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Precalculus with Trigonometry

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Prove or disprove the following Identities:

cos(-x) - sin(-x) = cos(x) + sin (x)

sin raised to the 4 (theta) - cos raised to the 4 (theta) = sin squared (theta) - cos squared (theta)

cos (x+(pi)/(6)) + sin (x - (pi)/(3)) = 0

cos(x+y)cos(x-y) = cos squared (x) - sin squared (y)

Sorry if you don't understand anything

  • Precalculus with Trigonometry - ,

    For the first problem:

    cos(-x) - sin(-x) = cos(x) + sin(x)

    A few identities for negatives:

    cos(-x) = cos(x)
    sin(-x) = -sin(x)

    Therefore:

    cos(x) - [-sin(x)] = cos(x) + sin(x)

    ==================================

    For your last problem:

    cos(x+y)cos(x-y) = cos^2(x) - sin^2(y)

    Some identities:
    cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
    cos(x-y) = cos(x)cos(y) + sin(x)sin(y)

    Multiplying using both identities:

    [cos^2(x) cos^2(y)] - [sin^2(x) sin^2(y)]

    Next, use the identity:
    sin^2(x) + cos^2(x) = 1

    [1-sin^2(x)][1-sin^2(y)] - [sin^2(x) sin^2(y)]

    Multiply [1-sin^2(x)][1-sin^2(y)]:

    1 - sin^2(y) - sin^2(x) + [sin^2(x) sin^2(y)] - [sin^2(x) sin^2(y)]

    We are left with this:
    1 - sin^2(y) - sin^2(x)

    Which equals this:
    cos^2(x) - sin^2(y)

    ===================

    I'll stop there. I hope this helps.

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