How to calculate phase difference between $\sin^2(x)$ and $\cos^2(x)$?

Question

Question: What is Phase difference $(\Delta\phi)$ between $\sin^{2}(x)$ and $\cos^{2}(x)$ ?

My attempt:

An example, $\Delta\phi$ for $\sin(x)$ and $\sin(x+\pi)$ is $\pi$.

Reducing $\sin^{2}(x)$ and $\cos^{2}(x)$.

$$\sin^{2}(x) =\frac{1-\cos(2x)}{2}$$ $$\cos^{2}(x) =\frac{1+\cos(2x)}{2}$$

So, the question boils down to phase difference between $\cos(2x)$ and $-\cos(2x)$.

Simplifying, $-\cos(2x)=cos(2x+\pi) \ or \ \cos(\pi-2x)$;

Now, comparing $\cos(2x+\pi)$ with $\cos(2x)$, can we say $\Delta\phi =\pi$?

But from graph $\Delta\phi =\frac{\pi}{2}$.

Obviously graph is correct, so where am i doing wrong? Thanks in advance.

Answer 1

You are very close.

Notice that we have

$$-\cos (2x) = \cos(2x + \pi) = \cos\left(2\left(x + \frac{\pi}{2}\right)\right)$$

We have $\Delta \phi = \frac{\pi}{2}$.

Answer 2

$\sin^2 x = \cos^2 (x - \frac \pi2)$