Solve the equation $\sin \left( \theta \right)+\sin \left( 2\theta \right)+\sin \left( 3\theta \right)+\sin \left( 4\theta \right)=0$.

Hint: We will apply the formula of trigonometry given by $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ to solve the equation in the question. Here, A and B are the given angles.

Complete step-by-step answer:

We will consider the equation given by $\sin \left( \theta \right)+\sin \left( 2\theta \right)+\sin \left( 3\theta \right)+\sin \left( 4\theta \right)=0...(i)$.

We will now interchange the values $\sin \left( \theta \right)$ and $\sin \left( 2\theta \right)$ with each other and $\sin \left( 2\theta \right)$ and $\sin \left( 4\theta \right)$ with each other. Therefore, we have $\left( \sin \left( \theta \right)+\sin \left( 3\theta \right) \right)+\left( \sin \left( 2\theta \right)+\sin \left( 4\theta \right) \right)...(ii)$.

Now, we will apply the trigonometric formula given by $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and apply it to the equation (ii). Thus, we get

$\begin{align}

  & \left( \sin \left( \theta \right)+\sin \left( 3\theta \right) \right)+\left( \sin \left( 2\theta \right)+\sin \left( 4\theta \right) \right)=0 \\

 & \Rightarrow 2\sin \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right)+2\sin \left( \dfrac{2\theta +4\theta }{2} \right)\cos \left( \dfrac{2\theta -4\theta }{2} \right)=0 \\

 & \Rightarrow 2\sin \left( \dfrac{4\theta }{2} \right)\cos \left( \dfrac{-2\theta }{2} \right)+2\sin \left( \dfrac{6\theta }{2} \right)\cos \left( \dfrac{-2\theta }{2} \right)=0 \\

 & \Rightarrow 2\sin \left( 2\theta \right)\cos \left( -\theta \right)+2\sin \left( 3\theta \right)\cos \left( -\theta \right)=0 \\

\end{align}$

Now, we will take the common terms $2\cos \left( -\theta \right)$ and make it separated from the equation. Thus, we get $2\cos \left( -\theta \right)\left[ \sin \left( 2\theta \right)+\sin \left( 3\theta \right) \right]=0$. Therefore, we can write $\cos \left( -\theta \right)$ and $\sin \left( 2\theta \right)+\sin \left( 3\theta \right)=0$.

Now, we will consider $\cos \left( -\theta \right)=0$.

As the value of $\cos \left( \dfrac{\pi }{2} \right)=0$. Therefore, we can write $\cos \left( -\theta \right)=\cos \left( \dfrac{\pi }{2} \right)$. Therefore, by the formula $\cos \left( x \right)=\cos \left( y \right)$ we have $x=y\pm 2n\pi $ where, n = 0, -1, +1, -2, +2, ... .

$\Rightarrow \cos \left( -\theta \right)=\cos \left( \dfrac{\pi }{2} \right)$ where, n = ..., -1, 0, -1, ... .

$\Rightarrow \left( -\theta \right)=\dfrac{\pi }{2}+2n\pi $ where, n = ..., -1, 0, 1, ... .

Or, $\theta =\dfrac{-\pi }{2}-2n\pi $ where, n = ..., -1, 0, 1, ... .

Now, we will consider $\sin \left( 2\theta \right)+\sin \left( 3\theta \right)=0$. Now, we will apply the formula given by $\begin{align}

  & \sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) \\

 & \Rightarrow \sin \left( 2\theta \right)+\sin \left( 3\theta \right)=0 \\

 & \Rightarrow 2\sin \left( \dfrac{2\theta +3\theta }{2} \right)\cos \left( \dfrac{2\theta -3\theta }{2} \right)=0 \\

 & \Rightarrow 2\sin \left( \dfrac{5\theta }{2} \right)cis\left( \dfrac{-\theta }{2} \right)=0 \\
\end{align}$

Now, we have that $\sin \left( \dfrac{5\theta }{2} \right)=0$ and $\cos \left( \dfrac{-\theta }{2} \right)=0$. We will first consider $\cos \left( \dfrac{-\theta }{2} \right)=0$. At this stage we will apply the formula $\cos x=\cos y$. Therefore, we have

 $\begin{align}

  & \cos \left( \dfrac{-\theta }{2} \right)=0 \\

 & \Rightarrow \cos \left( \dfrac{-\theta }{2} \right)=\cos \left( \dfrac{\pi }{2} \right) \\

\end{align}$

This is because we know that the volume of $\cos \left( \dfrac{\pi }{2} \right)=0$. And by the formula $\cos x=\cos y$ we have $x=y\pm n\pi $ where, n = ..., -1, 0, 1, .... .

$\Rightarrow \dfrac{-\theta }{2}=\dfrac{\pi }{2}+2n\pi $ where, n = ..., -1, 0, 1, .... .

$\dfrac{-\theta }{2}=-\dfrac{\pi }{2}-2n\pi $ where, n = ..., -1, 0, 1, .... .

$\Rightarrow \theta =-\pi -n\pi $ where, n = ..., -1, 0, 1, .... .

Now, we will consider $\sin \left( \dfrac{5\theta }{2} \right)=0$. Since, we know that the value of $\sin \left( \pi \right)=0$. Thus, we can write $\sin \left( \dfrac{5\theta }{2} \right)=\sin \left( 0 \right)$.

Now, by the formula $\sin x=\sin y$ we have $x=n\pi $ because we are considering $\sin y=0.$

$\begin{align}

  & \Rightarrow \sin \left( \dfrac{5\theta }{2} \right)=\sin \left( 0 \right) \\

 & \Rightarrow \dfrac{5\theta }{2}=n\pi \\

 & \Rightarrow \theta =\dfrac{2n\pi }{5} \\

\end{align}$

Where n = ..., -1, 0, 1, .... .

Hence, the value of $\theta $ in the equation $\sin \left( \theta \right)+\sin \left( 2\theta \right)+\sin \left( 3\theta \right)+\sin \left( 4\theta \right)=0$ is given by $\theta =\dfrac{-\pi }{2}-2n\pi \,\,,\,\,\theta =2n\pi \,\,,\,\,\theta =\dfrac{-\pi }{2}-2n\pi \,\,,\,\,\theta =-\pi -n\,\,,\,\,\theta =\dfrac{2n\pi }{5}$.


Note: We could have applied the formula $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ to the expression $\sin \left( \theta \right)+\sin \left( 2\theta \right)$ instead of $\sin \left( \theta \right)+\sin \left( 3\theta \right)$. But we use it for $\sin \left( \theta \right)+\sin \left( 3\theta \right)$ so that when we get $2\sin \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right)$ we could easily solve \[\sin \left( \dfrac{\theta +3\theta }{2} \right)\] as $\sin \left( \dfrac{4\theta }{2} \right)$ or $\sin \left( 2\theta \right)$ which is not at all a complex term. Alternative method of solving $\cos \left( \dfrac{-\theta }{2} \right)=\cos \left( \dfrac{\pi }{2} \right)$ by using the fact that since $\cos \left( \dfrac{\pi }{2} \right)$ is positive in the first and fourth quadrants. So, by considering the first quadrant we have,

$\begin{align}

  & \cos \left( \dfrac{-\theta }{2} \right)=cos\left( \dfrac{\pi }{2} \right) \\

 & \Rightarrow \left( \dfrac{-\theta }{2} \right)=\dfrac{\pi }{2} \\


 & \Rightarrow -\theta =\pi \\

\end{align}$

  Where, n = ..., -1, 0, 1, ... .