Question

Some one please help me .Find the exact value by using a half-angle identity.


`Tan( (7 \pi )/(8) )`


show work please

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}\\\\ -------------------------------\\\\ \cfrac{7}{8}\cdot 2\implies \cfrac{7}{4}\qquad therefore\qquad \cfrac{(7)/(4)}{2}\implies \cfrac{7}{8}\qquad thus`


`\bf tan\left( (7\pi )/(8) \right)\implies tan\left( \cfrac{(7\pi )/(4)}{2} \right)=\cfrac{sin\left( (7\pi )/(4) \right)}{1+cos\left( (7\pi )/(4) \right)} \\\\\\ tan\left( \cfrac{(7\pi )/(4)}{2} \right)=\cfrac{-(√(2))/(2)}{1+(√(2))/(2)}\implies tan\left( \cfrac{(7\pi )/(4)}{2} \right)=\cfrac{-(√(2))/(2)}{(2+√(2))/(2)}`


`\bf tan\left( \cfrac{(7\pi )/(4)}{2} \right)=-\cfrac{√(2)}{\underline{2}}\cdot \cfrac{\underline{2}}{2+√(2)}\implies tan\left( \cfrac{(7\pi )/(4)}{2} \right)=-\cfrac{√(2)}{2+√(2)}`

now, let's rationalize the denominator, by multiplying top and bottom by the denominator's conjugate,


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