What is the period of $(\sin x)^3 (\sin(3x))$?

Question

If I do it in the general way of L.CM I'm getting $2\pi$ as answer I don't know where I'm making mistake because the answer is given as $\pi$

Answer 1

Hint:

Linearise first this trigonometric polynomial: as $\;\sin 3x=3\sin x -4\sin ^3x$, we deduce that $$\sin^3x=\frac14(3\sin x-\sin 3x),$$ whence, using the standard linearisation formulæ: \begin{align}\sin^3x\sin 3x&=\frac14(3\sin x\sin 3x-\sin^2 3x)=\frac 38(\cos 2x-\cos 4x)-\frac18(1-\cos 6x)\\ &=\frac18(\cos 6x- 3\cos 4x+3\cos 2x-1). \end{align}

Answer 2

Correct if wrong :

$\sin ^3 x( \sin2x \cos x +\cos 2x \sin x)=$

$(1/2)(\sin^2 x) \sin^2 2x +$

$ (\sin^4 x) \cos 2x.$

Left to do:

Find the basic periods of the individual functions above.

(What is the basic period of $\sin^2 x$, and of $\sin^4 x$ ?)

And then?

Answer 3

Your LCM method will give you a largest possible period but it could be shorter. You need to examine the particular functions involved. Consider the period of $sin^2(x)$. Is it $2\pi$? No, it is only $\pi$. It can be rewritten in terms of $\sin(2x)$. Here's an even simpler example. If $f(x)$ and $g(x)$ both have period $2\pi$ then what is the period of $f(x) + g(x)$? How about if $f(x) = \sin(x)$ and $g(x) = - \sin(x)$?