Question

What are the roots of the equation $1 - \cos \theta = \sin \theta \cdot \sin \dfrac{\theta }{2}$?
A. $k\pi ,k \in I$
B. $2k\pi ,k \in I$
C. $\dfrac{{k\pi }}{2},k \in I$
D. None of these

Answer 1

Hint: First we will apply half-angle trigonometry identity to solve the given equation. Then apply the general solution of $\sin \theta $ to find the solution of the given equation.

Formula Used:
Half-angle trigonometry identity:
(a) $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
(b) $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
The general solution of the $\sin \theta = 0$ is $\theta = n\pi ,\,\,n \in I$.

Complete step by step solution:
Given equation is $1 - \cos \theta = \sin \theta \cdot \sin \dfrac{\theta }{2}$.
Now apply the half-angle trigonometry identity (a) on the left side expression
$ \Rightarrow 2{\sin ^2}\dfrac{\theta }{2} = \sin \theta \cdot \sin \dfrac{\theta }{2}$
Again, apply the half-angle trigonometry identity (b) on the right-side expression
$ \Rightarrow 2{\sin ^2}\dfrac{\theta }{2} = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2} \cdot \sin \dfrac{\theta }{2}$
$ \Rightarrow 2{\sin ^2}\dfrac{\theta }{2} = 2{\sin ^2}\dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
Now subtract $2{\sin ^2}\dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ from both sides
$ \Rightarrow 2{\sin ^2}\dfrac{\theta }{2} - 2{\sin ^2}\dfrac{\theta }{2}\cos \dfrac{\theta }{2} = 2{\sin ^2}\dfrac{\theta }{2}\cos \dfrac{\theta }{2} - 2{\sin ^2}\dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
$ \Rightarrow 2{\sin ^2}\dfrac{\theta }{2} - 2{\sin ^2}\dfrac{\theta }{2}\cos \dfrac{\theta }{2} = 0$
Take common $2{\sin ^2}\dfrac{\theta }{2}$ from the left side expression.
$ \Rightarrow 2{\sin ^2}\dfrac{\theta }{2}\left( {1 - \cos \dfrac{\theta }{2}} \right) = 0$
Now apply the half-angle trigonometry identity (a) on the left side expression
$ \Rightarrow 2{\sin ^2}\dfrac{\theta }{2} \cdot 2{\sin ^2}\dfrac{\theta }{4} = 0$
Equate each vector by zero.
Either,
$2{\sin ^2}\dfrac{\theta }{2} = 0$
$ \Rightarrow \sin \dfrac{\theta }{2} = 0$
Apply the general solution formula
$ \Rightarrow \dfrac{\theta }{2} = k\pi ,\,\,n \in I$
$ \Rightarrow \theta = 2k\pi ,\,\,n \in I$

Or,
$2{\sin ^2}\dfrac{\theta }{4} = 0$
$ \Rightarrow \sin \dfrac{\theta }{4} = 0$
Apply the general solution formula
$ \Rightarrow \dfrac{\theta }{4} = k\pi ,\,\,n \in I$
$ \Rightarrow \theta = 4k\pi ,\,\,n \in I$
Therefore, the solution of $1 - \cos \theta = \sin \theta \cdot \sin \dfrac{\theta }{2}$ are $2k\pi ,\,\,n \in I$ and $4k\pi ,\,\,n \in I$.

Option ‘B’ is correct

Note: The general solution of $\sin \theta = 0$ is $\theta = n\pi ,\,\,n \in I$. The general solution of $\sin \theta = 1$ is $\theta = n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{2}$ .