Period of $f(x) = \cos^2 ax - \cos^4 ax$ is $\frac{\pi}{4}$ what is $a$?

Question

I can not access the solutions I just know the correct answer which is $2$. I tried to solve the problem and I got $4$ instead here is my solution: $\cos^2 ax(1 - \cos^2 ax) = \cos^2 ax \sin^2 ax = 1/4 \sin^2 (2ax)$ and then to get the period $2\pi/|2a| = \pi/4$ which result that $a$ is $+4$ or $-4$.

Answer 1

Here's a hint: $a^2 - b^2 = (a + b)(a - b)$. If you see squared terms its generally better to assign them with a variable. Like, let $k = \cos(ax)$. Then you get, $k^2(1 - k^2)$. See how its more cleaner, and you can generally think of more ways of solving it in this format.

Edit: My bad for not understanding the question clearly.

To the find the period, you can try condensing the your multi-trigonometric function in to a linear form, and your already halfway there. The identities:

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

If you substitute these into your answer, you get: $$\frac{1}{2}(1-\cos(2x))\frac{1}{2}(1+\cos(2x))$$ $$\frac{1}{4}(1-\cos^2(2x))$$ $$\frac{1}{4}(\sin^2(2x))$$

Another identity is:

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

$\frac{1}{4}(\sin^2(2x))$ can be split into $\frac{1}{4}(\sin(2x)\sin(2x))$, where your $y$ would just be $x$. So it would become:

$$\frac{1}{4}(\frac{1}{2}(\cos(0)-\cos(4x))) = \frac{1-\cos(4x)}{8}$$

Hence, your period is $\frac{2\pi}{4} = \frac{\pi}{2}$. Your solution is affected by the period. As the functions are periodic, so will your solutions be periodic. They will repeat every period. (there is a solution for $x = 0$, and there will another solution for every $\pm \frac{\pi}{2})$. (i.e. there will be a solution at $-\frac{\pi}{2}, \frac{\pi}{2}, \pi$ and so on...). Therefore, solutions from $0$ are every +$\frac{\pi n}{2}$, where $n \in \mathbb{Z}$.