Question

How do you use the half-angle formulas to find the exact value of $\tan \left( {\dfrac{{3\pi }}{8}} \right)$ ?

Answer 1

Hint : Here we have to use half-angle formula to of $\tan \left( \theta \right)$ to find the value of $\tan \left( {\dfrac{{3\pi }}{8}} \right)$ .
We can substitute the value of $\tan \left( {\dfrac{{3\pi }}{8}} \right)$ with any variable after applying the formula, so that we can form a quadratic equation and then we can find the roots of that equation. Also, we know that the value of $\tan \left( {\dfrac{{3\pi }}{4}} \right)\,\,is\, - 1$.
Formula used: $\tan \left( {2\theta } \right) = \dfrac{{2\tan \left( \theta \right)}}{{1 - {{\tan }^2}\theta }}$

Complete step-by-step answer :
In the given question, let $\tan \left( {\dfrac{{3\pi }}{8}} \right)\, = \,t$
Now, applying the formula $\tan \left( {2\theta } \right) = \dfrac{{2\tan \left( \theta \right)}}{{1 - {{\tan }^2}\theta }}$
$\tan \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{{2\tan \left( {\dfrac{{3\pi }}{8}} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{{3\pi }}{8}} \right)}}$
Now putting the value of $\tan \left( {\dfrac{{3\pi }}{8}} \right)\, = \,t\,\,and\,\,\tan \left( {\dfrac{{3\pi }}{4}} \right)\, = \, - 1$
$- 1 = \dfrac{{2t}}{{1 - {t^2}}}$
On cross-multiplication, we get
$- 1\left( {1 - {t^2}} \right) = 2t$
${t^2} - 1 = 2t$
On transposing, we get
${t^2} - 2t - 1 = 0$
Now using the quadratic formula to find the roots of above equation
$t\, = \,\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now, compare the above equation with $a{x^2} + bx + c = 0$
Therefore, $a\, = \,1\,,\,b = \, - 2\,,\,c\, = \, - 1$
Now, on putting the values
$\Rightarrow t\, = \,\dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$
On simplification, we get
$\Rightarrow t\, = \,\dfrac{{2 \pm \sqrt {4 + 4} }}{2}$
On adding, we get
$\Rightarrow t\, = \,\dfrac{{2 \pm \sqrt 8 }}{2}$
Taking the square root of $8$
$\Rightarrow t\, = \,\dfrac{{2 \pm 2\sqrt 2 }}{2}$
On dividing, we get
$\Rightarrow t\, = \,1 \pm \sqrt 2$
So, there are two values of t as $1 - \sqrt 2 \,and\,1 + \sqrt 2$ but we have to neglect one value of t i.e. $1 - \sqrt 2$ because it is a negative value but the value of $\tan \left( {\dfrac{{3\pi }}{8}} \right)$ would be positive because angle lies in the first quadrant and the value of $\tan \left( \theta \right)$ is positive in the first quadrant.
Therefore, the value of $\tan \left( {\dfrac{{3\pi }}{8}} \right)$ is $1 + \sqrt 2$.
So, the correct answer is “$1 + \sqrt 2$”.

Note : This is a very standard method of finding the values of trigonometric quantities having small angles. It is very important to remember the trigonometric formulas. Without formulas, trigonometry is incomplete. Half-angle formulas are one among those formulas. The negative value is neglected based on the quadrant in which the given tan function lies; if it was lying in the 3rd quadrant then $1 - \sqrt 2$ will be the solution.