Question #a927e

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Question #a927e

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Answer 1 · 2017-03-12T18:27:30

We know

$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ$ $cos2theta=cos^2theta-sin^2theta$

$cos 2 θ = {cos}^{2} θ - (1 - {cos}^{2} θ)$ $cos2theta=cos^2theta-(1-cos^2theta)$

$cos 2 θ = 2 {cos}^{2} θ - 1$ $cos2theta=2cos^2theta-1$

${cos}^{2} θ = \frac{1}{2} \cdot (1 + cos 2 θ)$ $cos^2theta=1/2*(1+cos2theta)$

$cos θ = \sqrt{\frac{1}{2} (1 + cos 2 \cdot θ)}$ $costheta=sqrt(1/2(1+cos2*theta))$

putting $θ = 22.5$ $theta=22.5$ we get

$cos (22.5) = \sqrt{\frac{1}{2} (1 + {cos 45}^{o})}$ $cos(22.5)=sqrt(1/2(1+cos45^o)$

$cos (22.5) = \sqrt{\frac{1}{2} (1 + \frac{1}{\sqrt{2}})}$ $cos(22.5)=sqrt(1/2(1+1/sqrt2)$

Answer 2 · 2017-03-12T22:30:11

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

Explanation:

Call cos (22.5) = cos t
$cos 2 t = cos (45) = \frac{\sqrt{2}}{2}$ $cos 2t = cos (45) = sqrt2/2$
Use trig identity:
$2 {cos}^{2} t = 1 + cos 2 t = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$ $2cos^2 t = 1 + cos 2t = 1 + sqrt2/2 = (2 + sqrt2)/2$
${cos}^{2} t = \frac{2 + \sqrt{2}}{4}$ $cos^2 t = (2 + sqrt2)/4$
$cos t = \pm \frac{\sqrt{2 + \sqrt{2}}}{2}$ $cos t = +- sqrt(2 + sqrt2)/2$
Since cos 22.5 is positive, then, take the positive answer.
$cos (22.5) = cos t = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $cos (22.5) = cos t = sqrt(2 +sqrt2)/2$

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