How do you evaluate $cos (\frac{13 π}{8})$ $cos((13pi)/8)$ ?

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How do you evaluate $cos (\frac{13 π}{8})$ $cos((13pi)/8)$ ?

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Answer 1 · 2016-05-14T14:43:15

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

Explanation:

using formula $cos θ = \sqrt{\frac{1}{2} (1 + cos (2 θ))}$ $costheta=sqrt(1/2(1+cos(2theta)))$
$cos (\frac{13 π}{8})$ $cos((13pi)/8)$
$= \sqrt{\frac{1}{2} (1 + cos (2 \cdot \frac{13 π}{8}))}$ $=sqrt(1/2(1+cos(2*(13pi)/8)))$

$= \sqrt{\frac{1}{2} (1 + cos (\frac{13 π}{4}))}$ $=sqrt(1/2(1+cos((13pi)/4)))$

$= \sqrt{\frac{1}{2} (1 + cos (3 π + \frac{π}{4}))}$ $=sqrt(1/2(1+cos(3pi+pi/4))$

$= \sqrt{\frac{1}{2} (1 - cos (\frac{π}{4}))}$ $=sqrt(1/2(1-cos(pi/4)))$

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

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

$= \sqrt{\frac{2 - \sqrt{2}}{2 \times 2}}$ $=sqrt((2-sqrt2)/(2xx2)$

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

Answer 2 · 2016-05-14T22:31:48

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

Explanation:

$cos (\frac{13 π}{8}) = cos (\frac{3 π}{8} + cos (\frac{16 π}{8})) = cos (\frac{3 π}{8} + 2 π) =$ $cos ((13pi)/8) = cos ((3pi)/8 + cos ((16pi)/8)) = cos ((3pi)/8 + 2pi) =$
$= cos (\frac{3 π}{8}) .$ $= cos ((3pi)/8).$
Evaluate $cos (\frac{3 π}{8})$ $cos ((3pi)/8)$ by applying the trig identity:
$cos 2 a = 2 {cos}^{2} a - 1$ $cos 2a = 2cos^2 a - 1$ .
In this case, we get -->
$2 {cos}^{2} (\frac{3 π}{8}) - 1 = cos (\frac{6 π}{8}) = cos (\frac{3 π}{4}) = - \frac{\sqrt{2}}{2}$ $2cos^2 ((3pi)/8) - 1 = cos ((6pi)/8) = cos ((3pi)/4) = -sqrt2/2$
$2 {cos}^{2} (\frac{3 π}{8}) = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$ $2cos^2 ((3pi)/8) = 1 - sqrt2/2 = (2 - sqrt2)/2$
${cos}^{2} (\frac{3 π}{8}) = \frac{2 - \sqrt{2}}{4}$ $cos^2 ((3pi)/8) = (2 - sqrt2)/4$
$cos (\frac{3 π}{8}) = \pm \frac{\sqrt{2 - \sqrt{2}}}{2}$ $cos ((3pi)/8) = +- sqrt(2 - sqrt2)/2$
$cos (\frac{13 π}{8}) = cos (\frac{3 π}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $cos ((13pi)/8) = cos ((3pi)/8) = sqrt(2 - sqrt2)/2$ , because
$cos (\frac{3 π}{8})$ $cos ((3pi)/8)$ is positive.

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How do you evaluate $cos (\frac{13 π}{8})$ $cos((13pi)/8)$ ?

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