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

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.