ 𝟑𝟎′ = √ (√𝟐 + 𝟏) 𝟐√𝟐 (iii) 𝐭𝐚𝐧𝟐(𝟐𝟐∘ 𝟑𝟎′) = 𝐬𝐢𝐧𝟐(𝟐𝟐∘𝟑𝟎′) 𝐜𝐨𝐬𝟐(𝟐𝟐∘𝟑𝟎’) = (√𝟐−𝟏) (𝟐√𝟐) × (𝟐√𝟐) (√𝟐+𝟏) = √𝟐−𝟏 √𝟐+𝟏 = (√𝟐−𝟏) (√𝟐+𝟏) × (√𝟐−𝟏) (√𝟐−𝟏) = (√𝟐 − 𝟏) 𝟐 𝐭𝐚𝐧 𝟐𝟐∘ 𝟑𝟎′ = (√𝟐 − 𝟏) Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar