Can someone explain to me how you would solve this question: cos23(pi) ------- 12 please and thank you!
Last edited by skeske1234; Mar 25th 2009 at 12:03 PM.
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Originally Posted by skeske1234 Can someone explain to me how you would solve this question: cos2(pi) ------- 12 please and thank you! Hi Do you mean $\displaystyle \cos \frac{2\pi}{12}$ or $\displaystyle \frac{\cos 2\pi}{12}$ or anything else ?
Originally Posted by running-gag Hi Do you mean $\displaystyle \cos \frac{23\pi}{12}$ or $\displaystyle \frac{\cos 23\pi}{12}$ or anything else ? I mean: $\displaystyle \cos \frac{23\pi}{12}$ please, thanks
Originally Posted by skeske1234 I mean: $\displaystyle \cos \frac{23\pi}{12}$ please, thanks $\displaystyle \cos \frac{23\pi}{12} = \cos \left(-\frac{\pi}{12}\right) = \cos \frac{\pi}{12}$ Then using $\displaystyle \cos 2a = 2 \cos^2 a - 1$ and $\displaystyle \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ you can compute $\displaystyle \cos \frac{\pi}{12}$
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