Question

How do you find the exact values of $\tan \dfrac{{3\pi }}{8}$ using the half-angle formula?

Answer 1

Hint:In the above question, we were asked to find the exact value of $\tan \dfrac{{3\pi }}{8}$ using the half-angle formula. We will use the formula $\tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 - \cos \theta }}{{\sin \theta }}$ , this is the half-angle formula for $\tan $. We will substitute $\dfrac{\theta }{2}$ with $\dfrac{{3\pi }}{8}$. Substituting the value then we will operate the equation and simplify accordingly to find the required value of our problem. So, let’s see how we solve the problem.

Complete step by step answer:
The half-angle formula for tangent can be written as:
$\tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 - \cos \theta }}{{\sin \theta }}$
We will use this formula to solve the above problem.
Let us take, $\theta = \dfrac{{3\pi }}{4}$.
Now, substituting this value in the above formula, we get,
$\tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{1 - \cos \left( {\dfrac{{3\pi }}{4}} \right)}}{{\sin \dfrac{{3\pi }}{4}}}$
We know, value of $\cos \dfrac{{3\pi }}{4}$ is $ - \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{{3\pi }}{4}$ is $\dfrac{1}{{\sqrt 2 }}$.

Now, using these values in the above equation, we get,
So, $\tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{1 - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}$
$ \Rightarrow \tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{1 + \dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}}$
Taking LCM in the numerator, we get,
$ \Rightarrow \tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}}$
Now, converting the fraction in its simplest form, we get,
$ \Rightarrow \tan \left( {\dfrac{{3\pi }}{8}} \right) = \dfrac{{\sqrt 2 + 1}}{1}$

Therefore, the value of $\tan \dfrac{{3\pi }}{8}$ by the half-angle formula is $\sqrt 2 + 1$.

Note: In the above solution, we have used a half-angle formula for the tangent.There is a half-angle formula for $\sin $ and $\cos $ as well. These are,
$\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}} $ and $\cos \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 + \cos \theta }}{2}} $
Also, $\cos \left( {2\theta } \right) = {\cos ^2}\theta - {\sin ^2}\theta = 1 - 2{\sin ^2}\theta = 2{\cos ^2}\theta - 1$, $\sin \left( {2\theta } \right) = 2\sin \theta \cos \theta $ and $\tan \left( {2\theta } \right) = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$. The parent formulas for the half angle formulas are the formulas with $2\theta $. In these formulas, substituting $2\theta $ by $\theta $ and $\theta $ by $\dfrac{\theta }{2}$ gives the resulting half angle formulas. All these formulas are very useful and sometimes they are converted according to the problem statement and are used accordingly.