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Math Help - Some Trig Help: Find the exact value of cos(−5pi/12)

  1. #1
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    Some Trig Help: Find the exact value of cos(−5pi/12)

    I've used the sum and difference identity for cos and here is my answer:

    <br />
\frac{\sqrt{3}+\sqrt{2}}{4}<br />

    Cos -5pi/12 I translate as -75 degrees, so I used my known angles which were:

    pi/-6 and -pi/4.

    I also know that -5pi/12 is in the 4th region so the value should be positive (All Students Take Calculus).

    From the Cosine Sum formula we have:

    Cos(a+b)=Cos(a)Cos(b)-Sin(a)Sin(b)
    Cos(-pi/4 + -pi/6)= Cos(-pi/4)Cos(-pi/6)-Sin(-pi/4)Sin(-pi/6)
    Cos(-pi/4 + -pi/6)=(sqrt2/2)(sqrt3/2)-(sqrt2/2)(1/2)

    .....


    So what Did I do wrong guys?
    Last edited by mvho; July 3rd 2009 at 03:57 PM.
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  2. #2
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    \cos \left(-\frac{5\pi}{12}\right)
    \begin{aligned}<br />
&= \cos \left(-\frac{\pi}{4} - \frac{\pi}{6}\right) \\<br />
&= \cos \left(-\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(-\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right) \\<br />
&= \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) \\<br />
&= \frac{\sqrt{6} - \sqrt{2}}{4}<br />
\end{aligned}

    Not sure how you got your answer...


    01
    Last edited by mr fantastic; July 3rd 2009 at 06:49 PM.
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  3. #3
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    Thanks for the explanation, looks like I rushed through it.
    Last edited by mr fantastic; July 3rd 2009 at 06:50 PM.
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