## Trigonometry (11th Edition) Clone

$$\tan\Big(-\frac{\pi}{8}\Big)=1-\sqrt2$$ 8 should be matched with E.
$$\tan\Big(-\frac{\pi}{8}\Big)$$ - 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.