Question

How do you solve $\sin \left( {\dfrac{x}{2}} \right) = 1 - \cos x?$

Answer 1

In this question, we are going to solve the given equation to find the value of $x.$ We can apply the identity for $\cos x$ and $1$ then substitute in the given trigonometric equation. Let us sum and get the equivalent equation. Next solve the trigonometric equation for $x.$
Hence, we can get the required result.

Complete step by step solution:
In this we are going to find the value of $x$ by solving the equation.
First write the given trigonometric equation
$\sin \left( {\dfrac{x}{2}} \right) = 1 - \cos x$ $\left( 1 \right)$
By using trigonometric formula,
$\cos x$ Can be written in the form as,
$\cos x = {\cos ^2}\left( {\dfrac{x}{2}} \right) - {\sin ^2}\left( {\dfrac{x}{2}} \right)$ $\left( 2 \right)$
And$1$can be written in the form as,
$1 = {\sin ^2}\left( {\dfrac{x}{2}} \right) + {\cos ^2}\left( {\dfrac{x}{2}} \right)$ $\left( 3 \right)$
Let us now substitute equation $\left( 2 \right)$ and $\left( 3 \right)$in equation $\left( 1 \right)$we get,
$\sin \left( {\dfrac{x}{2}} \right) = {\sin ^2}\left( {\dfrac{x}{2}} \right) + {\cos ^2}\left( {\dfrac{x}{2}} \right) - {\cos ^2}\left( {\dfrac{x}{2}} \right) + {\sin ^2}\left( {\dfrac{x}{2}} \right)$
Cancelling positive and negative of cosine in the above term we get,
$\sin \left( {\dfrac{x}{2}} \right) = {\sin ^2}\left( {\dfrac{x}{2}} \right) + {\sin ^2}\left( {\dfrac{x}{2}} \right)$
Adding sine terms we get,
$\sin \left( {\dfrac{x}{2}} \right) = 2{\sin ^2}\left( {\dfrac{x}{2}} \right)$
Now rewrite the above equation in a quadratic form we get,
$2{\sin ^2}\left( {\dfrac{x}{2}} \right) - \sin \left( {\dfrac{x}{2}} \right) = 0$
Taking common terms outside we get,
$\sin \left( {\dfrac{x}{2}} \right)\left( {2\sin \left( {\dfrac{x}{2}} \right) - 1} \right) = 0$
Equating both term equal to zero we get,
$\Rightarrow \sin \dfrac{x}{2} = 0$
From the unit circle we get the result
$\Rightarrow \dfrac{x}{2} = k\pi$
We just cross multiply,
$\Rightarrow x = 2k\pi$
$\left( {2\sin \left( {\dfrac{x}{2}} \right) - 1} \right) = 0$
From the unit circle we can get the result,
$\Rightarrow \sin \dfrac{x}{2} = \dfrac{1}{2}$
$\Rightarrow \dfrac{x}{2} = \dfrac{\pi }{6} + 2k\pi$
$or\dfrac{x}{2} = \dfrac{5}{6}\pi + 2k\pi$
$\Rightarrow x = \dfrac{\pi }{3} + 2k\pi$
$or{\text{ }}x = \dfrac{5}{3}\pi + 2k\pi$
Therefore by solving the equation we get,
$x = 2k\pi$ Or $x = \dfrac{\pi }{3} + 2k\pi$
or $x = \dfrac{5}{3}\pi + 2k\pi$
Hence we get the required result.

Note:
We can get so many different formulas for trigonometric function. So, we need to be very careful while choosing the formula. So find the right formula by deriving it from the closest easy one.
Always know where to apply addition formula and double angle formula or half angle formula. We had to memorize the reference angles so it will be very easy for us to find the solution.
$\sin \dfrac{\alpha }{2} = \pm \sqrt {\dfrac{{1 - \cos \alpha }}{2}}$
There are many applications of trigonometry half angle formula to science and engineering with respect to light and sound.