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# Trig identity help with LHS?

Trig identity - prove the equation as an identity

1) (cos 2x + cos 2y) / (sin x + cos y) = 2 cos y - 2 sin x

2) sin^4 = (1 - cot^2 x + cos^2 x cot^2 x) / csc^2x

to clarify it's ...cost^2 x times cot^2 x / csc^2x

### 1 Answer

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- ireadlotsLv 61 decade agoFavorite Answer
1) LHS

= (cos 2x + cos 2y) /(sinx+cosy)

= (1-2sin^2x + 2cos^2y - 1)/(sinx + cosy)

= (2cos^2y- 2sin^2x)/(sinx + cosy)

= 2(cosy - sinx)(cosy + sinx)/(sinx + cosx)

= 2cosy - 2 sinx

= RHS

RHS

=(1-cos^2x/Sin^2x + cos^2x * cos^2x/Sin^2x) * sin^2x

= Sin^2x - Cos^2x + cos^4x

= sin^2x- (1-SIn^2x) + cos^4x

= 2sin^2x+ cos^4x - 1

= 2sin^2x + (1-Sin^2x)^2 - 1

= 2sin^2x + 1 - 2Sin^2x + SIn^4x -1

= sin^4x

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