JEE Main202127 Jul 2021Morning ShiftMathematicsTrigonometric Ratios & IdentitiesActual
If sin θ + cos θ = 1 2 , then 16 sin 2 θ + cos 4 θ + sin 6 θ is equal to:
Options
- A23
- B- 27
- C- 23
- D27
Correct answer
C. - 23
Step-by-step solution
sin θ + cos θ = 1 2 sin 2 θ + cos 2 θ + 2 sin θ cos θ = 1 4 ⇒ sin 2 θ = - 3 4 Now: cos 4 θ = 1 - 2 sin 2 2 θ = 1 - 2 - 3 4 2 = 1 - 2 × 9 16 = - 1 8 And sin 6 θ = 3 sin 2 θ - 4 sin 3 2 θ = 3 - 4 sin 2 2 θ · sin 2 θ = 3 - 4 9 16 · - 3 4 = 3 4 × - 3 4 = - 9 16 So, 16 sin 2 θ + cos 4 θ + sin 6 θ = 16 - 3 4 - 1 8 - 9 16 = - 23