Question

Evaluate:

1. Tan pi/8
2. Sin17pi/8

Answer 1

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{following}\mathrm{trigonometric}\mathrm{value}. \\ \tan \frac{\mathrm{\pi }}{8} \\ \mathrm{Use}\mathrm{half}\mathrm{angle}\mathrm{udenttiy}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{above}\mathrm{value}. \\ \tan \frac{\mathrm{\pi }}{8}=\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{8}}}{\cos {\displaystyle \frac{\mathrm{\pi }}{8}}} \\ \mathrm{Multiply}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{on}\mathrm{RHS}\hspace{0.17em}\mathrm{by}2\cos \frac{\mathrm{\pi }}{8} \\ \tan \frac{\mathrm{\pi }}{8}=\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}}}{2\cos {\displaystyle \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}}} \\ =\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}}}{2\cos {\displaystyle {}^2\frac{\mathrm{\pi }}{8}}} \\ \mathrm{Now}\mathrm{use}\mathrm{the}\mathrm{half}\mathrm{angle}\mathrm{formulas} \\ 2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}=\mathrm{sinx} \\ 2{\cos }^2\frac{\mathrm{x}}{2}=1+\mathrm{cosx} \\ \mathrm{Here}\mathrm{x}\mathrm{is}\mathrm{half}\mathrm{of}\frac{\mathrm{\pi }}{8}\mathrm{that}\mathrm{is}\frac{\mathrm{\pi }}{4},\mathrm{so}\mathrm{it}\mathrm{implies}\mathrm{that} \\ \tan \frac{\mathrm{\pi }}{8}=\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{4}}}{1+\cos {\displaystyle \frac{\mathrm{\pi }}{4}}} \\ =\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{1+{\displaystyle \frac{1}{\sqrt{2}}}} \\ =\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{\sqrt{2}+1}{\sqrt{2}}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}+1}} \end{gathered}$$


$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{trigonometrci}\mathrm{value}. \\ \sin \frac{17\mathrm{\pi }}{8} \\ \mathrm{This}\mathrm{can}\mathrm{be}\mathrm{rewritten}\mathrm{as}, \\ \sin \frac{17\mathrm{\pi }}{8}=\sin \left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{8}\right) \\ =\sin \frac{\mathrm{\pi }}{8}(\mathrm{since}\mathrm{sine}\mathrm{function}\mathrm{has}\mathrm{period}2\mathrm{\pi }) \\ \mathrm{By}\mathrm{half}\mathrm{angle}\mathrm{identity}, \\ 2{\sin }^2\frac{\mathrm{x}}{2}=1-\mathrm{cosx}\mathrm{which}\mathrm{implies}\mathrm{that}\sin \frac{\mathrm{x}}{2}=\sqrt{\frac{1-\mathrm{cosx}}{2}} \\ \mathrm{Take}\mathrm{x}=\frac{\mathrm{\pi }}{4}\mathrm{to}\mathrm{get} \\ \sin \frac{\mathrm{\pi }}{8}=\sqrt{\frac{1-\cos {\displaystyle \frac{\mathrm{\pi }}{4}}}{2}} \\ =\sqrt{\frac{1-{\displaystyle \frac{{\displaystyle 1}}{\sqrt{2}}}}{2}} \\ =\sqrt{\frac{{\displaystyle \frac{{\displaystyle \sqrt{2}-1}}{\sqrt{2}}}}{2}} \\ =\sqrt{\frac{{\displaystyle \sqrt{2}-1}}{2\sqrt{2}}} \\ =\sqrt{\frac{{\displaystyle \left(\sqrt{2}-1\right)\times \sqrt{2}}}{2\sqrt{2}\times \sqrt{2}}} \\ =\sqrt{\frac{{\displaystyle 2-\sqrt{2}}}{4}} \\ =\frac{\sqrt{2-\sqrt{2}}}{2} \\ \mathrm{Thus}\mathrm{the}\mathrm{required}\mathrm{value}\mathrm{of}\sin \frac{17\mathrm{\pi }}{8}\mathrm{is}\frac{\sqrt{\mathbf{2}\mathbf{-}\sqrt{\mathbf{2}}}}{\mathbf{2}} \end{gathered}$$