Dylan Chandler

2023-03-11

How to use the half-angle identity to find the exact value of cos (11pi/8)?

Bronchialyo2

$\mathrm{cos}\left(\frac{11\pi }{8}\right)=\mathrm{cos}\left(\frac{3\pi }{8}+\pi \right)=-\frac{\mathrm{cos}\left(3\pi \right)}{8}$= cos t
employ trig identity $\mathrm{cos}2t=2{\mathrm{cos}}^{2}t-1$
$\mathrm{cos}2t=\mathrm{cos}\left(\frac{6\pi }{8}\right)=\mathrm{cos}\left(\frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}$
$-\frac{\sqrt{2}}{2}=2{\mathrm{cos}}^{2}t-1$
$2{\mathrm{cos}}^{2}t=1-\frac{\sqrt{2}}{2}=\frac{2-\sqrt{2}}{2}$
${\mathrm{cos}}^{2}t=\frac{2-\sqrt{2}}{4}$
$\mathrm{cos}t=\mathrm{cos}\left(\frac{11\pi }{8}\right)=±\frac{\sqrt{2-\sqrt{2}}}{2}.$
Only the negative answer is accepted because the arc (11pi/8 is in Quadrant II.
$\mathrm{cos}\left(\frac{11\pi }{8}\right)=-\frac{\sqrt{2-\sqrt{2}}}{2}$
Check by calculator.
$\mathrm{cos}\left(\frac{11\pi }{8}\right)=\mathrm{cos}247.5$ deg = - 0.382
$-\frac{\sqrt{2-\sqrt{2}}}{2}=$ - 0.382.

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