Question

Please try to prove it...............

Answer 1

cosθ = 1 - cos²θ

⇒ cosθ = sin²θ

given: sin¹²θ + 3sin¹⁰θ + 3sin⁸θ + sin⁶θ + 2sin⁴θ - 2cos²θ

but cosθ = sin²θ

⇒ cos⁶θ + 3cos⁵θ + 3cos⁴θ + cos³θ + 2cos²θ - 2cos²θ

⇒ the above expression reduces to :

cos⁶θ + 3cos⁵θ + 3cos⁴θ + cos³θ

taking cos³θ common from the above expression

we get cos³θ (cos³θ + 3cos²θ + 3cosθ + 1)

we know that (a + b)³ = a³ + 3ab² + 3a²b + b³

⇒ cos³θ + 3cos²θ + 3cosθ + 1 = (cosθ + 1)³

⇒ the above expression is cos³θ(cosθ + 1)³

⇒ cos³θ(cosθ + 1)³ = {cosθ(cosθ +1)}³

⇒ (cos²θ + cosθ)³

⇒but given cos²θ + cosθ = 1

⇒ (cos²θ + cosθ)³ = 1³ = 1