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If sinθ + sin2θ = 1, then the value of cos12θ + 3cos10θ + 3cos8θ + cos6θ – 1 is (SSC CGL 1st Sit. 2012)
- a)0
- b)1
- c)–1
- d)2
Correct answer is option 'A'. Can you explain this answer?
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If sinθ + sin2θ = 1, then the value of cos12θ + 3cos...
sinθ + sin2θ = 1
⇒ sinθ = 1 – Sin2θ
⇒ sinθ = cos2θ
∴ cos12θ + 3cos10θ + 3cos8θ + cos6θ – 1
= (cos4θ + cos2θ)3 – 1
= (sin2θ + cos2θ)3 – 1 = 1 – 1 = 0
⇒ sinθ = 1 – Sin2θ
⇒ sinθ = cos2θ
∴ cos12θ + 3cos10θ + 3cos8θ + cos6θ – 1
= (cos4θ + cos2θ)3 – 1
= (sin2θ + cos2θ)3 – 1 = 1 – 1 = 0
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