How do you use half angle identity to find exact value of $ \cos {{22.5}^{\circ }} $ ?

Hint: By using trigonometric functions, we can apply the trigonometric ratios for the particular angle and find its value. So for this question, we should know about the trigonometric ratios for different angles. Additionally, we should know even and odd functions. We mainly use the half-angle formula of cosine because it is mentioned in the question.
Half angle formula of cosine is:
 $ \Rightarrow {{\cos }^{2}}\left( x \right)=\dfrac{1}{2}\left( 1+\cos \left( 2x \right) \right) $

Complete step by step answer:
Let us get started with the solution by knowing how we get half-angle formula:
If we say cos2x, can we write it as cos(x + x) and it’s formula is:
 $ \Rightarrow $ cos2x = cos(x + x) = cosx.cosx - sinx.sinx
Which will become:
 $ \Rightarrow $ cos2x = $ {{\cos }^{2}}x-{{\sin }^{2}}x.....(i) $
As we know that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ . So take out the value of $ {{\sin }^{2}}x $ :
 $ \Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x $
Now, put this value in equation(i), we get:
 $ \Rightarrow $ cos2x = $ {{\cos }^{2}}x-(1-{{\cos }^{2}}x) $
After opening the bracket, we get:
 $ \Rightarrow $ cos2x = $ {{\cos }^{2}}x-1+{{\cos }^{2}}x $
 $ \Rightarrow $ cos2x = $ 2{{\cos }^{2}}x-1 $
To derive half angle formula we have to let 2x = $ \theta $ and we can also say that x = $ \dfrac{\theta }{2} $ . By substituting these values in above equation, we get:
 $ \Rightarrow \cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}-1 $
Now take 1 to the other side:
 $ \Rightarrow 1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2} $
Now, divide the whole equation by , we get:
 $ \Rightarrow \dfrac{1}{2}\left( 1+\cos \theta \right)={{\cos }^{2}}\dfrac{\theta }{2} $
Now, again put $ \theta $ = 2x, we get:
 $ \Rightarrow {{\cos }^{2}}\left( x \right)=\dfrac{1}{2}\left( 1+\cos \left( 2x \right) \right) $

As we know, the half-angle formula is:
 $ \Rightarrow {{\cos }^{2}}\left( x \right)=\dfrac{1}{2}\left( 1+\cos \left( 2x \right) \right) $
As given in question, the value of angle is $ {{22.5}^{\circ }} $ so x = $ {{22.5}^{\circ }} $ and 2x = $ 2\times {{22.5}^{\circ }}={{45}^{\circ }} $
Now put the values in the formula. We get:
 $ \Rightarrow {{\cos }^{2}}\left( {{22.5}^{\circ }} \right)=\dfrac{1}{2}\left( 1+\cos \left( {{45}^{\circ }} \right) \right) $
Let’s make a table of trigonometric ratios for basic trigonometric functions i.e. sin, cos, tan, cot, sec, and cosec.




Now put the value of $ \cos {{45}^{\circ }} $ from the table above and solve the equation.
 $ \Rightarrow {{\cos }^{2}}\left( {{22.5}^{\circ }} \right)=\dfrac{1}{2}\left( 1+\dfrac{1}{\sqrt{2}} \right) $
The value of $ \dfrac{1}{\sqrt{2}} $ is 0.7071. Now put the value and solve:
 $ \Rightarrow {{\cos }^{2}}\left( {{22.5}^{\circ }} \right)=\dfrac{1}{2}\left( 1+0.7071 \right) $
By adding the terms inside the bracket, we get:
 $ \Rightarrow {{\cos }^{2}}\left( {{22.5}^{\circ }} \right)=\dfrac{1}{2}\left( 1.7071 \right) $
Dividing the term 1.7071 by 2 will give us:
 $ \Rightarrow {{\cos }^{2}}\left( {{22.5}^{\circ }} \right)=0.85355 $
Now, taking the square root on both sides of the equation, we will get:
 $ \Rightarrow \cos \left( {{22.5}^{\circ }} \right)=\sqrt{0.85355} $
Simplifying further, we get:
 $ \therefore \cos \left( {{22.5}^{\circ }} \right)=0.9238 $
So, this is the final answer.

Note:
After putting all the values, be aware of calculation mistakes. To find any half-angle, we just need to multiply each of the terms of the formula by half. There is no specific formula for finding half angles.