If sinθ + sin²θ = 1, then the value of

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If sinθ + sin²θ = 1, then the value of cos² θ + 3 cos¹⁰ θ + 3 cos⁸ θ + cos⁶ θ – 1 is

sinθ + sin²θ = 1
⇒ sinθ = 1 – sin²θ = cos²θ
Now, cos¹²θ + 3 cos¹⁰q + 3 cos⁸θ + cos⁶θ – 1
= (cos⁴θ + cos2θ)³ – 1
= (sin²θ + cos²2θ)³ – 1
= 1 – 1 = 0