chururiakop

2023-03-11

How to find the exact values of cos(11pi/12) using the half angle formula?

Metafune7re

Call $\mathrm{cos}\left(\frac{11\pi }{12}\right)=\mathrm{cos}t$
$\mathrm{cos}2t=\mathrm{cos}\left(\frac{22\pi }{12}\right)=\mathrm{cos}\left(\frac{11\pi }{6}\right)=\mathrm{cos}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$
Put the trig identity to use: $\mathrm{cos}2t=2{\mathrm{cos}}^{2}t-1$
$\mathrm{cos}2t=\frac{\sqrt{3}}{2}=2{\mathrm{cos}}^{2}t-1$
$2{\mathrm{cos}}^{2}t=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}$
${\mathrm{cos}}^{2}t=\frac{2+\sqrt{3}}{4}$
$\mathrm{cos}t=\mathrm{cos}\left(\frac{11\pi }{12}\right)=±\frac{\sqrt{2+\sqrt{3}}}{2}.$
Since the arc ((11pi)/12) is located in Quadrant II, only the negative answer is accepted.
$\mathrm{cos}\left(\frac{11\pi }{12}\right)=-\frac{\sqrt{2+\sqrt{3}}}{2}$
Check by calculator.
$Arc\left(\frac{11\pi }{12}\right)=165$ deg-> $\mathrm{cos}\left(\frac{11\pi }{12}\right)=\mathrm{cos}165=-0.97.$
$-\left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)=-0.97$.

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