Question

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

$Tan( \frac{7 \pi }{8} )$


show work please

Answer 1

$\bf tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \cfrac{1-cos({{ \theta}})}{sin({{ \theta}})} \end{cases}\\\\ -------------------------------\\\\ \cfrac{7}{8}\cdot 2\implies \cfrac{7}{4}\qquad therefore\qquad \cfrac{\frac{7}{4}}{2}\implies \cfrac{7}{8}\qquad thus$

$\bf tan\left( \frac{7\pi }{8} \right)\implies tan\left( \cfrac{\frac{7\pi }{4}}{2} \right)=\cfrac{sin\left( \frac{7\pi }{4} \right)}{1+cos\left( \frac{7\pi }{4} \right)} \\\\\\ tan\left( \cfrac{\frac{7\pi }{4}}{2} \right)=\cfrac{-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}\implies tan\left( \cfrac{\frac{7\pi }{4}}{2} \right)=\cfrac{-\frac{\sqrt{2}}{2}}{\frac{2+\sqrt{2}}{2}}$

$\bf tan\left( \cfrac{\frac{7\pi }{4}}{2} \right)=-\cfrac{\sqrt{2}}{\underline{2}}\cdot \cfrac{\underline{2}}{2+\sqrt{2}}\implies tan\left( \cfrac{\frac{7\pi }{4}}{2} \right)=-\cfrac{\sqrt{2}}{2+\sqrt{2}}$

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

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