## Trigonometry (11th Edition) Clone

$$\cos\Big(\frac{\pi}{8}\Big)=\frac{\sqrt{2+\sqrt2}}{2}$$ 7 should be matched with D.
$$\cos\Big(\frac{\pi}{8}\Big)$$ Examine $\frac{\pi}{8}$: $$\frac{\pi}{8}=\frac{\pi}{4\times2}=\frac{1}{2}\times\frac{\pi}{4}$$ Therefore, $$\cos\Big(\frac{\pi}{8}\Big)=\cos\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)$$ - Recall the half-angle identities: $$\cos\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1+\cos A}{2}}$$ So, if we replace $A=\frac{\pi}{4}$, we would have $$\cos\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\pm\sqrt{\frac{1+\cos\frac{\pi}{4}}{2}}$$ $$\cos\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\pm\sqrt{\frac{1+\frac{\sqrt2}{2}}{2}}$$ $$\cos\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\pm\sqrt{\frac{\frac{2+\sqrt2}{2}}{2}}$$ $$\cos\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\pm\sqrt{\frac{2+\sqrt2}{4}}$$ $$\cos\Big(\frac{1}{2}\times\frac{\pi}{4}\Big)=\frac{\pm\sqrt{2+\sqrt2}}{2}$$ Thus, $$\cos\Big(\frac{\pi}{8}\Big)=\frac{\pm\sqrt{2+\sqrt2}}{2}$$ Now about the sign, it depends on the sign of $\cos\Big(\frac{\pi}{8}\Big)$. $\frac{\pi}{8}$ lies in quadrant I, where $\cos\theta\gt0$. So, $\cos\Big(\frac{\pi}{8}\Big)\gt0$. We need to select the positive square root. $$\cos\Big(\frac{\pi}{8}\Big)=\frac{\sqrt{2+\sqrt2}}{2}$$ 7 should be matched with D.