Find all $(a,b)$, such that $y = \cos^2 x + \cos^2 (x+a) + 2\cos x\cos (x+a)\cos b$ is constant for all $x \in \mathbb{R}$.

Question

I've tried using trig identities and $\frac{\partial y}{\partial x} = 0$ to no avail...

I've proved that $\{(a,b)|a = b+\pi\}$ works, but I would like to know whether those are the only real solutions...

Answer 1

Hint:

As you noted, if $y(x)$ is a constant, $y'(x)=0$. Differentiating that wrt $x$ should give you the condition $\cos a + \cos b = 0$...

Answer 2

Your intuition is correct. We can rewrite the derivative as $y'(x) = -2(cos(a) + cos(b))sin(a + 2x)$ to observe the necessary condition $cos(a) = -cos(b)$. The expression $\cos^2(x) + \cos^2(x + a) - 2\cos(x)\cos(x + a)\cos(a)$ can be rewritten as $\sin^2(a)$.

Answer 3

HINT:

$$S=\cos^2x+\cos^2(x+a)+2\cos x\cos(x+a)\cos b$$

$$=1+\cos^2(x+a)-\sin^2x+2\cos x\cos(x+a)\cos b$$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$S=1+\cos a\cos(2x+a)+\{\cos(2x+a)+\cos a\}\cos b$$

$$=1+\cos a\cos b+(\cos b+\cos a)\cos(2x+a)$$

Can you take it from here?