The value of $$\cos^{2}\dfrac{\pi}{8} + \cos^{2}\dfrac{3\pi}{8} + \cos^{2}\dfrac{5\pi}{8} + \cos^{2}\dfrac{7\pi}{8}$$ is

$$\cos^{2}\dfrac{\pi}{8} + \cos^{2}\dfrac{3\pi}{8} + \cos^{2}\dfrac{5\pi}{8} + \cos^{2}\dfrac{7\pi}{8}$$

$$\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\left(\pi-\dfrac{3\pi}{8}\right)+\cos^2\left(\pi-\dfrac{\pi}{8}\right)$$

We know that, $$\cos\left(\pi-\theta\ \right)=-\cos\left(\theta\right)$$, therefore -

$$\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{\pi}{8}$$

$$2\cos^2\dfrac{\pi}{8}+2\cos^2\dfrac{3\pi}{8}$$

$$2\cos^2\left(\dfrac{\pi}{2}-\dfrac{3\pi}{8}\right)+2\cos^2\dfrac{3\pi}{8}$$

We know that, $$\cos\left(\dfrac{\pi}{2}-\theta\ \right)=\sin\left(\theta\right)$$, therefore - 

$$2\sin^2\dfrac{3\pi}{8}+2\cos^2\dfrac{3\pi}{8}$$

$$2\left(\sin^2\dfrac{3\pi}{8}+\cos^2\dfrac{3\pi}{8}\right)$$

$$2\left(1\right)=2$$