Question

Solve the following equation to get a simpler form $\sin \theta +\sin 2\theta +\sin 3\theta =0$.

Answer 1

Hint: We will apply the formula of trigonometry $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ to the equation given in the question to simplify it and solve it. We will apply it to the terms $\sin \theta +\sin 3\theta$.

Complete step-by-step answer:

Considering the equation,
$\sin \theta +\sin 2\theta +\sin 3\theta =0....(1)$
We will first replace $\sin 2\theta$ and $\sin 3\theta$ with each other. Therefore, we have,
$\sin \theta +\sin 3\theta +\sin 2\theta =0$.
Now we will apply the formula of trigonometry,
$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ in the expression $\sin \theta +\sin 3\theta$. Therefore, we have $\sin \theta +\sin 3\theta =2\sin \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right)$.
Now we will substitute the value of the expression $\sin \theta +\sin 3\theta$ in the equation $\sin \theta +\sin 3\theta +\sin 2\theta =0$.
Thus we get,

& 2\sin \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right)+\sin 2\theta =0 \\\ & 2\sin \left( \dfrac{4\theta }{2} \right)\cos \left( \dfrac{-2\theta }{2} \right)+\sin 2\theta =0 \\\ & 2\sin 2\theta \cos \left( -\theta \right)+\sin 2\theta =0 \\\ & \sin 2\theta \left[ 2\cos \left( -\theta \right)+1 \right]=0 \\\ \end{aligned}$$ $$\Rightarrow \sin 2\theta =0$$ or $$2\cos \left( -\theta \right)+1=0$$. We know that $$\cos \left( -\theta \right)=\cos \theta $$. Therefore we have $$\sin 2\theta =0$$ or $$2\cos \theta +1=0$$. Considering the expressions we get $$\sin 2\theta =0$$. $$\Rightarrow \sin 2\theta =\sin 0\Rightarrow sin2\theta =sin\left( n\pi \right)$$ where $$n=0,\pm 1,\pm 2....$$ This is because the value of $$\sin 0=sin\left( n\pi \right)=0$$. $$\Rightarrow \sin 2\theta =0$$ results in $$2\theta =0$$ or $$\theta =0$$. Also, $$\sin 2\theta =\sin \left( n\pi \right)$$ is the expression resulting into $$\Rightarrow 2\theta =n\pi \Rightarrow \theta =\dfrac{n\pi }{2}$$ where $$n=0,\pm 1,\pm 2....$$ Now we will consider the expression, $$2\cos \theta +1=0$$. $$\begin{aligned} & \Rightarrow 2\cos \theta =-1 \\\ & \Rightarrow \cos \theta =\dfrac{-1}{2} \\\ \end{aligned}$$ Since the value of $$\dfrac{1}{2}=\cos \dfrac{\pi }{3}$$. Hence we can write, $$\cos \theta =-\cos \dfrac{\pi }{3}$$. As we know that the value of cos is negative in the second and third quadrant. Therefore, we first consider the second quadrant. In this quadrant, $$\begin{aligned} & -\cos \left( \dfrac{\pi }{3} \right)=\cos \left( \pi -\dfrac{\pi }{3} \right) \\\ & \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right) \\\ & \cos \theta =\cos \left( \dfrac{3\pi -\pi }{3} \right) \\\ & \cos \theta =\cos \left( \dfrac{2\pi }{3} \right) \\\ & \Rightarrow \theta =\dfrac{2\pi }{3} \\\ \end{aligned}$$ $$\Rightarrow \theta =\dfrac{2\pi }{3}+2n\pi $$ where $$n=0,\pm 1,\pm 2....$$ Now we will consider the third quadrant. In this quadrant the value of $$-\cos \left( \dfrac{\pi }{3} \right)=\cos \left( \pi +\dfrac{\pi }{3} \right)$$. $$\begin{aligned} & \Rightarrow \cos \theta =\cos \left( \pi +\dfrac{\pi }{3} \right) \\\ & \Rightarrow \cos \theta =\cos \left( \dfrac{4\pi }{3} \right) \\\ & \Rightarrow \theta =\dfrac{4\pi }{3} \\\ \end{aligned}$$ $$\Rightarrow \theta =\dfrac{4\pi }{3}+2n\pi $$ where $$n=0,\pm 1,\pm 2....$$ Hence the value $$\theta $$ of the expression $$\sin \theta +\sin 2\theta +\sin 3\theta =0$$ is given by $$\theta =0,\dfrac{\pi }{2},\dfrac{2\pi }{3},\dfrac{4\pi }{3}$$ when n = 0. Note: We can also substitute the formula $$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$$ in the expression $$\sin \theta +\sin 2\theta $$ but the only difference was that when we get $$\sin \theta +\sin 2\theta =2\sin \left( \dfrac{\theta +2\theta }{2} \right)\cos \left( \dfrac{\theta -2\theta }{2} \right)$$ which is further equal to $$2\sin \left( \dfrac{3\theta }{2} \right)\cos \left( \dfrac{-\theta }{2} \right)$$ we come to know here that $$\theta $$ is in fraction. Since we are asked about the basic solutions instead of general solutions, that is why we write $$\dfrac{\pi }{2}=\theta ,\dfrac{2\pi }{3}=\theta $$ and so on instead of $$\theta =\dfrac{\pi }{2}+2n\pi ,\theta =\dfrac{2\pi }{3}+2n\pi $$.