Question 3b(i)

SCH_CALC L3CALC

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$$\tan \left(\frac{\pi }{8}\right)=\sqrt{2}-1$$ $$\tan \left(\frac{3\pi }{8}\right)=?$$

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$$\tan \left(3\cdot \frac{\pi }{8}\right)=\tan \left(\frac{\pi }{8}\cdot (2+1)\right)=\tan \left(\frac{2\pi }{8}+\frac{\pi }{8}\right)$$

Using compound angles:

$$\tan \left(\frac{2\pi }{8}+\frac{\pi }{8}\right)=\frac{\tan \frac{2\pi }{8}+\tan \frac{\pi }{8}}{1-\tan \frac{2\pi }{8}\cdot \tan \frac{\pi }{8}}$$ $$=\frac{\tan \left(2\cdot \frac{\pi }{8}\right)+\tan \frac{\pi }{8}}{1-\tan \left(2\cdot \frac{\pi }{8}\right)\cdot \tan \frac{\pi }{8}}$$ $$=\frac{1+\tan \frac{\pi }{8}}{1-1\cdot \tan \frac{\pi }{8}}$$

Since $\tan \frac{\pi }{8}=\sqrt{2}-1$:

$$=\frac{1+\sqrt{2}-1}{1-(\sqrt{2}-1)}$$ $$=\frac{\sqrt{2}}{2-\sqrt{2}}$$ $$=1+\sqrt{2}$$