# How do you use the sum and difference formula to simplify cos((17pi)/12)?

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Nghi N. Share
Mar 23, 2017

$- \frac{\sqrt{2 - \sqrt{3}}}{2}$

#### Explanation:

$\cos \left(\frac{17 \pi}{12}\right) = \cos \left(\frac{- 7 \pi}{12} + \frac{24 \pi}{12}\right) = \cos \left(\frac{- 7 \pi}{12} + 2 \pi\right) =$
$= \cos \left(\frac{- 7 \pi}{12}\right) = \cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{12} + \frac{\pi}{2}\right) = - \sin \left(\frac{\pi}{12}\right)$
Find $\sin \left(\frac{\pi}{12}\right)$ by using trig identity:
$2 {\sin}^{2} \left(\frac{\pi}{12}\right) = 1 - \cos \left(\frac{\pi}{6}\right) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$
${\sin}^{2} \left(\frac{\pi}{12}\right) = \frac{2 - \sqrt{3}}{4}$
$\sin \left(\frac{\pi}{12}\right) = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$.
Since $\sin \left(\frac{\pi}{12}\right)$ is positive, then take the positive value.
$\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$
Finally,
$\cos \left(\frac{17 \pi}{12}\right) = - \sin \left(\frac{\pi}{12}\right) = - \frac{\sqrt{2 - \sqrt{3}}}{2}$

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Nghi N. Share
Mar 3, 2017

$- \cos \left(\frac{5 \pi}{12}\right)$

#### Explanation:

Simplify -->
$\cos \left(\frac{17 \pi}{12}\right) = \cos \left(\frac{5 \pi}{12} + \pi\right) = - \cos \left(\frac{5 \pi}{12}\right)$

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dk_ch Share
Oct 29, 2016

$\cos \left(\frac{17 \pi}{12}\right)$

$\cos \left(\frac{24 \pi - 7 \pi}{12}\right)$

$= \cos \left(2 \pi - \frac{7 \pi}{12}\right)$

$= \cos \left(\frac{7 \pi}{12}\right)$

$= \cos \left(\frac{4 \pi + 3 \pi}{12}\right)$

$= \cos \left(\frac{4 \pi}{12} + \frac{3 \pi}{12}\right)$

$= \cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right)$

$= \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)$

$= \frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}$

$= \frac{1 - \sqrt{3}}{2 \sqrt{2}}$

$= \frac{1}{4} \left(\sqrt{2} - \sqrt{6}\right)$

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Sep 15, 2016

$\cos \left(\frac{17 \pi}{12}\right) = \frac{1 - \sqrt{3}}{2 \sqrt{2}}$

#### Explanation:

$\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$

Let $A = \frac{8 \pi}{12} = 2 \frac{\pi}{3}$ and $B = \frac{9 \pi}{12} = \frac{3 \pi}{4}$, then $A + B = \frac{17 \pi}{12}$

Now $\cos A = \cos \left(\frac{2 \pi}{3}\right) = \cos \left(\pi - \frac{\pi}{3}\right) = - \cos \left(\frac{\pi}{3}\right) = - \frac{1}{2}$;

$\sin A = \sin \left(\frac{2 \pi}{3}\right) = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$;

$\cos B = \cos \frac{3 \pi}{4} = \cos \left(\pi - \frac{\pi}{4}\right) = - \cos \frac{\pi}{4} = - \frac{1}{\sqrt{2}}$; and

$\sin B = \sin \left(\frac{3 \pi}{4}\right) = \sin \left(\pi - \frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Hence, $\cos \left(\frac{17 \pi}{12}\right) = \cos \left(A + B\right)$

= $\cos A \cos B - \sin A \sin B$

= $\left(- \frac{1}{2}\right) \times \left(- \frac{1}{\sqrt{2}}\right) - \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{1}{\sqrt{2}}\right)$

= $\frac{1}{2 \sqrt{2}} - \frac{\sqrt{3}}{2 \sqrt{2}} = \frac{1 - \sqrt{3}}{2 \sqrt{2}}$

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