Class 11_Chapter 3_Lecture_4.pdf

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Class XI
Mathematics
Chapter- 3
Trigonometric Functions
Lecture - 4
Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.
Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar

youtube.com/@MathematicsOnlineLectures
For all 𝒙, 𝒚 ∈ 𝑹, we have
(𝒂) 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒚 = 𝐬𝐢𝐧 (𝒙 + 𝒚) + 𝐬𝐢𝐧 (𝒙 − 𝒚)
(𝒃) 𝟐 𝐜𝐨𝐬 𝒙 𝐬𝐢𝐧 𝒚 = 𝐬𝐢𝐧 (𝒙 + 𝒚) − 𝐬𝐢𝐧 (𝒙 − 𝒚)
(𝒄) 𝟐 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒚 = 𝐜𝐨𝐬 (𝒙 + 𝒚) + 𝐜𝐨𝐬 (𝒙 − 𝒚)
(𝒅) 𝟐 𝐬𝐢𝐧 𝒙 𝐬𝒊𝐧𝐲 = 𝐜𝐨𝐬 (𝒙 − 𝒚) − 𝐜𝐨𝐬 (𝒙 + 𝒚)
We know that 𝐬𝐢𝐧 (𝒙 + 𝒚) = 𝐬𝒊𝒏 𝒙 𝐜𝐨𝐬 𝐲 + 𝐜𝐨𝐬 𝒙 𝐬𝐢𝐧 𝒚 (i)
𝐬𝐢𝐧 (𝒙 − 𝒚) = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒚 − 𝐜𝐨𝐬 𝒙 𝐬𝐢𝐧 𝒚 (ii)
(a) Adding (i) and (ii), we get 𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒚 = 𝐬𝐢𝐧 (𝒙 + 𝒚) + 𝐬𝐢𝐧 (𝒙 − 𝒚) .
(b) Subtracting (ii) from (i), we get 𝟐 𝐜𝐨𝐬 𝒙 𝐬𝐢𝐧 𝒚 = 𝐬𝐢𝐧 (𝒙 + 𝒚) − 𝐬𝐢𝐧 (𝒙 − 𝒚) .
Further, we know that
𝐜𝐨𝐬 (𝒙 + 𝒚) = 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒚 − 𝐬𝐢𝐧 𝒙 𝐬𝒊𝐧𝒚 (iii)
𝐜𝐨𝐬 (𝒙 − 𝒚) = 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒚 + 𝐬𝐢𝐧 𝒙 𝐬𝒊𝐧𝐲 (iv)
(c) Adding (iii) and (iv), we get 𝟐 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝒚 = 𝐜𝐨𝐬 (𝒙 + 𝒚) + 𝐜𝐨𝐬 (𝒙 − 𝒚) .
(d) Subtracting (iii) from (iv), 𝟐 𝐬𝐢𝐧 𝒙 𝐬𝐢𝐧 𝒚 = 𝐜𝐨𝐬 (𝒙 − 𝒚) − 𝐜𝐨𝐬 (𝒙 + 𝒚) .

For all 𝒙, 𝒚 ∈ 𝑹, we have
(𝒊) 𝐬𝐢𝐧 𝒙 + 𝐬𝐢𝐧 𝒚 = 𝟐𝐬𝒊𝒏 (
𝒙+𝒚
𝟐
) 𝐜𝐨𝐬 (
𝒙−𝐲
𝟐
)
(𝒊𝒊) 𝐬𝐢𝐧 𝒙 − 𝐬𝐢𝐧 𝒚 = 𝟐 𝐜𝐨𝐬 (
𝒙+𝒚
𝟐
) 𝐬𝐢𝐧 (
𝒙−𝒚
𝟐
)
(iii) 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝒚 = 𝟐 𝐜𝐨𝐬 (
𝒙+𝒚
𝟐
) 𝐜𝐨𝐬 (
𝒙−𝒚
𝟐
)
(𝒊𝒗) 𝐜𝐨𝐬 𝒙 − 𝐜𝐨𝐬 𝒚 = −𝟐 𝐬𝐢𝐧 (
𝒙+𝒚
𝟐
) 𝐬𝐢𝐧 (
𝒙−𝒚
𝟐
)
Putting 𝒙 = (𝒂 + 𝒃) and 𝒚 = (𝒂 − 𝒃) , we get 𝒂 = (
𝒙+𝒚
𝟐
) and 𝒃 = (
𝒙−𝒚
𝟐
) .
(i) 𝐬𝒊𝐧𝒙 + 𝐬𝐢𝐧 𝒚 = 𝐬𝐢𝐧 (𝒂 + 𝒃) + 𝐬𝐢𝐧 (𝒂 − 𝒃) = 𝟐 𝐬𝐢𝐧 𝒂 𝐜𝐨𝐬 𝒃
= 𝟐𝐬𝒊𝒏 (
𝒙+𝒚
𝟐
) 𝐜𝐨𝐬 (
𝒙−𝐲
𝟐
) .
(ii) 𝐬𝒊𝐧𝒙 − 𝐬𝐢𝐧 𝒚 = 𝐬𝐢𝐧 (𝒂 + 𝒃) − 𝐬𝐢𝐧 (𝒂 − 𝒃) = 𝟐 𝐜𝐨𝐬 a 𝐬𝐢𝐧 𝒃
= 𝟐 𝐜𝐨𝐬 (
𝒙+𝒚
𝟐
) 𝐬𝐢𝐧 (
𝒙−𝒚
𝟐
) .
(iii) 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝒚 = 𝐜𝐨𝐬 (𝒂 + 𝒃) + 𝐜𝐨𝐬 (𝒂 − 𝒃) = 𝟐 𝐜𝐨𝐬 a 𝐜𝐨𝐬 𝒃
= 𝟐 𝐜𝐨𝐬 (
𝒙+𝒚
𝟐
) 𝐜𝐨𝐬 (
𝒙−𝒚
𝟐
) .
(iv) 𝐜𝐨𝐬 𝒙 − 𝐜𝐨𝐬 𝒚 = 𝐜𝐨𝐬 (𝒂 + 𝒃) − 𝐜𝐨𝐬 (𝒂 − 𝒃) = −𝟐 𝐬𝐢𝐧 𝒂 𝐬𝐢𝐧 𝒃
= −𝟐𝐬𝒊𝒏 (
𝒙 + 𝒚
𝟐
) 𝐬𝐢𝐧 (
𝒙 − 𝐲
𝟐
)

