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,
\[\begin{align}
  & 2\sin \left( \dfrac{\theta +3\theta }{2} \right)\cos \left( \dfrac{\theta -3\theta }{2} \right)+\sin 2\theta =0 \\
 & 2\sin \left( \dfrac{4\theta }{2} \right)\cos \left( \dfrac{-2\theta }{2} \right)+\sin 2\theta =0 \\
 & 2\sin 2\theta \cos \left( -\theta \right)+\sin 2\theta =0 \\
 & \sin 2\theta \left[ 2\cos \left( -\theta \right)+1 \right]=0 \\
\end{align}\]
\[\Rightarrow \sin 2\theta =0\] 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, \[2\cos \theta +1=0\].
\[\begin{align}
  & \Rightarrow 2\cos \theta =-1 \\
 & \Rightarrow \cos \theta =\dfrac{-1}{2} \\
\end{align}\]
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,
\[\begin{align}
  & -\cos \left( \dfrac{\pi }{3} \right)=\cos \left( \pi -\dfrac{\pi }{3} \right) \\
 & \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right) \\
 & \cos \theta =\cos \left( \dfrac{3\pi -\pi }{3} \right) \\
 & \cos \theta =\cos \left( \dfrac{2\pi }{3} \right) \\
 & \Rightarrow \theta =\dfrac{2\pi }{3} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{2\pi }{3}+2n\pi \] where \[n=0,\pm 1,\pm 2....\]
Now we will consider the third quadrant. In this quadrant the value of \[-\cos \left( \dfrac{\pi }{3} \right)=\cos \left( \pi +\dfrac{\pi }{3} \right)\].
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \left( \pi +\dfrac{\pi }{3} \right) \\
 & \Rightarrow \cos \theta =\cos \left( \dfrac{4\pi }{3} \right) \\
 & \Rightarrow \theta =\dfrac{4\pi }{3} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{4\pi }{3}+2n\pi \] 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 \[\theta =\dfrac{\pi }{2}+2n\pi ,\theta =\dfrac{2\pi }{3}+2n\pi \].