Question

How do you simplify $\sin \left( {\dfrac{\theta }{2}} \right)$ using the double angle identities?

Answer 1

Hint : In order to find the solution to this particular numerical, we will use the double angle formula of cosine function in the form of sine function i.e., $\cos 2\theta = 1 - 2{\sin ^2}\theta$ .From the formula, first of all we will find the value of $\sin \theta$ .After that we will replace $\theta$ by $\dfrac{\theta }{2}$ and simplify the expression. After simplification we will get the required result.

Complete step-by-step answer :
We have given $\sin \left( {\dfrac{\theta }{2}} \right)$ and we have to simplify it using a double angle identity.
In order to simplify $\sin \left( {\dfrac{\theta }{2}} \right)$ we will first apply the double angle formula of cosine function in the form of sine function.
The double angle formula is as follows:
$\cos 2\theta = 1 - 2{\sin ^2}\theta {\text{ }} - - - \left( i \right)$
Now, we will find out the value of $\sin \theta$
So, from equation $\left( i \right)$ we have,
$\cos 2\theta - 1 = - 2{\sin ^2}\theta {\text{ }}$
Now, on dividing by $- 2$ on both sides, we get
$\dfrac{{\cos 2\theta - 1}}{{ - 2}} = {\sin ^2}\theta {\text{ }}$
On multiplying by $- 1$ on the left-hand side of the above expression, we get
$\dfrac{{ - 1 \cdot \left( {\cos 2\theta - 1} \right)}}{{ - 1 \cdot \left( { - 2} \right)}} = {\sin ^2}\theta {\text{ }}$
On simplification, we get
$\dfrac{{1 - \cos 2\theta }}{2} = {\sin ^2}\theta {\text{ }}$
Now, on taking square root on both the sides, we get
$\sqrt {\dfrac{{1 - \cos 2\theta }}{2}} = \sqrt {{{\sin }^2}\theta {\text{ }}}$
$\Rightarrow \pm \sqrt {\dfrac{{1 - \cos 2\theta }}{2}} = \sin \theta$
Thus, we get the value of $\sin \theta$ as $\pm \sqrt {\dfrac{{1 - \cos 2\theta }}{2}}$
Now, we will replace $\theta$ by $\dfrac{\theta }{2}$
Therefore, we get
$\sin \dfrac{\theta }{2} = {\text{ }} \pm \sqrt {\dfrac{{1 - \cos 2\dfrac{\theta }{2}}}{2}}$
On simplifying it, we get
$\sin \dfrac{\theta }{2} = {\text{ }} \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}}$
Hence, we get our required result.
So, the correct answer is “$\sin \dfrac{\theta }{2} = {\text{ }} \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}}$”.

Note : This question can also be solved by using another double angle formula of cosine function which is in the form of cosine itself.
The formula is as follows:
$\cos 2\theta = 2{\cos ^2}\theta - 1$
First of all, let us assume $\sin \dfrac{\theta }{2} = x{\text{ }} - - - \left( 1 \right)$
We know that ${\sin ^2}x + {\cos ^2}x = 1$
Therefore, ${\sin ^2}\dfrac{\theta }{2} + {\cos ^2}\dfrac{\theta }{2} = 1$
$\Rightarrow {\cos ^2}\dfrac{\theta }{2} = 1 - {\sin ^2}\dfrac{\theta }{2}$
Put the value of $\sin \dfrac{\theta }{2}$ we get
$\Rightarrow {\cos ^2}\dfrac{\theta }{2} = 1 - {x^2}$
Now, using the formula, i.e., $\cos 2\theta = 2{\cos ^2}\theta - 1$
On replacing $\theta$ by $\dfrac{\theta }{2}$ we get
$\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$
Now, on substituting the value of ${\cos ^2}\dfrac{\theta }{2}$ we get,
$\cos \theta = 2\left( {1 - {x^2}} \right) - 1$
$\Rightarrow \cos \theta = 2 - 2{x^2} - 1$
$\Rightarrow \cos \theta = 1 - 2{x^2}$
After taking $x$ terms on the left-hand side and constants term of the right-hand side, we get
$2{x^2} = 1 - \cos \theta$
On dividing by $2$ we get
${x^2} = \dfrac{{1 - \cos \theta }}{2}$
$\Rightarrow x = \sqrt {\dfrac{{1 - \cos \theta }}{2}}$
From $\left( 1 \right)$ , we have
$\Rightarrow \sin \dfrac{\theta }{2} = \sqrt {\dfrac{{1 - \cos \theta }}{2}}$
Hence, we get the required result.