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Math Help - another trig identity question

  1. #1
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    another trig identity question

    Hi
    I need help the following:

    Given 0 < \alpha,\gamma < \pi and cos(\alpha) = \frac{2}{3}, cos(\gamma) = \frac{-1}{3} calculate exact expression for cos(\alpha - \gamma)?

    This is what i have done however it is incorrect:

    we know that cos(\alpha) = \frac{2}{3}, cos(\gamma) = \frac{-1}{3},<br />
sin(\alpha) = \frac{3}{\sqrt{13}} and sin(\gamma) = \frac{-26}{81}

    i sub it into: cos(\alpha)cos(\gamma) + sin(\alpha)sin(\gamma)

    so i get:  \frac{-\sqrt{13}}{27} + \frac{-78}{81\sqrt{13}}

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help the following:

    Given 0 < \alpha,\gamma < \pi and cos(\alpha) = \frac{2}{3}, cos(\gamma) = \frac{-1}{3} calculate exact expression for cos(\alpha - \gamma)?

    This is what i have done however it is incorrect:

    we know that cos(\alpha) = \frac{2}{3}, cos(\gamma) = \frac{-1}{3},<br />
sin(\alpha) = \frac{3}{\sqrt{13}} and sin(\gamma) = \frac{-26}{81}

    i sub it into: cos(\alpha)cos(\gamma) + sin(\alpha)sin(\gamma)

    so i get:  \frac{-\sqrt{13}}{27} + \frac{-78}{81\sqrt{13}}

    P.S
    Dear Paymemoney,

    cos\alpha = \frac{2}{3}

    sin\alpha=\sqrt{1-cos^2\alpha}=\frac{\sqrt{5}}{3}

    cos\gamma = \frac{-1}{3}

    sin\gamma=\sqrt{1-cos^2\gamma}=\frac{\sqrt{8}}{3}

    Can you continue from here??
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  3. #3
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    yep i got the answer
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