Prove that (𝒊) 𝐜𝐨𝐬 (
𝝅
𝟒
+ 𝒙) + 𝐜𝐨𝐬 (
𝝅
𝟒
− 𝒙) = √𝟐 𝐜𝐨𝐬 𝒙
(𝒊𝒊) 𝐜𝐨𝐬 (
𝟑𝝅
𝟒
+ 𝒙) − 𝐜𝐨𝐬 (
𝟑𝝅
𝟒
− 𝒙) = −√𝟐 𝐬𝐢𝐧 𝒙
(i) LHS = 𝐜𝐨𝐬 (
𝝅
𝟒
+ 𝒙) + 𝐜𝐨𝐬 (
𝝅
𝟒
− 𝒙)
= 𝟐 𝐜𝐨𝐬
𝝅
𝟒
𝐜𝐨𝐬 𝒙 [ 𝐜𝐨𝐬 (𝑨 + 𝑩) + 𝐜𝐨𝐬 (𝑨 − 𝐁) = 𝟐 𝐜𝐨𝐬 𝑨 𝐜𝐨𝐬 𝑩]
= (𝟐 ×
𝟏
√𝟐
𝐜𝐨𝐬 𝒙) = √𝟐 𝐜𝐨𝐬 𝒙 = 𝐑𝐇𝐒.
(ii) 𝐋𝐇𝐒 = 𝐜𝐨𝐬 (
𝟑𝝅
𝟒
+ 𝒙) − 𝐜𝐨𝐬 (
𝟑𝝅
𝟒
− 𝒙)
= −𝟐𝐬𝒊𝒏
𝟑𝝅
𝟒
𝐬𝐢𝐧 𝒙[ 𝐜𝐨𝐬 (𝑨 + 𝐁) − 𝐜𝐨𝐬 (𝑨 − 𝑩) = −𝟐 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩]
= −𝟐𝐬𝒊𝒏 (𝝅 −
𝝅
𝟒
) 𝐬𝐢𝐧 𝒙 = −𝟐𝐬𝒊𝒏
𝝅
𝟒
𝐬𝐢𝐧 𝒙 [⋅.⋅ 𝐬𝐢𝐧 (𝝅 −
𝝅
𝟒
) = 𝐬𝐢𝐧
𝝅
𝟒
]
= (−𝟐 ×
𝟏
√𝟐
) 𝐬𝐢𝐧 𝒙 = −√𝟐 𝐬𝐢𝐧 𝒙 = 𝐑𝐇𝐒.

Express each of the following as an algebraic sum of sines or cosines:
(i) 𝟐 𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬 𝟐𝒙 (ii) 𝟐 𝐜𝐨𝐬 𝟒𝒙 𝐬𝐢𝐧 𝟐𝒙
(iii) 𝟐 𝐜𝐨𝐬 𝟔𝒙 𝐜𝐨𝐬 𝟒𝒙 (iv) 𝟐 𝐬𝐢𝐧 𝟑𝒙 𝐬𝐢𝐧 𝟓𝒙
(i) We know that 𝟐 𝐬𝐢𝐧 𝑨 𝐜𝐨𝐬 𝑩 = 𝐬𝐢𝐧 (𝑨 + 𝑩) + 𝐬𝐦
̇ (𝑨 − 𝑩) .
𝟐 𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬 𝟐𝒙 = 𝐬𝐢𝐧 (𝟑𝒙 + 𝟐𝒙) + 𝐬𝐢𝐧 (𝟑𝒙 − 𝟐𝒙) = 𝐬𝐢𝐧 𝟓𝒙 + 𝐬𝐢𝐧 𝒙.
(ii) We know that 𝟐 𝐜𝐨𝐬 𝑨 𝐬𝐢𝐧 𝐁 = 𝐬𝐢𝐧 (𝑨 + 𝑩) − 𝐬𝐢𝐧 (𝑨 − 𝑩) .
𝟐 𝐜𝐨𝐬 𝟒𝒙 𝐬𝐢𝐧 𝟐𝒙 = 𝐬𝐢𝐧 (𝟒𝒙 + 𝟐𝒙) − 𝐬𝐦
̇ (𝟒𝒙 − 𝟐𝒙) = 𝐬𝐢𝐧 𝟔𝒙 − 𝐬𝒊𝒏 𝟐𝒙.
(iii) We know that 𝟐 𝐜𝐨𝐬 𝑨 𝐜𝐨𝐬 𝑩 = 𝐜𝐨𝐬 (𝑨 + 𝑩) + 𝐜𝐨𝐬 (𝑨 − 𝑩) .
𝟐 𝐜𝐨𝐬 𝟔𝒙 𝐜𝐨𝐬 𝟒𝒙 = 𝐜𝐨𝐬 (𝟔𝒙 + 𝟒𝒙) + 𝐜𝐨𝐬 (𝟔𝒙 − 𝟒𝒙) = 𝐜𝐨𝐬 𝟏𝟎𝒙 + 𝐜𝐨𝐬 𝟐𝒙.
(iv) We know that 𝟐 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝐁 = 𝐜𝐨𝐬 (𝑨 − 𝑩) − 𝐜𝐨𝐬 (𝑨 + 𝑩) .
𝟐 𝐬𝐢𝐧 𝟑𝒙 𝐬𝐢𝐧 𝟓𝒙 = 𝟐 𝐬𝐢𝐧 𝟓𝒙 𝐬𝐢𝐧 𝟑𝒙 = 𝐜𝐨𝐬 (𝟓𝒙 − 𝟑𝒙) − 𝐜𝐨𝐬 (𝟓𝒙 + 𝟑𝒙)
= 𝐜𝐨𝐬 𝟐𝒙 − 𝐜𝐨𝐬 𝟖𝒙.

