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.