Thread: Some Trig Help: Find the exact value of cos(−5pi/12)
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Old July 3rd, 2009, 06:02 PM
yeongil yeongil is offline
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\cos \left(-\frac{5\pi}{12}\right)
\begin{aligned} &= \cos \left(-\frac{\pi}{4} - \frac{\pi}{6}\right) \\ &= \cos \left(-\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(-\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right) \\ &= \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) \\ &= \frac{\sqrt{6} - \sqrt{2}}{4} \end{aligned}

Not sure how you got your answer...


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Last edited by mr fantastic; July 3rd, 2009 at 08:49 PM.
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