Express each of the following as a product of sines or cosines or sine and cosine:
(i) 𝐜𝐨𝐬 𝟓𝒙 + 𝐜𝐨𝐬 𝟑𝒙 (ii) 𝐜𝐨𝐬 𝟓𝒙 − 𝐜𝐨𝐬 𝟕𝒙
(iii) 𝐬𝐢𝐧 𝟕𝒙 + 𝐬𝐢𝐧 𝟑𝒙 (iv) 𝐬𝐢𝐧 𝟓𝒙 − 𝐬𝐢𝐧 𝟑𝒙
(i) 𝐜𝐨𝐬 𝐂 + 𝐜𝐨𝐬 𝑫 = 𝟐 𝐜𝐨𝐬 (
𝐂+𝐃
𝟐
) 𝐜𝐨𝐬 (
𝐂−𝑫
𝟐
) .
𝐜𝐨𝐬 𝟓𝒙 + 𝐜𝐨𝐬 𝟑𝒙 = 𝟐 𝐜𝐨𝐬 (
𝟓𝒙 + 𝟑𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟓𝒙 − 𝟑𝒙
𝟐
) = 𝟐 𝐜𝐨𝐬 𝟒𝒙 𝐜𝐨𝐬 𝒙.
(ii) 𝐜𝐨𝐬 𝐂 − 𝐜𝐨𝐬 𝐃 = −𝟐𝐬𝐦
̇ (
𝐂+𝑫
𝟐
) 𝐬𝐢𝐧 (
𝐂−𝑫
𝟐
) .
𝐜𝐨𝐬 𝟓𝒙 − 𝐜𝐨𝐬 𝟕𝒙 = −𝟐 𝐬𝐢𝐧 (
𝟓𝒙 + 𝟕𝒙
𝟐
) 𝐬𝐢𝐧 (
𝟓𝒙 − 𝟕𝒙
𝟐
) = 𝟐𝐬𝒊𝒏 𝟔𝒙 𝐬𝐢𝐧 𝒙.
(iii) 𝐬𝐢𝐧 𝐂 + 𝐬𝒊𝒏 𝑫 = 𝟐 𝐬𝐢𝐧 (
𝐂+𝐃
𝟐
) 𝐜𝐨𝐬 (
𝐂−𝑫
𝟐
) .
𝐬𝐢𝐧 𝟕𝒙 + 𝐬𝐢𝐧 𝟑𝒙 = 𝟐 𝐬𝐢𝐧 (
𝟕𝒙+𝟑𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟕𝒙−𝟑𝒙
𝟐
) = 𝟐𝐬𝒊𝒏 𝟓𝒙 𝐜𝐨𝐬 𝟐𝒙.
(iv) 𝐬𝐢𝐧 𝐂 − 𝐬𝐢𝐧 𝐃 = 𝟐 𝐜𝐨𝐬 (
𝐂+𝐃
𝟐
) 𝐬𝐢𝐧 (
𝐂−𝑫
𝟐
) .
𝐬𝐢𝐧 𝟓𝒙 − 𝐬𝒊𝒏 𝟑𝒙 = 𝟐 𝐜𝐨𝐬 (
𝟓𝒙+𝟑𝒙
𝟐
) 𝐬i𝐧 (
𝟓𝒙−𝟑𝒙
𝟐
) = 𝟐 𝐜𝐨𝐬 𝟒𝒙 𝐬𝐢𝐧 𝒙.

Prove that
𝐜𝐨𝐬 𝟔𝒙+ 𝐜𝐨𝐬 𝟒𝒙
𝐬𝐢𝐧 𝟔𝒙− 𝐬𝐢𝐧 𝟒𝒙
= 𝐜𝐨𝐭 𝒙.
LHS =
𝐜𝐨𝐬 𝟔𝒙+ 𝐜𝐨𝐬 𝟒𝒙
𝐬𝐢𝐧 𝟔𝒙−𝐬𝐦
̇ 𝟒𝒙
=
𝟐 𝐜𝐨𝐬 (
𝟔𝒙+𝟒𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟔𝒙−𝟒𝒙
𝟐
)
𝟐 𝐜𝐨𝐬 (
𝟔𝒙+𝟒𝒙
𝟐
) 𝐬𝐢𝐧 (
𝟔𝒙−𝟒𝒙
𝟐
)
{
𝐜𝐨𝐬𝐂 + 𝐜𝐨𝐬𝐃 = 𝟐𝐜𝐨𝐬 (
𝐂 + 𝐃
𝟐
) 𝐜𝐨𝐬 (
𝐂 − 𝑫
𝟐
)
𝐬𝐢𝐧𝐂 − 𝐬𝐢𝐧𝐃 = 𝟐𝐜𝐨𝐬 (
𝑪 + 𝐃
𝟐
) 𝐬𝐢𝐧 (
𝐂 − 𝑫
𝟐
)
}
=
𝟐 𝐜𝐨𝐬 𝟓𝒙 𝐜𝐨𝐬 𝒙
𝟐 𝐜𝐨𝐬 𝟓𝒙 𝐬𝐢𝐧 𝒙
=
𝐜𝐨𝐬 𝒙
𝐬𝐦
̇ 𝒙
= 𝐜𝐨𝐭 𝒙 = 𝐑𝐇𝐒.

