How do you find the exact values of cos 22.5 degrees using the half angle formula?

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Answer 1 · 2015-08-06T01:45:23

The half angle identity for cosine can be derived (since I don't recall it off-hand):

${cos}^{2} (x) = \frac{1 + cos (2 x)}{2}$ $cos^2(x) = (1+cos(2x))/2$

By inference:
${cos}^{2} (\frac{x}{2}) = \frac{1 + cos x}{2}$ $cos^2(x/2) = (1+cosx)/2$

Square root to get:

$cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + cos x}{2}}$ $cos(x/2) = pmsqrt((1+cosx)/2)$
$+$ $+$ if quadrant I or IV
$-$ $-$ if quadrant II or III

${22.5}^{o}$ $22.5^o$ is quadrant I, so it is positive.

$cos (\frac{45^{o}}{2}) = \sqrt{\frac{1 + {cos 45}^{o}}{2}}$ $cos(45^o/2) = sqrt((1+cos45^o)/2)$

$= \sqrt{\frac{1 + (\frac{\sqrt{2}}{2})}{2}}$ $= sqrt((1+(sqrt2/2))/2)$

$= \sqrt{\frac{(\frac{2 + \sqrt{2}}{2})}{2}}$ $= sqrt((((2+sqrt2)/2))/2)$

$= \sqrt{\frac{2 + \sqrt{2}}{4}}$ $= sqrt((2+sqrt2)/4)$

$= \frac{\sqrt{2 + \sqrt{2}}}{2}$ $= color(blue)(sqrt(2+sqrt2)/2)$

or $\approx 0.9238795$ $~~0.9238795$