Question

Solve the following expression $\sin mx+\sin nx=0$

Answer 1

Hint: We will first simplify the given expression $\sin mx+\sin nx=0$ using the formula $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and then we will solve both the factors separately to get two values of x.

Complete step-by-step answer:
The expression mentioned in the question is $\sin mx+\sin nx=0.........(1)$
Now we know the formula $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and hence applying this in equation (1) we get,
$\Rightarrow 2\sin \left( \left( \dfrac{m+n}{2} \right)x \right)\cos \left( \left( \dfrac{m-n}{2} \right)x \right)=0........(2)$
Now separating the terms and equating it to 0 in equation (2) we get,
$\Rightarrow \sin \left( \left( \dfrac{m+n}{2} \right)x \right)=0.......(3)$ and $\cos \left( \left( \dfrac{m-n}{2} \right)x \right)=0.......(4)$
Now we can write 0 in the right hand side of equation (3) as sin 0 and hence we get,
$\Rightarrow \sin \left( \left( \dfrac{m+n}{2} \right)x \right)=\sin 0.......(5)$
Now we know the general formula when $\sin \theta =\sin \alpha$ and it implies $\theta =n\pi +{{(-1)}^{n}}\alpha$. So applying this in equation (5) we get,
$\Rightarrow (m+n)\dfrac{x}{2}=n\pi .......(6)$
Now rearranging and solving for x in equation (6) we get,
$\Rightarrow x=\dfrac{2n\pi }{m+n}.....(7)$
Now solving equation (4) to find the other value of x and substituting $\cos \dfrac{\pi }{2}$ in place of 0 we get,
$\Rightarrow \cos \left( \left( \dfrac{m-n}{2} \right)x \right)=\cos \dfrac{\pi }{2}.......(8)$
Now we know the general formula when $\cos \theta =\cos \alpha$ and it implies $\theta =2n\pi \pm \alpha$. So applying this in equation (8) we get,
$\Rightarrow (m-n)\dfrac{x}{2}=(2n+1)\dfrac{\pi }{2}.......(9)$
Now rearranging and solving for x in equation (9) we get,
$\Rightarrow x=\dfrac{(2n+1)\pi }{m-n}.......(10)$
Hence from equation (7) and equation (10) we get two values of x which are $x=\dfrac{2n\pi }{m+n}$ and $x=\dfrac{(2n+1)\pi }{m-n}$.

Note: Remembering the basic trigonometry formulas is the key here. Also we should remember the general formulas $\theta =n\pi +{{(-1)}^{n}}\alpha$ when $\sin \theta =\sin \alpha$ and $\theta =2n\pi \pm \alpha$ when $\cos \theta =\cos \alpha$. We in a hurry can make a mistake in solving equation (6) and equation (9) and hence we need to be careful while doing these steps.