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.
 A: Notice that $\frac{15\pi}{8} = 2\pi - \frac{\pi}{8}$. So $\sin \frac{15\pi}{8} = -\sin \frac{\pi}{8}$. Your expression then reduces/evaluates down to: 
$$\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} = 1$$
A: 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
$$\therefore\cos\left(\frac{\pi}8\right)^2+\sin\left(\frac{15\pi}8\right)^2=\boxed{1}\,.$$
A: Using $\cos2A=2\cos^2A-1=1-2\sin^2A$
$$2\cos^2\dfrac\pi8=1+\cos\dfrac\pi4$$
$$2\sin^2\dfrac{15\pi}4=1-\cos\dfrac{15\pi}4$$
Now $\cos\dfrac{15\pi}4=\cos\left(4\pi-\dfrac\pi4\right)=\cos\dfrac\pi4$
