Question

Find the exact value of tan π/12.

Answer 1

`\textit{Half-Angle Identities}\qquad tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}~~\leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\\\ 2\cdot \cfrac{\pi }{12}\implies \cfrac{\pi }{6}~\hspace{10em} \cfrac{~~ (\pi )/(6)~~}{2}\implies \cfrac{\pi }{6}\cdot \cfrac{1}{2}\implies \cfrac{\pi }{12}`

so let's use the double of the angle in the half-angle identities, check the picture below for the cosine, sine pairs

`tan\left( \cfrac{~~ (\pi )/(6)~~}{2} \right)=\cfrac{sin\left( (\pi )/(6) \right)}{1+cos\left( (\pi )/(6) \right)}\implies tan\left( \cfrac{~~ (\pi )/(6)~~}{2} \right)=\cfrac{~~ (1)/(2)~~}{1+(√(3))/(2)}`

`tan\left( \cfrac{~~ (\pi )/(6)~~}{2} \right)=\cfrac{~~ (1)/(2)~~}{~~(2+√(3))/(2)~~}\implies tan\left( \cfrac{~~ (\pi )/(6)~~}{2} \right)=\cfrac{1}{2}\cdot \cfrac{2}{2+√(3)} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill tan\left( \cfrac{~~ (\pi )/(6)~~}{2} \right)=\cfrac{1}{2+√(3)}~\hfill`