Question

the period of sinx.sin(2pi by 3+x).sin(2 pi by 3 -x)​

Answer 1

We need to find the period of,

$\longrightarrow f(x)=\sin x\sin\left(\dfrac{2\pi}{3}+x\right)\sin\left(\dfrac{2\pi}{3}-x\right)$

Let us simplify it first.

Recall the following product - to - sum formulas,

• $\sin A\sin B=\dfrac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]\quad\quad\dots(i)$

• $\cos A\sin B=\dfrac{1}{2}\left[\sin(A+B)-\sin(A-B)\right]\quad\quad\dots(ii)$

So,

$\longrightarrow f(x)=\sin x\sin\left(\dfrac{2\pi}{3}+x\right)\sin\left(\dfrac{2\pi}{3}-x\right)$

By (i),

$\longrightarrow f(x)=\dfrac{1}{2}\sin x\left[\cos\left(2x\right)-\cos\left(\dfrac{4\pi}{3}\right)\right]$

$\longrightarrow f(x)=\dfrac{1}{2}\sin x\left[\cos\left(2x\right)+\dfrac{1}{2}\right]$

$\longrightarrow f(x)=\dfrac{1}{2}\cos\left(2x\right)\sin x+\dfrac{1}{4}\sin x$

By (ii),

$\longrightarrow f(x)=\dfrac{1}{4}\left[\sin\left(3x\right)-\sin x\right]+\dfrac{1}{4}\sin x$

$\longrightarrow f(x)=\dfrac{1}{4}\sin\left(3x\right)-\dfrac{1}{4}\sin x+\dfrac{1}{4}\sin x$

$\longrightarrow f(x)=\dfrac{1}{4}\sin\left(3x\right)$

Now, the period of the function $f(x)$ is a positive real number $T\neq0$ such that,

$\longrightarrow f(T+x)=f(x)$

$\longrightarrow\dfrac{1}{4}\sin\left(3(T+x)\right)=\dfrac{1}{4}\sin\left(3x\right)$

$\longrightarrow\sin\left(3T+3x\right)=\sin\left(3x\right)$

$\longrightarrow\sin(3T)\cos(3x)+\cos(3T)\sin(3x)=\sin\left(3x\right)$

$\longrightarrow\sin(3T)\cos(3x)+\left[\cos(3T)-1\right]\sin(3x)=0$

Since $\sin(3x)$ and $\cos(3x)$ are two distinct dependent variables wrt $x,$ their coefficient should be zero in order to satisfy this equation, i.e.,

• $\sin(3T)=0$

• $\cos(3T)=1$

This implies,

$\longrightarrow 3T=2n\pi,\quad n\in\mathbb{Z}$

For smaller possible value of $T,$ let $n=1.$ Then,

$\longrightarrow 3T=2\pi$

$\longrightarrow\underline{\underline{T=\dfrac{2\pi}{3}}}$

This is the period.