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I'm having trouble finding the exact value of $\cos^2\left(\frac{5\pi}{12}\right)$ in radians.

I was able to figure out that:

$$ \begin{align} \cos\left(\frac{7\pi}{12}\right) &= \cos\left(\frac{3\pi}{12} + \frac{4\pi}{12}\right) &\text{Split into known unit circle values.}\\ &= \cos\left(\frac{\pi}{4} + \frac{\pi}{3}\right) &\text{Reduce.}\\ &= \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) &\text{Use Cosine Identity.}\\ &= \cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{3}\right) &\text{Substitute values.}\\ &= \left(\frac{\sqrt2}{2} \times \frac{1}{2}\right) - \left(\frac{\sqrt2}{2} \times \frac{\sqrt3}{2}\right) &\text{Evaluate.}\\ &= \frac{\sqrt2}{4} - \frac{\sqrt6}{4}\\ &= \frac{\sqrt2-\sqrt6}{4}\\ \end{align} $$

I know that $\cos^2\left(\frac{5\pi}{12}\right)$ is not all that different, but the square exponent ($^2$) is throwing me for a loop.

Can anyone break it down for me?

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Hint: Try the double angle identity $$\cos^2\alpha = \frac{1}{2}(1+\cos 2\alpha).$$

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  • $\begingroup$ So would I just plug in 5pi/12 in for α and work it out from there? $\endgroup$ – Analytic Lunatic Sep 3 '15 at 20:07
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    $\begingroup$ I think you mean $\cos^2(a)=\frac{1}{2}(1+\cos(2a))$. $\endgroup$ – Eric Sep 3 '15 at 20:07
  • $\begingroup$ @Eric, same question: Would I then just plug in 5pi/12 in for α and work it out from there? $\endgroup$ – Analytic Lunatic Sep 3 '15 at 20:12
  • $\begingroup$ @AnalyticLunatic Yes. Try it. $\endgroup$ – rogerl Sep 3 '15 at 20:14
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    $\begingroup$ @AnalyticLunatic I think you probably do, since $2\cdot \frac{5\pi}{12} = \frac{5\pi}{6}$. $\endgroup$ – rogerl Sep 3 '15 at 20:18
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Since $\cos\alpha=-\cos(\pi-\alpha)$, $$ \cos\frac{5\pi}{12}=-\cos\left(\pi-\frac{5\pi}{12}\right)= -\cos\frac{7\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4} $$ Then $$ \cos^2\frac{5\pi}{12}=\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^2 =\frac{2-\sqrt{3}}{4} $$

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