Solving trig equations + double angle formulae
- 1. Block 3 Solving Trig Equations and Compound Angle Formulas
- 2. What is to be learned? • How to use compound angle formulas to solve more difficult trig equations
- 3. a0180 – a 180 + a 360 - a iii iii iv CT ASsin2x+ cosx = 0 2sinxcosx + cosx = 0 (2sinx + 1) = 0 cosx = 0 2sinx + 1 = 0 2sinx = -1 sinx = - ½ sin-1 ( ½ ) = 300 x = 180+30 or 360 – 30 x = 900 or 2700 cosx x = 2100 or 3300 x = 900 , 2100 , 2700 , 3300
- 4. a0180 – a 180 + a 360 - a iii iii iv CT ASsin2x + sinx = 0 2sinxcosx + sinx = 0 (2cosx + 1) = 0 sinx = 0 2cosx + 1 = 0 2cosx = -1 cosx = - ½ cos-1 ( ½ ) = 600 x = 180 – 60 or 180 + 60 x = 00 , 1800 or 3600 sinx x = 1200 or 2400 x = 00 , 1200 , 1800 , 2400 , 3600
- 5. Trig Equations and Double Angles • Use double angle formula • Get one side to zero • Factorise • Solve mini trig equations (exact angles and wee trig graphs handy)
- 6. a0180 – a 180 + a 360 - a iii iii iv CT ASsin2x = sinx 2sinxcosx – sinx = 0 (2cosx – 1) = 0 sinx = 0 2cosx – 1 = 0 2cosx = 1 cosx = ½ cos-1 ( ½ ) = 600 x = 600 or 360 – 60 x = 00, 1800 or 3600 sinx 2sinxcosx = sinx x = 00 , 600 , 1800 , 3000 , 3600 x = 600 or 3000
- 7. a0180 – a 180 + a 360 - a iii iii iv CT ASsin2x+ cosx = 0 2sinxcosx + cosx = 0 (2sinx + 1) = 0 cosx = 0 2sinx + 1 = 0 2sinx = -1 sinx = - ½ sin-1 ( ½ ) = 300 x = 180+30 or 360 – 30 x = 900 or 2700 cosx x = 2100 or 3300 x = 900 , 2100 , 2700 , 3300 Key Question
- 8. cos2x + cosx = 0 Three choices!!!! Cos2A = 1 – 2Sin2 A = 2Cos= 2Cos22 A – 1A – 1Cos2A Cos2A = Cos2 A – Sin2 A
- 9. cos2x + cosx = 0 2cos2 x – 1 + cosx = 0 2cos2 x + cosx – 1 = 0 2a2 + a – 1 (2a – 1)(a + 1)(2cosx – 1)(cosx + 1) = 0 2cosx–1 = 0 cosx + 1 = 0 cosx = ½ cosx = -1 cos-1 ( ½ ) = 600 x = 1800 x = 60 or 300 x = 600 , 1800 or 3000 Three choices!!!! Cos2A = 1 – 2Sin2 A = 2Cos= 2Cos22 A – 1A – 1Cos2A Cos2A = Cos2 A – Sin2 A
- 10. cos2x – cosx = 0 2cos2 x – 1 – cosx = 0 2cos2 x – cosx – 1 = 0 2a2 – a – 1 (2a + 1)(a – 1)(2cosx + 1)(cosx – 1) = 0 2cosx+1 = 0 cosx – 1 = 0 cosx = -½ cosx = 1 cos-1 ( ½ ) = 600 x = 0 ,3600 x = 120 or 240 x = 00 , 1200 , 2400 or 3600
- 11. For cos2x substitute the formula that will leave equation all sine or all cos.
- 12. cos2x – sinx = 0 1 – 2sin2 x – sinx = 0 2sin2 x + sinx – 1 = 0 2a2 + a – 1 (2a – 1)(a + 1)(2sinx – 1) (sinx + 1) = 0 2sinx – 1 = 0 Sinx + 1 = 0 sinx = ½ sinx = -1 sin-1 ( ½ ) = 300 x = 2700 x = 300 or 1500 x = 300 , 1500 or 2700 rearrange and multiply by -1
- 13. cos2x + sinx + 2 = 0 1 – 2sin2 x + sinx + 2 = 0 2sin2 x – sinx – 3 = 0 2a2 – a – 3 (2a – 3)(a + 1)(2sinx – 3) (sinx + 1) = 0 2sinx – 3 = 0 Sinx + 1 = 0 sinx = 3 /2 sinx = -1 No solutions x = 2700 x = 2700 rearrange and multiply by -1 Key Question