Question

How do you use the half-angle formula to find the exact values of $\sin {22.5^ \circ }$?

Answer 1

Hint: For solving this question we will use the identity $\cos 2\theta = 1 - {\sin ^2}\theta $. Then, make this identity equal to ${\sin ^2}\theta $ and put ${22.5^ \circ }$ in the place of $\theta $ in the whole identity. Then, put the value of $\cos {45^ \circ }$ equal to $\dfrac{1}{{\sqrt 2 }}$.

Complete Step by Step Solution:
Half-angle identities are a set of equations that help you translate the trigonometric values of unfamiliar angles into more familiar values. There are many half-angle identities but we use them according to the necessity of the question.
We have to find the value of $\sin {22.5^ \circ }$ where angle is ${22.5^ \circ }$. As already stated in the question we have to use the half angle formula, so, we will use the formula of $\cos 2\theta $. So, we know that –
$\cos 2\theta = 1 - {\sin ^2}\theta $
By transposition method making this formula equal to ${\sin ^2}\theta $, we get –
$ \Rightarrow {\sin ^2}\theta = \dfrac{1}{2}\left[ {1 - \cos 2\theta } \right]$
We have to find the value of $\sin {22.5^ \circ }$ where the angle is ${22.5^ \circ }$. Putting this value of angle in the place of $\theta $ in the above identity, we get –
$
   \Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {1 - \cos 2 \times 22.5} \right] \\
   \Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {1 - \cos 45} \right] \\
 $
Putting the value of $\cos {45^ \circ }$ on the above equation, we get –
$ \Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {1 - \dfrac{1}{{\sqrt 2 }}} \right]$
Now, further solving, we get –
$
   \Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} \right] \\
   \Rightarrow {\sin ^2}22.5 = \dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }} \\
 $
Now, doing the square root on the both sides, we get –
$ \Rightarrow \sin 22.5 = \pm \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} $
But we will discard the negative value of the $\sin 22.5$ and considering only positive value, we get –
$\sin 22.5 = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} $

Hence, the value of $\sin 22.5$ is $\sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} $

So, this is the required value of $\sin 22.5$.

Note: Use the half-angle formula only which has $\sin \theta $ to solve this question. Many students can make mistakes while taking the value of $\sin 22.5$ only positive and not negative as the value of $\sin $ is only positive from the angle $0$ to ${90^ \circ }$.