Question

How do you find the exact value of $\tan \left( {\dfrac{{ - \pi }}{{12}}} \right)$ ?

Answer 1

Hint: To get a better understanding of the angle, we shall convert the angle from radians to degree. If the angle is not one of the standard values we remember from the trigonometric table, we split the angle into two standard angles whose tangent value we know, such that the sum or difference of two angles results in the given angle.

Complete step-by-step answer:
To know if the given angle is from the table with the values of sine, cosine, and tangent functions for some standard trigonometric angles, we shall convert the given angle into degrees.
As we know, to convert an angle from radian to degree, we must multiply the angle with the factor $\dfrac{{180}}{\pi }$ .
$\Rightarrow {\text{Degrees = }}\dfrac{{180}}{\pi } \times \dfrac{{ - \pi }}{{12}}$
Converting to factors and rearranging the equation,
$\Rightarrow {\text{Degrees = }}\dfrac{{ - \pi \times 15 \times 12}}{{\pi \times 12}}$
$\Rightarrow {\text{Degrees = }} - 15^\circ$
Hence, we can write $\dfrac{{ - \pi }}{{12}}$ in degrees as $- 15^\circ$ .
But, as we know $- 15^\circ$ is not one of the standard angles whose tangent value we remember.
Hence we need to split $- 15^\circ$ in two different angles whose tangent value is known to us such that the sum or difference of the two angles should result in $- 15^\circ$ .
Let us consider here the angels $60^\circ$ and $45^\circ$ .
If we subtract $60^\circ$ from $45^\circ$ , we get the result as $- 15^\circ$, and also the tangent value of $45^\circ$ and $60^\circ$ is known to us.
Hence, we can write tangent of $- 15^\circ$ as
$\Rightarrow \tan ( - 15^\circ ) = \tan (45^\circ - 60^\circ )$
Let’s consider this equation as an Equation $(1)$ .
Here, we need to use the trigonometric identity for the tangent of the sum of two angles or the difference of two angles, which is as follows
$\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
$\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$
Here, in the second identity, if we replace $x$ by $45^\circ$ and $y$ by $60^\circ$, we can obtain the value of $\tan ( - 15^\circ )$ , as shown
$\Rightarrow \tan (45^\circ - 60^\circ ) = \dfrac{{\tan 45^\circ - \tan 60^\circ }}{{1 + \tan 45^\circ \tan 60^\circ }}$
Now, we know the values of the angles as
$\Rightarrow$$\tan 45^\circ = 1$
$\Rightarrow$$\tan 60^\circ = \sqrt 3$
Substituting these values in the equation
$\Rightarrow \tan (45^\circ - 60^\circ ) = \dfrac{{1 - \sqrt 3 }}{{1 + (1)\left( {\sqrt 3 } \right)}}$
$\Rightarrow \tan (45^\circ - 60^\circ ) = \dfrac{{1 - \sqrt 3 }}{{1 + \sqrt 3 }}$
Substituting the value from the equation $(1)$ ,
$\Rightarrow \tan ( - 15^\circ ) = \dfrac{{1 - \sqrt 3 }}{{1 + \sqrt 3 }}$
Now, rationalizing the answer by multiplying the numerator and the denominator by $1 - \sqrt 3$
$\Rightarrow \tan ( - 15^\circ ) = \dfrac{{1 - \sqrt 3 }}{{1 + \sqrt 3 }} \times \dfrac{{1 - \sqrt 3 }}{{1 - \sqrt 3 }}$
Multiplying the numerators and the denominators
$\Rightarrow \tan ( - 15^\circ ) = \dfrac{{(1 - \sqrt 3 )(1 - \sqrt 3 )}}{{(1 + \sqrt 3 )(1 - \sqrt 3 )}}$
Simplifying the equation by opening the brackets,
$\Rightarrow \tan ( - 15^\circ ) = \dfrac{{1 \times 1 - 1 \times \sqrt 3 - \sqrt 3 \times 1 + \sqrt 3 \times \sqrt 3 }}{{1 \times 1 - 1 \times \sqrt 3 + \sqrt 3 \times 1 - \sqrt 3 \times \sqrt 3 }}$
$\Rightarrow \tan ( - 15^\circ ) = \dfrac{{1 - \sqrt 3 - \sqrt 3 + 3}}{{1 - \sqrt 3 + \sqrt 3 - 3}}$
Rearranging the numerator and denominator,
$\Rightarrow \tan ( - 15^\circ ) = \dfrac{{1 + 3 - \sqrt 3 - \sqrt 3 }}{{1 - 3 - \sqrt 3 + \sqrt 3 }}$
$\Rightarrow \tan ( - 15^\circ ) = \dfrac{{4 - 2\sqrt 3 }}{{ - 2}}$
Eliminating the common factor $\; - 2$ from numerator and denominator
$\Rightarrow \tan ( - 15^\circ ) = \sqrt 3 - 2$

Hence the exact value of $\tan \left( {\dfrac{{ - \pi }}{{12}}} \right)$ is $\sqrt 3 - 2$.

Note:
The pair of angles considered here to get the angle $- 15^\circ$ is not mandatory. You can take any other pair whose sum or difference leads to $- 15^\circ$ and the sine, cosine, and tangent values of both the angles should be known.