How do you evaluate $\sin 165^{\circ}$?

Hint:
For solving the given problem, we have to know about the angles and degrees of and . We will be using the formula of $\sin (\alpha - \beta )$. $\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \sin \beta \cos \alpha $ . Also, you can see that the angle 165 is closed from 180. So we will subtract 165 from 180. Let’s see how we can solve this problem.

Complete step by step solution:
We have to calculate the value of sin(165). We know that
 $ \Rightarrow \sin (\theta ) = \sin ({180^ \circ } - \theta )$ .
 $ \Rightarrow \sin (\alpha - \beta ) = \sin \alpha \cos \beta - \sin \beta \cos \alpha $
Using this formula we will solve this problem
 $ \Rightarrow \sin ({165^ \circ }) = \sin ({180^ \circ } - {165^ \circ })$
After subtracting 165 with 180 we get,
 $ = \sin ({15^ \circ })$
This can also be written as
 $ = \sin ({45^ \circ } - {30^ \circ })$
After applying the formula, we will get,
 \[ = \sin {45^ \circ }\cos {30^ \circ } - \sin {30^ \circ }\cos {45^ \circ }\] ------ (i)
So now, we have to find the value of sin45, cos45, sin30 and cos30.
We know that,
 $ \Rightarrow \sin {45^ \circ } = \cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ , $\sin {30^ \circ } = \dfrac{1}{2}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Putting all the values in equation (i)
 $ = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2}.\dfrac{1}{{\sqrt 2 }}$
After taking common we get,
 $ = \dfrac{1}{4}(\sqrt 6 - \sqrt 2 )$

Therefore, sin(165) $ = \dfrac{1}{4}(\sqrt 6 - \sqrt 2 )$.

Additional Information:
The formula being used here of compound angle, $\sin (\alpha - \beta )$ which is $\sin \alpha \cos \beta - \sin \beta \cos \alpha $ . The formula of $\sin (\alpha + \beta )$ is $\sin \alpha \cos \beta + \sin \beta \cos \alpha $ . We have a compound angle formula for all $\cos \theta $ and $\tan \theta $ as well. We will see those formulae later.

Note:
There is an alternative method for solving this problem. Let’s see the other process.
165 can also be written as 90 + 75
 $ \Rightarrow \sin ({165^ \circ }) = \sin ({90^ \circ } + {75^ \circ })$
By applying the formula of $\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \sin \beta \cos \alpha $ , we will get
 $ = \sin {90^ \circ }\cos {75^ \circ } + \cos {90^ \circ }\sin {75^ \circ }$
Since cos90 is 0, therefore cos90 sin75 will be 0. So now we get,
 $ = \sin {90^ \circ }.\cos {75^ \circ }$
Now, as sin90$^ \circ $ is 1, therefore we get
$ = \cos {75^ \circ }$
75 can also be written as 45 + 30
 $ = \cos ({45^ \circ } + {30^ \circ })$
The formula of $\cos (\alpha + \beta )$ is $\cos \alpha \cos \beta - \sin \alpha \sin \beta $
Therefore, using this formula we will get,
 $ = \cos {45^ \circ }.\cos {30^ \circ } - \sin {45^ \circ }.\sin {30^ \circ }$
 $ = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}$
After taking common we get,
$ = \dfrac{1}{4}(\sqrt 6 - \sqrt 2 )$