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

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  • 1 decade ago
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    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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