Answer
$$\tan\Big(-\frac{\pi}{8}\Big)=1-\sqrt2$$
8 should be matched with E.
Work Step by Step
$$\tan\Big(-\frac{\pi}{8}\Big)$$
- From negative-angle Identities: $\tan(-\theta)=-\tan\theta$. Therefore,
$$\tan\Big(-\frac{\pi}{8}\Big)=-\tan\frac{\pi}{8}$$
Examine $\frac{\pi}{8}$: $$\frac{\pi}{8}=\frac{\pi}{4\times2}=\frac{1}{2}\times\frac{\pi}{4}$$
Therefore, $$\tan\Big(-\frac{\pi}{8}\Big)=-\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)$$
- Recall the half-angle identities:
$$\tan\Big(\frac{A}{2}\Big)=\frac{1-\cos A}{\sin A}$$
So, if we replace $A=\frac{\pi}{4}$, we would have
$$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{1-\cos\frac{\pi}{4}}{\sin\frac{\pi}{4}}$$
$$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{1-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}$$
$$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{\frac{2-\sqrt2}{2}}{\frac{\sqrt2}{2}}$$
$$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{2-\sqrt2}{\sqrt2}$$
$$\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\sqrt2-1$$
Thus, $$\tan\Big(-\frac{\pi}{8}\Big)=-\tan\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=-(\sqrt2-1)=1-\sqrt2$$
8 should be matched with E.