Prove that
𝐬𝐢𝐧 𝟑𝒙−𝐬𝒊𝒏 𝒙
𝐜𝐨𝐬 𝒙− 𝐜𝐨𝐬 𝟑𝒙
= 𝐜𝐨𝐭 𝟐𝒙.
LHS =
𝐬𝒊𝒏 𝟑𝒙− 𝐬𝐢𝐧 𝒙
𝐜𝐨𝐬 𝒙− 𝐜𝐨𝐬 𝟑𝒙
=
𝟐 𝐜𝐨𝐬 (
𝟑𝒙+𝒙
𝟐
) 𝐬𝐢𝐧 (
𝟑𝒙−𝒙
𝟐
)
−𝟐 𝐬𝐢𝐧 (
𝒙+𝟑𝒙
𝟐
) 𝐬𝐢𝐧 (
𝒙−𝟑𝒙
𝟐
)
{
𝐬𝐢𝐧 𝐂 − 𝐬𝐢𝐧 𝐃 = 𝟐 𝐜𝐨𝐬 (
𝑪 + 𝐃
𝟐
) 𝐬𝐢𝐧 (
𝑪 − 𝑫
𝟐
)
𝐜𝐨𝐬 𝐂 − 𝐜𝐨𝐬 𝑫 = −𝟐 𝐬𝐢𝐧 (
𝐂 + 𝐃
𝟐
) 𝐬𝐢𝐧 (
𝐂 − 𝑫
𝟐
)
}
=
𝐜𝐨𝐬 𝟐𝒙 𝐬𝐢𝐧 𝒙
− 𝐬𝐢𝐧 𝟐𝒙 𝐬𝐢𝐧(−𝐱)
=
𝐜𝐨𝐬 𝟐𝒙 𝐬𝐢𝐧 𝒙
𝐬𝐢𝐧 𝟐𝒙 𝐬𝐢𝐧 𝒙
= 𝐜𝐨𝐭 𝟐𝒙 = 𝐑𝐇𝐒.
Prove that
𝐬𝐢𝐧 𝟑𝒙−𝐬𝒊𝒏 𝒙
𝐜𝐨𝐬 𝟐𝒙
= 𝟐 𝐬𝐢𝐧 𝒙.
LHS =
𝐬𝒊𝒏 𝟑𝒙− 𝐬𝐢𝐧 𝒙
=
𝟐 𝐜𝐨𝐬 (
𝟑𝒙+𝒙
𝟐
) 𝐬𝐢𝐧 (
𝟑𝒙−𝒙
𝟐
)
𝒄𝒐𝒔 𝟐𝒙
=
𝟐 𝐜𝐨𝐬 𝟐𝒙 𝐬𝐢𝐧 𝒙
= 𝟐 𝐬𝐢𝐧 𝒙 = 𝐑𝐇𝐒.

Prove that
𝐬𝐢𝐧 𝟓𝒙−𝟐 𝐬𝐢𝐧 𝟑𝒙+ 𝐬𝐢𝐧 𝒙
𝐜𝐨𝐬 𝟓𝒙− 𝐜𝐨𝐬 𝒙
= 𝐜𝐨𝐬𝐞𝐜 𝟐𝒙 − 𝐜𝐨𝐭 𝟐𝒙.
We have
LHS =
( 𝐬𝐢𝐧 𝟓𝒙+ 𝐬𝐢𝐧 𝒙)−𝟐 𝐬𝐢𝐧 𝟑𝒙
𝐜𝐨𝐬 𝟓𝒙− 𝐜𝐨𝐬 𝒙
=
𝟐𝐬𝒊𝒏 (
𝟓𝒙+𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟓𝒙−𝒙
𝟐
)−𝟐 𝐬𝐢𝐧 𝟑𝒙
−𝟐 𝐬𝐢𝐧 (
𝟓𝒙+𝒙
𝟐
) 𝐬𝐢𝐧 (
𝟓𝒙−𝒙
𝟐
)
[⋅.⋅ 𝐬𝐢𝐧 𝐂 + 𝐬𝐢𝐧 𝐃 = 𝟐 𝐬𝐢𝐧 (
𝐂+𝑫
𝟐
) 𝐜𝐨𝐬 (
𝐂−𝑫
𝟐
) and
𝐜𝐨𝐬 𝐂 − 𝐜𝐨𝐬 𝑫 = −𝟐𝐬𝐦
̇ (
𝑪 + 𝑫
𝟐
) 𝐬𝐢𝐧 (
𝑪 − 𝐃
𝟐
)]
=
𝟐𝐬𝒊𝒏 𝟑𝒙 𝐜𝐨𝐬 𝟐𝒙−𝟐 𝐬𝐢𝐧 𝟑𝒙
−𝟐 𝐬𝐢𝐧 𝟑𝒙 𝐬𝐢𝐧 𝟐𝒙
=
𝟐 𝐬𝐢𝐧 𝟑𝒙( 𝐜𝐨𝐬 𝟐𝒙−𝟏)
−𝟐 𝐬𝐢𝐧 𝟑𝒙 𝐬𝐢𝐧 𝟐𝒙
=
(𝟏− 𝐜𝐨𝐬 𝟐𝒙)
𝐬𝐢𝐧 𝟐𝒙
=
𝟏
−
= 𝐜𝐨𝐬𝐞𝐜 𝟐𝒙 − 𝐜𝐨𝐭 𝟐𝒙 = 𝐑𝐇𝐒.

