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Math Help - Trig identity

  1. #1
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    Trig identity

    I have partially solved a question.

    Prove that:

    (sin x)^2 - (sin y)^2 = (sin x cos y)^2 - (cos x sin y)^2
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  2. #2
    Junior Member Rachel.F's Avatar
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    Quote Originally Posted by Stuck Man View Post
    I have partially solved a question.

    Prove that:

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

    \sin^2 x - \sin^2 y = \sin^2 x \cos^2 y - \cos^2 x \sin^2 y

    \sin^2 x - \sin^2 y = (1-cos^2 x)\cos^2 y - \cos^2 x (1-\cos^2 y)

    \sin^2 x - \sin^2 y = \cos^2 y-cos^2 x\cos^2 y - \cos^2 x + \cos^2 x\cos^2 y

    \sin^2 x - \sin^2 y = \cos^2 y - \cos^2 x

    \sin^2 x - \sin^2 y = 1-\sin^2 y -(1- \sin^2 x)

    \sin^2 x - \sin^2 y = 1-\sin^2 y -1+\sin^2 x

    \sin^2 x - \sin^2 y = \sin^2 x-\sin^2 y
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  3. #3
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    Thanks. I didn't know the formula used twice by the first line. Is there a standard formula for that?
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  4. #4
    Junior Member Rachel.F's Avatar
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    Quote Originally Posted by Stuck Man View Post
    Thanks. I didn't know the formula used twice by the first line. Is there a standard formula for that?
    \sin^2 x + \cos^2 x = 1
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  5. #5
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    I don't know what that has to do with it. I meant is the a formula for (sin x cos y)^2 = (sin x)^2 (cos y)^2?
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  6. #6
    Super Member General's Avatar
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    Quote Originally Posted by Stuck Man View Post
    I don't know what that has to do with it. I meant is the a formula for (sin x cos y)^2 = (sin x)^2 (cos y)^2?
    It is not a formula.
    Its one of the exponents' laws.
    (ab)^c=a^c \;\ b^c
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  7. #7
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    Thanks.
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