Question

if cos theta + cos²theta = 1,then sin¹²theta+3 sin¹⁰theta+3 sin⁸theta+sin⁶theta =
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Answer 1

Answer:

Step-by-step explanation:

cosθ=1−cos2θ

⇒cosθ=sin2θ   [sin²x+cos²x=1]

cos2θ=sin4θ

1−sin2θ=sin4θ    [sin²x+cos²x=1]

sin4θ+sin2θ=1 →equation (1)

Now cube on both sides ;

⇒sin12θ+sin6θ+3sin4θsin2θ(sin4θ+sin2θ)=1

⇒sin12θ+sin6θ+3sin10θ+3sin8θ=1

To obtain above result we add and subtract 2 on LHS side ;

⇒sin12θ+sin6θ+3sin10θ+3sin8θ+2(1)−2=1

From equation (1), 1=sin4θ+sin2θ

⇒sin12θ+sin6θ+3sin10θ+3sin8θ+2(sin4θ+sin2θ)−2=1

⇒sin12θ+3sin10θ+3sin8θ+sin6θ+2sin4θ+2sin2θ−2=1

Hence proved