sinxsecx=tanx????? cos^2(2x)-sin^2(2x)=cos(4x)???? please help with these two problems?

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  • 1 decade ago
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    LHS = sinx secx

    = sinx (1/cosx)

    = sinx / cosx

    = tanx

    = RHS

    so sinx secx = tanx

    cos^2 (2x) - sin^2 (2x) = cos(4x)

    For this one I'm going to use the trig identities:

    cos(2x) = 2cos^2 x - 1 which can be rewritten cos^2 x = (1/2)[cos(2x) + 1]

    and cos(2x) = 1 - 2sin^2 x which can be rewritten sin^2 x = (1/2)[1 - cos(2x)]

    and substituting 2x for x in the identities like this:

    LHS = cos^2(2x) - sin^2(2x)

    = (1/2) [cos(4x) + 1] - (1/2)[1 - cos(4x)]

    = (1/2)cos(4x) + 1/2 - 1/2 + (1/2)cos(4x)

    = cos(4x)

    = RHS

    so cos^2 (2x) - sin^2 (2x) = cos(4x)

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