Question

How do you find the exact values of $\tan 165{}^\circ$ using the half angle formula?

Answer 1

This question belongs to the topic of trigonometry. In this question, first we will double the angle and find the value of tan function at that double angle. After that, we will use here a trigonometric formula or we can say trigonometric identity. After using that formula, we will solve the further solution and get the value of $\tan 165{}^\circ$.

Complete step-by-step solution:
Let us solve this question.
In this question, we are going to find the exact values of $\tan 165{}^\circ$ using the half angle formula of tan function.
Here, we will first find the value of tan function at an angle which is double of 165 degrees. So, the double of 165 degrees will be 330 degrees.
So, we can say that
$\tan 2\left( 165 \right){}^\circ =\tan 330{}^\circ$
The above equation can also be written as
$\Rightarrow \tan 2\left( 165 \right){}^\circ =\tan \left( 360{}^\circ -30{}^\circ \right)$
As we know that $\tan \left( 360{}^\circ -\theta \right)=-\tan \theta$, so we can write the above equation as
$\Rightarrow \tan 2\left( 165{}^\circ \right)=-\tan 30{}^\circ$
As we know that the value of $\tan 30{}^\circ$ is $\dfrac{1}{\sqrt{3}}$
So, we can write the above equation as
$\Rightarrow \tan 2\left( 165{}^\circ \right)=-\dfrac{1}{\sqrt{3}}$
Now, the identity of tan function is going to be used here in the solution is:
$\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
This formula is half angle formula.
By putting the value of $\theta$ as $165{}^\circ$in the above formula, we get
$\tan 2\left( 165{}^\circ \right)=\dfrac{2\tan 165{}^\circ }{1-{{\tan }^{2}}165{}^\circ }$
As we have found in the above that the value of $\tan 2\left( 165{}^\circ \right)$is$-\dfrac{1}{\sqrt{3}}$.
So, we can write
$\Rightarrow -\dfrac{1}{\sqrt{3}}=\dfrac{2\tan 165{}^\circ }{1-{{\tan }^{2}}165{}^\circ }$
Let us write the term $\tan 165{}^\circ$ as x. So, the above equation can also be written as
$\Rightarrow -\dfrac{1}{\sqrt{3}}=\dfrac{2x}{1-{{x}^{2}}}$
The above equation can also be written as
$\Rightarrow -1=\dfrac{2\sqrt{3}x}{1-{{x}^{2}}}$
The above equation can also be written as
$\Rightarrow -\left( 1-{{x}^{2}} \right)=2\sqrt{3}x$
The above equation can also be written as
$\Rightarrow -1+{{x}^{2}}=2\sqrt{3}x$
The above equation can also be written as
$\Rightarrow {{x}^{2}}-2\sqrt{3}x-1=0$
According to Sridharacharya method, the value of x will be
$x=\dfrac{-2\sqrt{3}\pm \sqrt{{{\left( 2\sqrt{3} \right)}^{2}}-4\times 1\times \left( -1 \right)}}{2\times \left( -1 \right)}$
The above equation can also be written as
$x=\dfrac{-2\sqrt{3}\pm \sqrt{12+4}}{-2}$
We can write the above equation as
$x=\dfrac{-2\sqrt{3}\pm 4}{-2}=\dfrac{-2\sqrt{3}}{-2}\pm \dfrac{4}{-2}=\sqrt{3}\mp 2$
Hence, we can write the value of x as $\sqrt{3}+2$ and $\sqrt{3}-2$.
As we have taken $\tan 165{}^\circ$ as x.
So, $\tan 165{}^\circ =\sqrt{3}\pm 2$
As we can see that the angle 165 degrees is between 135 degrees and 180 degrees. So, we can say that the value of tan function at 165 degrees will be between -1 and 0.
Therefore, we can say that the exact value of $\tan 165{}^\circ$ is only $\sqrt{3}-2$ and not $\sqrt{3}+2$ because it is greater than 1.
We can take reference from the following figure for the above solution.

Note: We should have a better knowledge in the topic trigonometry to solve this type of question.
Don’t forget the formulas and identities like:
$\tan \left( 360{}^\circ -\theta \right)=-\tan \theta$
$\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$
Half angle formula: $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
And, also remember that, if the quadratic equation is given as $a{{x}^{2}}+bx+c=0$, then according to Sridharacharya rule the value of x will be :
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
The above formulas and identities should be kept remembered to solve this type of question easily.