How do you find \[{\mathop{\rm tan}\nolimits} 22.5\] using the half-angle formula?

Hint: In the given question, we have been asked to find the value of a trigonometric function. Now, the argument of the given trigonometric function is not in the range of the known values of the trigonometric functions as given in the standard table. We only know the values of five standard angles - \[0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ \]. But we can calculate that by using the formula of half-angle.

Formula Used:
We are going to use the formula of half-angle, which is:
\[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]

Complete step-by-step answer:
Here, we have to calculate the value of \[\tan \left( {\dfrac{\pi }{8}} \right)\].
If \[\dfrac{\theta }{2} = \dfrac{\pi }{8}\],
then \[\theta = \dfrac{\pi }{4}\], which is a standard angle in the first quadrant.
\[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
Now, since \[\dfrac{\pi }{8} > \dfrac{\pi }{2}\], it lies in the first quadrant, and hence, we only have to consider the positive solution. The negative solution is extraneous.
Thus,
\[\tan \left( {\dfrac{\pi }{4}} \right) = \dfrac{{2\tan \left( {\pi /8} \right)}}{{1 - {{\tan }^2}\left( {\pi /8} \right)}}\]
Let \[\tan \left( {\pi /8} \right) = x\], so
$\Rightarrow$ \[1 = \dfrac{{2x}}{{1 - {x^2}}}\]
Doing cross-multiplication, we get,
$\Rightarrow$ \[1 - {x^2} = 2x\]
Taking all the elements to one side,
$\Rightarrow$ \[{x^2} + 2x - 1 = 0\]
Taking the determinant formula to solve,
\[x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times \left( { - 1} \right)} }}{2} = \dfrac{{ - 2 \pm \sqrt {4 + 4} }}{2} = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2} = - 1 \pm \sqrt 2 \]
Since, \[\tan \left( {\pi /8} \right) > 0\],
Hence, \[\tan \left( {22.5^\circ } \right) = \sqrt 2 - 1\].

Note: In the given question, we had to find the value of a trigonometric function whose value is not given in the standard trigonometric table. It lies in the range of the values (\[0^\circ \] to \[90^\circ \]), but is not a known one. We found the value of the function using the formula of half-angle. Then, after applying the formula, it was no effort at all. So, it is really important that we know the formulae and where, when and how to use them so that we can get the correct result.