Question

I'm confused! Please explain!

Answer 1

First, use the fact that $\tan x$ is $\pi$-periodic. This means $\tan(x+\pi)=\tan x$ for all $x$, and in particular

$\tan\dfrac{9\pi}8=\tan\left(\dfrac\pi8+\pi\right)=\tan\dfrac\pi8$

Now, recall the half-angle identities,

$\sin^2\dfrac x2=\dfrac{1-\cos x}2$

$\cos^2\dfrac x2=\dfrac{1+\cos x}2$

$\implies\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{1-\cos x}{1+\cos x}$

When we take the square root, we get two possible values, but we know $\tan x>0$ for $0$, so we know to take the positive square root:

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$

So we have

$\tan\dfrac\pi8=\sqrt{\dfrac{1-\cos\frac\pi4}{1+\cos\frac\pi4}}=\sqrt{\dfrac{1-\frac1{\sqrt2}}{1+\frac1{\sqrt2}}}=\sqrt{(\sqrt2-1)^2}$

$\boxed{\tan\dfrac\pi8=\sqrt2-1}$