How do you find the exact values of $ \sin {165^0} $ using the half angle formulae ?

Hint: First we will evaluate the right-hand of the equation and then further the left-hand side of the equation. We will use the following formula to evaluate $ \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} $ and then we will further this expression to $ \sin x $ form and hence evaluate the value of the term.

Complete step-by-step answer:
We will start off by using the formula $ \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} $ .
Hence, the equation will become,
\[
   = \sin 165 \\
  \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} \\
  \sin \left( {\dfrac{{330}}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos 330}}{2}} \\
  \sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (360 - 330)}}{2}} \\
  \sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (30)}}{2}} \;
 \]
Now we know that the value of $ \cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2} $
So now here, the value will be,
\[
  \sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (30)}}{2}} \\
  \sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{2}} \\
  \sin \left( {165} \right) = \pm \sqrt {\dfrac{{2 - \sqrt 3 }}{4}} \\
  \sin \left( {165} \right) = \pm \dfrac{{\sqrt {2 - \sqrt 3 } }}{2} \;
 \]
So, the correct answer is “ $ \pm \dfrac{{\sqrt {2 - \sqrt 3 } }}{2} $ ”.

Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities $ {\sin ^2}x + {\cos ^2}x = 1 $ and $ \cos 2x = 2{\cos ^2}x - 1 $ . While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that $ \tan x = \,\dfrac{{\sin x}}{{\cos x}} $ .
While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities.