Question

Evaluate tan(17pi/12) using the half angle formula

Answer 1

`\bf tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \cfrac{1-cos({{ \theta}})}{sin({{ \theta}})} \end{cases}\\\\ -----------------------------\\\\`


`\bf 2\cdot \cfrac{17}{12}\implies \cfrac{17}{6}\qquad thus\quad\cfrac{17\pi }{12}\iff\cfrac{(17\pi )/(6)}{2} \\\\\\ now\quad \cfrac{17}{6}\iff 2+\cfrac{5}{6}\quad or\quad 2\pi +\cfrac{5\pi }{6} \\\\\\ \textit{so the angle goes one way around the circle }2\pi \\ \textit{and has a co-terminal angle of }\cfrac{5\pi }{6}\\\\ \textit{so, let's use that one then}`


`\bf tan\left( \cfrac{(5\pi )/(6)}{2} \right)=\cfrac{sin\left( (5\pi )/(6) \right)}{1+cos\left( (5\pi )/(6) \right)}\implies tan\left( \cfrac{(5\pi )/(6)}{2} \right)=\cfrac{(1)/(2)}{1-(√(3))/(2)} \\\\\\ tan\left( \cfrac{(5\pi )/(6)}{2} \right)=\cfrac{(1)/(2)}{(2-√(3))/(2)}\implies tan\left( \cfrac{(5\pi )/(6)}{2} \right)=\cfrac{1}{2}\cdot \cfrac{2}{2-√(3)} \\\\\\ tan\left( \cfrac{(5\pi )/(6)}{2} \right)=\cfrac{1}{2-√(3)}`

now, let us use the conjugate of the denominator, to rationalize it


`\bf \cfrac{1}{2-√(3)}\cdot \cfrac{2+√(3)}{2+√(3)}\implies \cfrac{2+√(3)}{(2)^2-(√(3))^2}\implies \cfrac{2+√(3)}{4-3}\implies \boxed{2+√(3)}`