If
𝐜𝐨𝐬 (𝑨+𝑩)
𝐜𝐨𝐬 (𝑨−𝑩)
=
𝐬𝐢𝐧 (𝐂+𝐃)
𝐬𝐢𝐧 (𝐂−𝑫)
, prove that 𝐭𝐚𝐧 𝑨 𝐭𝐚𝐧 𝑩 𝐭𝐚𝐧 𝐂 + 𝐭𝐚𝐧 𝑫 = 𝟎.
𝐜𝐨𝐬 (𝑨+𝐁)
𝐜𝐨𝐬 (𝑨−𝑩)
=
𝐬𝒊𝒏 (𝐂+𝑫)
𝐬𝐢𝐧 (𝐂−𝑫)
⇒
𝐜𝐨𝐬 (𝑨+𝐁)+ 𝐜𝐨𝐬 (𝑨−𝑩)
𝐜𝐨𝐬 (𝑨+𝑩)− 𝐜𝐨𝐬 (𝑨−𝑩)
=
𝐬𝐢𝐧 (𝐂+𝐃)+𝐬𝒊𝒏 (𝐂−𝑫)
𝐬𝐢𝐧 (𝑪+𝐃)− 𝐬𝐢𝐧 (𝑪−𝑫)
⇒
𝟐 𝐜𝐨𝐬 𝑨 𝐜𝐨𝐬 𝑩
−𝟐 𝐬𝐢𝐧 𝑨 𝐬𝐢𝐧 𝑩
=
𝟐 𝐬𝐢𝐧 𝐂 𝐜𝐨𝐬 𝑫
𝟐 𝐜𝐨𝐬 𝑪 𝐬𝐢𝐧 𝑫
⇒ − 𝐜𝐨𝐭 𝑨 𝐜𝐨𝐭 𝑩 = 𝐭𝐚𝐧 𝐂 𝐜𝐨𝐭 𝑫
⇒ 𝐭𝐚𝐧 𝑨 𝐭𝐚𝐧 𝑩 𝐭𝐚𝐧 𝐂 = − 𝐭𝐚𝐧 𝑫 ⇒ 𝐭𝐚𝐧 𝑨 𝐭𝐚𝐧 𝑩 𝐭𝐚𝐧 𝐂 + 𝐭𝐚𝐧 𝑫 = 𝟎.
Hence, 𝐭𝐚𝐧 A 𝐭𝐚𝐧 𝑩 𝐭𝐚𝐧 𝐂 + 𝐭𝐚𝐧 𝑫 = 𝟎.

Prove that
𝐜𝐨𝐬 𝟒𝒙+ 𝐜𝐨𝐬 𝟑𝒙+ 𝐜𝐨𝐬 𝟐𝒙
𝐬𝐢𝐧 𝟒𝒙+ 𝐬𝐢𝐧 𝟑𝒙+ 𝐬𝐢𝐧 𝟐𝒙
= 𝐜𝐨𝐭 𝟑𝒙.
We have, LHS =
( 𝐜𝐨𝐬 𝟒𝒙+ 𝐜𝐨𝐬 𝟐𝒙)+ 𝐜𝐨𝐬 𝟑𝒙
( 𝐬𝐢𝐧 𝟒𝒙+ 𝐬𝐢𝐧 𝟐𝒙)+ 𝐬𝐢𝐧 𝟑𝒙
=
𝟐 𝐜𝐨𝐬 (
𝟒𝒙+𝟐𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟒𝒙−𝟐𝒙
𝟐
)+ 𝐜𝐨𝐬 𝟑𝒙
𝟐𝐬𝐦
̇ (
𝟒𝒙+𝟐𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟒𝒙−𝟐𝒙
𝟐
)+ 𝐬𝐢𝐧 𝟑𝒙
[using formulae for ( 𝐜𝐨𝐬 𝑪 + 𝐜𝐨𝐬 𝑫) and ( 𝐬𝐢𝐧 𝐂 + 𝐬𝒊𝒏 𝑫)]
=
𝟐 𝐜𝐨𝐬 𝟑𝒙 𝐜𝐨𝐬 𝒙+ 𝐜𝐨𝐬 𝟑𝒙
𝟐𝐬𝐦
̇ 𝟑𝒙 𝐜𝐨𝐬 𝒙+ 𝐬𝐢𝐧 𝟑𝒙
=
𝐜𝐨𝐬 𝟑𝒙(𝟐 𝐜𝐨𝐬 𝒙+𝟏)
𝐬𝐢𝐧 𝟑𝒙(𝟐 𝐜𝐨𝐬 𝒙+𝟏)
=
𝐜𝐨𝐬 𝟑𝒙
𝐬𝐢𝐧 𝟑𝒙
= 𝐜𝐨𝐭 𝟑𝒙 = 𝐑𝐇𝐒.

Prove that 𝐬𝐢𝐧 𝒙 + 𝐬𝒊𝒏 𝟑𝒙 + 𝐬𝐢𝐧 𝟓𝒙 + 𝐬𝐢𝐧 𝟕𝒙 = 𝟒 𝐬𝐢𝐧 𝟒𝒙 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬 𝒙.
We have 𝐋𝐇𝐒 = ( 𝐬𝐢𝐧 𝟕𝒙 + 𝐬𝐢𝐧 𝒙) + (𝐬𝒊𝒏 𝟓𝒙 + 𝐬𝐢𝐧 𝟑𝒙)
= 𝟐 𝐬𝐢𝐧 (
𝟕𝒙 + 𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟕𝒙 − 𝒙
𝟐
) + 𝟐 𝐬𝐢𝐧 (
𝟓𝒙 + 𝟑𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟓𝒙 − 𝟑𝒙
𝟐
)
= 𝟐 𝐬𝐢𝐧 𝟒𝒙 𝐜𝐨𝐬 𝟑𝒙 + 𝟐 𝐬𝐢𝐧 𝟒𝒙 𝐜𝐨𝐬 𝒙 = 𝟐𝐬𝒊𝒏 𝟒𝒙( 𝐜𝐨𝐬 𝟑𝒙 + 𝐜𝐨𝐬 𝒙)
= (𝟐 𝐬𝐢𝐧 𝟒𝒙) × 𝟐 𝐜𝐨𝐬 (
𝟑𝒙+𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟑𝒙−𝒙
𝟐
) = (𝟐 𝐬𝐢𝐧 𝟒𝒙) × 𝟐 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬 𝒙
= 𝟒𝐬𝒊𝒏 𝟒𝒙 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬 𝒙 = 𝐑𝐇𝐒.

