<s>[INST] Exact value of $\cos^2(\frac{\pi}{8})+\sin^2(\frac{15\pi}{8})?$ I tried separating it into $\cos^2(\frac{\pi}{8})+\sin^2(\frac{\pi}{8}+\frac{14\pi}{8})?$ and using the angle sum identity but it didn't help.
 [/INST] Notice that for all $x$ we have $\sin\left(2\pi-x\right)=\sin(-x)=-\sin(x)$ due to $2\pi$-periodicity of $\sin$ and the fact it is an odd function.
Applying, have
$$\sin\left(\frac{15\pi}{8}\right)^2=\left(\sin\left(2\pi-\frac{\pi}8\right)\right)^2=\left(\sin\left(-\frac{\pi}8\right)\right)^2=\sin\left(\frac{\pi}8\right)^2;$$
returning back to the original expression we get
$$\cos\left(\frac{\pi}8\right)^2+\sin\left(\frac{15\pi}8\right)^2=\cos\left(\frac{\pi}8\right)^2+\sin\left(\frac{\pi}8\right)^2,$$
which by the fundamental trigonometric identity $\sin(x)^2+\cos(x)^2=1$ easily gives
$$\therefore\cos\left(\frac{\pi}8\right)^2+\sin\left(\frac{15\pi}8\right)^2=\boxed{1}\,.$$
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