## Trigonometry (11th Edition) Clone

$$\cos^2\frac{\pi}{8}-\frac{1}{2}=\frac{\sqrt2}{4}$$
$$\cos^2\frac{\pi}{8}-\frac{1}{2}$$ Recall the Double-Angle Identity for cosine: $$2\cos^2A-1=\cos2A$$ So apparently, this case does not follow the already known Double-Angle Identity for cosine. That, nevertheless, does not mean we cannot apply the identity. What we need here are some transformations. $$\cos^2\frac{\pi}{8}-\frac{1}{2}=\Big(2\times\frac{1}{2}\times\cos^2\frac{\pi}{8}-\frac{1}{2}\Big)$$ $$\cos^2\frac{\pi}{8}-\frac{1}{2}=\frac{1}{2}\Big(2\cos^2\frac{\pi}{8}-1\Big)$$ Now $2\cos^2\frac{\pi}{8}-1$ can be applied with the identity $2\cos^2A-1=\cos2A$ for $A=\frac{\pi}{8}$. $$\cos^2\frac{\pi}{8}-\frac{1}{2}=\frac{1}{2}\Big[\cos\Big(2\times\frac{\pi}{8}\Big)\Big]$$ $$\cos^2\frac{\pi}{8}-\frac{1}{2}=\frac{1}{2}\Big(\cos\frac{\pi}{4}\Big)$$ $$\cos^2\frac{\pi}{8}-\frac{1}{2}=\frac{1}{2}\times\frac{\sqrt2}{2}$$ $$\cos^2\frac{\pi}{8}-\frac{1}{2}=\frac{\sqrt2}{4}$$