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Math Help - trigonometry problem????

  1. #1
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    trigonometry problem????

    using the exact values of the sine and cosine of both 1/6∏ and 3/4∏ and one of the sum and difference formulas, show that the exact value of cos(7/12∏) is 1/4(√2-√6)
    any help would be greatly welcomed...thanks
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    Quote Originally Posted by jordan1 View Post
    using the exact values of the sine and cosine of both 1/6∏ and 3/4∏ and one of the sum and difference formulas, show that the exact value of cos(7/12∏) is 1/4(√2-√6)
    Follow the instructions:

    You know that 7/12 = 3/4 - 1/6 = 9/12 - 2/12.

    You have memorized the exact values for the sine and cosine of the (1/4)(pi) and (1/6)(pi).

    You have been given some sum and difference identities.

    So restate the (7/12)(pi) in terms of (3/4)(pi) and (1/6)(pi), apply the appropriate identities, and then simplify the result, using what you know about the periodicity of the sine and cosine waves.

    If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!
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  3. #3
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    Quote Originally Posted by jordan1 View Post
    using the exact values of the sine and cosine of both 1/6∏ and 3/4∏ and one of the sum and difference formulas, show that the exact value of cos(7/12∏) is 1/4(√2-√6)
    any help would be greatly welcomed...thanks
    sin(\frac{\pi}{6}) = \frac{1}{2}

    cos(\frac{\pi}{6}) = \frac{\sqrt3}{2}

    sin(\frac{3\pi}{4}) = \frac{\sqrt2}{2}

    cos(\frac{3\pi}{4}) = -\frac{\sqrt2}{2}

    \frac{7\pi}{12} = \frac{3\pi}{4} - \frac{\pi}{6}

    cos(A-B) = cosAcosB + sinAsinB

    A = \frac{3\pi}{4} and B = \frac{\pi}{6}
    Last edited by e^(i*pi); April 6th 2009 at 07:11 AM. Reason: sign error
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