Question

(cos 5theta + i * sin 5theta)/((cos 3theta - i * sin 3theta) ^ 2) =
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Answer 1

Answer:

Factor.

1−(cos5θcos3θ+sin5θsin3θ)

=2sin2θ

Note that

cos A cos B+sin A sinnB=cos(A−B) .

1−(cos(5θ−3θ))=2sin2θ

1−cos(2θ)=2sin2θ

Now use

cos(A+B)=cosAcosB−sin A sin B

(because cos2θ=cos(θ+θ)) to find

an expansion for cos(2θ) .

1−(cos2θ−sin2θ)=2sin2θ

1−cos2θ+sin2θ=2sin2θ

Now apply sin2x+cos2x=1. This implies that sin2x=1−cos2x .

sin2θ+sin2θ=2sin2θ

2sin2θ=2sin2θ

LHS=RHS

Since both sides are equal for all values of

θ

, this identity has been proved.

Explanation:

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