Prove that 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬
𝒙
𝟐
− 𝐜𝐨𝐬 𝟑𝒙 𝐜𝐨𝐬
𝟗𝒙
𝟐
= −𝐬𝒊𝒏 𝟓𝒙 𝐬𝐢𝐧 (
𝟓𝒙
𝟐
) .
We have 𝐋𝐇𝐒 =
𝟏
𝟐
[𝟐 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬
𝒙
𝟐
− 𝟐 𝐜𝐨𝐬 𝟑𝒙 𝐜𝐨𝐬
𝟗𝒙
𝟐
]
=
𝟏
𝟐
[ 𝐜𝐨𝐬 (𝟐𝒙 +
𝒙
𝟐
) + 𝐜𝐨𝐬 (𝟐𝒙 −
𝒙
𝟐
)] −
𝟏
𝟐
[ 𝐜𝐨𝐬 (
𝟗𝒙
𝟐
+ 𝟑𝒙) + 𝐜𝐨𝐬 (
𝟗𝒙
𝟐
− 𝟑𝒙)]
=
𝟏
𝟐
( 𝐜𝐨𝐬
𝟓𝒙
𝟐
+ 𝐜𝐨𝐬
𝟑𝒙
𝟐
) −
𝟏
𝟐
( 𝐜𝐨𝐬
𝟏𝟓𝒙
𝟐
+ 𝐜𝐨𝐬
𝟑𝒙
𝟐
) =
𝟏
𝟐
( 𝐜𝐨𝐬
𝟓𝒙
𝟐
− 𝐜𝐨𝐬
𝟏𝟓𝒙
𝟐
)
=
𝟏
𝟐
[−𝟐 𝐬𝐢𝐧
(
𝟓𝒙
𝟐
+
𝟏𝟓𝒙
𝟐
)
𝟐
𝐬𝐢𝐧
(
𝟏𝟓𝒙
𝟐
−
𝟓𝒙
𝟐
)
𝟐
] = − 𝐬𝐢𝐧 𝟓𝒙 𝐬𝐢𝐧 (
𝟓𝒙
𝟐
) = 𝐑𝐇𝐒.

Prove that
𝐬𝒊𝒏 𝟖𝒙 𝐜𝐨𝐬 𝒙− 𝐬𝐢𝐧 𝟔𝒙 𝐜𝐨𝐬 𝟑𝒙
𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬 𝒙− 𝐬𝐢𝐧 𝟒𝒙 𝐬𝐢𝐧 𝟑𝒙
= 𝐭𝐚𝐧 𝟐𝒙.
We have, LHS =
𝟐 𝐬𝐢𝐧 𝟖𝒙 𝐜𝐨𝐬 𝒙−𝟐 𝐬𝐢𝐧 𝟔𝒙 𝐜𝐨𝐬 𝟑𝒙
𝟐 𝐜𝐨𝐬 𝟐𝒙 𝐜𝐨𝐬 𝒙−𝟐 𝐬𝐢𝐧 𝟒𝒙 𝐬𝐢𝐧 𝟑𝒙
=
[𝐬𝒊𝒏 (𝟖𝒙 + 𝒙) + 𝐬𝐢𝐧 (𝟖𝒙 − 𝒙)] − [ 𝐬𝐢𝐧 (𝟔𝒙 + 𝟑𝒙) + 𝐬𝐢𝐧 (𝟔𝒙 − 𝟑𝒙)]
[ 𝐜𝐨𝐬 (𝟐𝒙 + 𝒙) + 𝐜𝐨𝐬 (𝟐𝒙 − 𝒙)] − [ 𝐜𝐨𝐬 (𝟒𝒙 − 𝟑𝒙) − 𝐜𝐨𝐬 (𝟒𝒙 + 𝟑𝒙)]
[ 𝟐 𝐬𝐢𝐧 A 𝐜𝐨𝐬 𝑩 = 𝐬𝐢𝐧 (𝑨 + 𝑩) + 𝐬𝐢𝐧 (𝑨 − 𝑩) , etc.]
=
( 𝐬𝐢𝐧 𝟗𝒙 + 𝐬𝐦
̇ 𝟕𝒙) − ( 𝐬𝐢𝐧 𝟗𝒙 + 𝐬𝐢𝐧 𝟑𝒙)
( 𝐜𝐨𝐬 𝟑𝒙 + 𝐜𝐨𝐬 𝒙) − ( 𝐜𝐨𝐬 𝒙 − 𝐜𝐨𝐬 𝟕𝒙)
=
(𝐬𝒊𝒏 𝟕𝒙 − 𝐬𝐢𝐧 𝟑𝒙)
( 𝐜𝐨𝐬 𝟑𝒙 + 𝐜𝐨𝐬 𝟕𝒙)
=
𝟐 𝐜𝐨𝐬 (
𝟕𝒙+𝟑𝒙
𝟐
) 𝐬𝐢𝐧 (
𝟕𝒙−𝟑𝒙
𝟐
)
𝟐 𝐜𝐨𝐬 (
𝟕𝒙+𝟑𝒙
𝟐
) 𝐜𝐨𝐬 (
𝟕𝒙−𝟑𝒙
𝟐
)
[⋅.⋅ 𝐬𝐢𝐧 𝑪 − 𝐬i𝐧𝑫 = 𝟐 𝐜𝐨𝐬
(𝐂+𝑫)
𝟐
𝐬𝐢𝐧
(𝑪−𝑫)
𝟐
]
=
𝐜𝐨𝐬 𝟓𝒙 𝐬𝐢𝐧 𝟐𝒙
𝐜𝐨𝐬 𝟓𝒙 𝐜𝐨𝐬 𝟐𝒙
=
= 𝐭𝐚𝐧 𝟐𝒙 = 𝐑𝐇𝐒.
Prove that
( 𝐬𝐢𝐧 𝒙− 𝐬𝐢𝐧 𝒚)
( 𝐜𝐨𝐬 𝒙+ 𝐜𝐨𝐬 𝒚)
= 𝐭𝐚𝐧 (
𝒙−𝒚
𝟐
) .
We have LHS =
( 𝐬𝐢𝐧 𝒙− 𝐬𝐢𝐧 𝒚)
( 𝐜𝐨𝐬 𝒙+ 𝐜𝐨𝐬 𝒚)
=
𝟐 𝐜𝐨𝐬 (
𝒙+𝒚
𝟐
) 𝐬𝐢𝐧 (
𝒙−𝐲
𝟐
)
𝟐 𝐜𝐨𝐬 (
𝒙+𝒚
𝟐
) 𝐜𝐨𝐬 (
𝒙−𝒚
𝟐
)
= 𝐭𝐚𝐧 (
𝒙−𝐲
𝟐
) = 𝐑𝐇𝐒.

