How do you use the half angle formula to find $\sin {112.5^ \circ }$?

Hint: In this question we have to find the value of $\sin {112.5^ \circ }$. To find the value of $\sin {112.5^ \circ }$ we use trigonometric identity $\cos 2\theta = 1 - 2{\sin ^2}\theta $ and modify it to find required value$ \sin {112.5^ \circ }$ . Also use the fact that ${225^ \circ }$ can be written as ${225^ \circ } = 2({112.5^ \circ })$. Also we need to know in which quadrant each trigonometric function is positive and in which they are negative.

Complete step by step solution:
Let us try to solve this question in which we have to find the value of $\sin {112.5^ \circ }$. To solve this question we will first get half angle formula of cosine function from $\cos 2\theta $ and after which we will put the value ${112.5^ \circ }$in the formula to get the exact value of $\sin {112.5^ \circ }$ using half angle formula.
So here is the formula of $\cos 2\theta $
$\cos 2\theta = 1 - 2{\sin ^2}\theta $
And we know that ${225^ \circ }$ can be written as ${225^ \circ } = 2({112.5^ \circ })$
So we will assume that $2\theta = {225^ \circ }$ which means that value of $\theta = \dfrac{{{{225}^
\circ }}}{2}$
Implies that $\theta = {112.5^ \circ }$
Putting value of $\theta$ in the formula of $\cos 2\theta $ we get,
$\cos 2({112.5^ \circ }) = 1 - 2{\sin ^2}({112.5^ \circ })$
Now taking $2{\sin ^2}({112.5^ \circ })$ to LHS and $\cos 2({112.5^ \circ })$ to RHS we will get
$2{\sin ^2}({112.5^ \circ }) = 1 - \cos 2({112.5^ \circ })$
$2{\sin ^2}({112.5^ \circ }) = 1 - \cos {225^ \circ }$
Putting value of$\cos {225^ \circ } = - \dfrac{1}{{\sqrt 2 }}$, because $\cos ({180^ \circ } + {\phi ^ \circ })
= - \cos ({\phi ^ \circ })$ and$\cos {225^ \circ } = \cos ({180^ \circ } + {45^ \circ })$ in the above equation we get
\[
2{\sin ^2}({112.5^ \circ }) = \dfrac{1}{{\sqrt 2 }} + 1 \\
2{\sin ^2}({112.5^ \circ }) = \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }} \\
\\
\]
\[
{\sin ^2}({112.5^ \circ }) = \dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }} \\
\sin {112.5^ \circ } = \pm \sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} \\
\]
Now we know that the value of $\sin {112.5^ \circ }$ will be positive because ${112.5^ \circ }$ lies in the second quadrant and the value of the sine function in the first and second quadrant is positive.
Hence the value of $\sin {112.5^ \circ }$ is equal to
$\sin {112.5^ \circ } = \sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} $
So the value of $\sin {112.5^ \circ }$ using the half angle formula is equal to $\sqrt {\dfrac{{\sqrt 2 + 1}}{{2\sqrt 2 }}} $.

Note: While solving this type of question you need to be careful about the sign of final value. For this you have to check in which quadrant trigonometric function value will be positive and in which it is negative. Knowing trigonometric formulas is a must to solve this kind of problem.