# How can Sine, Cosine and Tangent be found?

## $\frac{5 \pi}{12}$

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#### Explanation

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#### Explanation:

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Mar 9, 2018

$\sin \left(\frac{5 \pi}{12}\right) = \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)$

$\cos \left(\frac{5 \pi}{12}\right) = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

$\tan \left(\frac{5 \pi}{12}\right) = 2 + \sqrt{3}$

#### Explanation:

Using the sum of angles formula for $\sin$ we find:

$\sin \left(\frac{5 \pi}{12}\right) = \sin \left(\frac{\pi}{4} + \frac{\pi}{6}\right)$

$\textcolor{w h i t e}{\sin \left(\frac{5 \pi}{12}\right)} = \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{4}\right)$

$\textcolor{w h i t e}{\sin \left(\frac{5 \pi}{12}\right)} = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{1}{2} \frac{\sqrt{2}}{2}$

$\textcolor{w h i t e}{\sin \left(\frac{5 \pi}{12}\right)} = \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)$

Using the sum of angles formula for $\cos$ we find:

$\cos \left(\frac{5 \pi}{12}\right) = \cos \left(\frac{\pi}{4} + \frac{\pi}{6}\right)$

$\textcolor{w h i t e}{\cos \left(\frac{5 \pi}{12}\right)} = \cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) - \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right)$

$\textcolor{w h i t e}{\cos \left(\frac{5 \pi}{12}\right)} = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \frac{1}{2}$

$\textcolor{w h i t e}{\cos \left(\frac{5 \pi}{12}\right)} = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

Then:

$\tan \left(\frac{5 \pi}{12}\right) = \sin \frac{\frac{5 \pi}{12}}{\cos} \left(\frac{5 \pi}{12}\right)$

$\textcolor{w h i t e}{\tan \left(\frac{5 \pi}{12}\right)} = \frac{\frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)}{\frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)}$

$\textcolor{w h i t e}{\tan \left(\frac{5 \pi}{12}\right)} = {\left(\sqrt{6} + \sqrt{2}\right)}^{2} / \left(\left(\sqrt{6} - \sqrt{2}\right) \left(\sqrt{6} + \sqrt{2}\right)\right)$

$\textcolor{w h i t e}{\tan \left(\frac{5 \pi}{12}\right)} = \frac{6 + 4 \sqrt{3} + 2}{6 - 2}$

$\textcolor{w h i t e}{\tan \left(\frac{5 \pi}{12}\right)} = 2 + \sqrt{3}$

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