Prove that ( 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝒚)𝟐
+ ( 𝐬𝐢𝐧 𝒙 + 𝐬𝐢𝐧 𝒚)𝟐
= 𝟒𝐜𝐨𝐬𝟐 (
𝒙−𝒚
𝟐
) .
We have 𝐋𝐇𝐒 = ( 𝐜𝐨𝐬 𝒙 + 𝐜𝐨𝐬 𝒚)𝟐
+ ( 𝐬𝐢𝐧 𝒙 + 𝐬𝐢𝐧 𝒚)𝟐
= {𝟐 𝐜𝐨𝐬 (
𝒙 + 𝒚
𝟐
) 𝐜𝐨𝐬 (
𝒙 − 𝒚
𝟐
)}
𝟐
+ {𝟐 𝐬𝐢𝐧 (
𝒙 + 𝒚
𝟐
) 𝐜𝐨𝐬 (
𝒙 − 𝒚
𝟐
)}
𝟐
= 𝟒𝐜𝐨𝐬𝟐 (
𝒙 + 𝒚
𝟐
) 𝐜𝐨𝐬𝟐 (
𝒙 − 𝒚
𝟐
) + 𝟒𝐬𝐢𝐧𝟐 (
𝒙 + 𝒚
𝟐
) 𝐜𝐨𝐬𝟐 (
𝒙 − 𝒚
𝟐
)
= 𝟒𝐜𝐨𝐬𝟐 (
𝒙 − 𝐲
𝟐
) ⋅ {𝐜𝐨𝐬𝟐 (
𝒙 + 𝒚
𝟐
) + 𝐬𝒊𝒏𝟐 (
𝒙 + 𝐲
𝟐
)} = 𝟒𝐜𝐨𝐬𝟐 (
𝒙 − 𝒚
𝟐
) = 𝐑𝐇𝐒.

Prove that
𝐜𝐨𝐬 𝜶 + 𝐜𝐨𝐬 𝜷 + 𝐜𝐨𝐬 𝜸 + 𝐜𝐨𝐬 (𝜶 + 𝜷 + 𝛄) = 𝟒 𝐜𝐨𝐬 (
𝜶+𝜷
𝟐
) 𝐜𝐨𝐬 (
𝜷+𝐘
𝟐
) 𝐜𝐨𝐬 (
𝜸+𝜶
𝟐
) .
We have
𝐋𝐇𝐒 = ( 𝐜𝐨𝐬 𝜶 + 𝐜𝐨𝐬 𝜷) + { 𝐜𝐨𝐬 (𝜶 + 𝜷 + 𝜸) + 𝐜𝐨𝐬 𝜸}
= 𝟐 𝐜𝐨𝐬 (
𝜶 + 𝜷
𝟐
) 𝐜𝐨𝐬 (
𝜶 − 𝜷
𝟐
) + 𝟐 𝐜𝐨𝐬 (
𝜶 + 𝜷 + 𝟐𝜸
𝟐
) 𝐜𝐨𝐬 (
𝜶 + 𝜷
𝟐
)
= 𝟐 𝐜𝐨𝐬 (
𝜶 + 𝜷
𝟐
) { 𝐜𝐨𝐬 (
𝜶 − 𝜷
𝟐
) + 𝐜𝐨𝐬 (
𝜶 + 𝜷 + 𝟐𝜸
𝟐
)}
= 𝟐 𝐜𝐨𝐬 (
𝜶 + 𝜷
𝟐
) {𝟐 𝐜𝐨𝐬 (
𝜸 + 𝜶
𝟐
) 𝐜𝐨𝐬 (
𝜷 + 𝜸
𝟐
)}
[using 𝐜𝐨𝐬 𝐂 + 𝐜𝐨𝐬 𝑫 = 𝟐 𝐜𝐨𝐬
(𝐂+𝑫)
𝟐
𝐜𝐨𝐬
(𝐂−𝑫)
𝟐
]
= 𝟒 𝐜𝐨𝐬 (
𝜶 + 𝜷
𝟐
) 𝟐 𝐜𝐨𝐬 (
𝜷 + 𝜸
𝟐
) 𝐜𝐨𝐬 (
𝜸 + 𝜶
𝟐
) = 𝐑𝐇𝐒.
Prove that 𝐬𝐢𝐧𝟐
𝟔𝒙 − 𝐬𝐢𝐧𝟐
𝟒𝒙 = 𝐬𝐢𝐧 𝟏𝟎𝒙 𝐬𝐢𝐧 𝟐𝒙.
LHS = 𝐬𝒊𝒏𝟐
𝟔𝒙 − 𝐬𝐢𝐧𝟐
𝟒𝒙 = 𝐬𝐢𝐧 (𝟔𝒙 + 𝟒𝒙) 𝐬𝐢𝐧 (𝟔𝒙 − 𝟒𝒙)
[ 𝐬𝐢𝐧𝟐
𝒙 − 𝐬𝐢𝐧𝟐
𝒚 = 𝐬𝐢𝐧 (𝒙 + 𝒚)𝐬𝒊𝐧(𝒙 − 𝒚)]= 𝐬𝐢𝐧 𝟏𝟎𝒙 𝐬𝐢𝐧 𝟐𝒙 = 𝐑𝐇𝐒.

Prove that 𝐜𝐨𝐬 𝟐𝟎∘
𝐜𝐨𝐬 𝟒𝟎∘
𝐜𝐨𝐬 𝟔𝟎∘
𝐜𝐨𝐬 𝟖𝟎∘
=
𝟏
𝟏𝟔
.
We have 𝐜𝐨𝐬 𝟐𝟎∘
𝐜𝐨𝐬 𝟖𝟎∘
=
𝟏
𝟐
𝐜𝐨𝐬 𝟒𝟎∘(𝟐 𝐜𝐨𝐬 𝟖𝟎∘
𝐜𝐨𝐬 𝟐𝟎∘)
=
𝟏
𝟐
×
𝟏
𝟐
⋅ { 𝐜𝐨𝐬 (𝟖𝟎∘
+ 𝟐𝟎∘) + 𝐜𝐨𝐬 (𝟖𝟎∘
− 𝟐𝟎∘)}
=
𝟏
𝟒
𝐜𝐨𝐬 𝟒𝟎∘( 𝐜𝐨𝐬 𝟏𝟎𝟎∘
+ 𝐜𝐨𝐬 𝟔𝟎∘) =
𝟏
𝟒
𝐜𝐨𝐬 𝟒𝟎∘ ( 𝐜𝐨𝐬 𝟏𝟎𝟎∘
+
𝟏
𝟐
)
=
𝟏
𝟖
(𝟐 𝐜𝐨𝐬 𝟏𝟎𝟎∘
𝐜𝐨𝐬 𝟒𝟎∘) +
𝟏
𝟖
=
𝟏
𝟖
[ 𝐜𝐨𝐬 (𝟏𝟎𝟎∘
+ 𝟒𝟎∘) + 𝐜𝐨𝐬 (𝟏𝟎𝟎∘
− 𝟒𝟎∘)] +
𝟏
𝟖
=
𝟏
𝟖
( 𝐜𝐨𝐬 𝟏𝟒𝟎∘
+ 𝐜𝐨𝐬 𝟔𝟎∘) +
𝟏
𝟖
=
𝟏
𝟖
𝐜𝐨𝐬 𝟏𝟒𝟎∘
+
𝟏
𝟏𝟔
+
𝟏
𝟖
=
𝟏
𝟖
𝐜𝐨𝐬 (𝟏𝟖𝟎∘
− 𝟒𝟎∘) +
𝟏
𝟏𝟔
+
𝟏
𝟖
= −
𝟏
𝟖
+
𝟏
𝟏𝟔
+
𝟏
𝟖
=
𝟏
𝟏𝟔
= 𝐑𝐇𝐒.

Prove that 𝐬𝐢𝐧 𝟏𝟎∘
𝐬𝐢𝐧 𝟓𝟎∘
𝐬𝐢𝐧 𝟔𝟎∘
𝐬𝐢𝐧 𝟕𝟎∘
=
√𝟑
𝟏𝟔
.
We have 𝐬𝐢𝐧 𝟏𝟎∘
𝐬𝐢𝐧 𝟓𝟎∘
𝐬𝐢𝐧 𝟕𝟎∘
=
𝟏
𝟐
𝐬𝐢𝐧 𝟏𝟎∘(𝟐 𝐬𝐢𝐧 𝟕𝟎∘
𝐬𝐢𝐧 𝟓𝟎∘)
=
𝟏
𝟐
×
√𝟑
𝟐
× 𝐬𝐢𝐧 𝟏𝟎∘
⋅ { 𝐜𝐨𝐬 (𝟕𝟎∘
− 𝟓𝟎∘) − 𝐜𝐨𝐬 (𝟕𝟎∘
+ 𝟓𝟎∘)}
=
√𝟑
𝟒
𝐬𝐢𝐧 𝟏𝟎∘
⋅ { 𝐜𝐨𝐬 𝟐𝟎∘
− 𝐜𝐨𝐬 𝟏𝟐𝟎∘}
=
√𝟑
𝟒
𝐬𝐢𝐧 𝟏𝟎∘ ( 𝐜𝐨𝐬 𝟐𝟎∘
+
𝟏
𝟐
) [⋅.⋅ 𝐜𝐨𝐬 𝟏𝟐𝟎∘
= −
𝟏
𝟐
]
=
√𝟑
𝟖
(𝟐 𝐜𝐨𝐬 𝟐𝟎∘
𝐬𝐢𝐧 𝟏𝟎∘) +
√𝟑
𝟖
𝐬𝐢𝐧 𝟏𝟎∘
=
√𝟑
𝟖
[ 𝐬𝐢𝐧 (𝟐𝟎∘
+ 𝟏𝟎∘) − 𝐬𝐢𝐧 (𝟐𝟎∘
− 𝟏𝟎∘)] +
√𝟑
𝟖
𝐬𝐢𝐧𝟏𝟎∘
=
√𝟑
𝟖
𝐬𝐢𝐧 𝟑𝟎∘
= (
√𝟑
𝟖
×
𝟏
𝟐
) =
√𝟑
𝟏𝟔
.

Class 11_Chapter 3_Lecture_4.pdf

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Class 11_Chapter 3_Lecture_4.pdf