Question

How do you use the half-angle identity to find the exact value of $\tan 165^\circ$?

Answer 1

We start solving the problem by recalling the half-angle formula for the sine function as $\tan \dfrac{x}{2} = \dfrac{{\sin x}}{{1 + \cos x}}$. We then find the value of x so that $\dfrac{x}{2}$ is equal to the value $165^\circ$. We then make use of the results $\cos \left( {2\pi - \theta } \right) = \cos \theta$ and $\sin \left( {2\pi - \theta } \right) = - \sin \theta$ to proceed through the problem. We then make the necessary arrangements inside the square root and make use of the fact that $165^\circ$ lies in the second quadrant and the tan function is negative in the second quadrant to get the required answer.

Complete step-by-step answer:
According to the problem, we are asked to find the value of $\tan 165^\circ$ using the half-angle formula. Let us recall the half-angle formula for the tangent function.
We know that the half-angle formula for the tangent function is defined as,
$\tan \dfrac{x}{2} = \dfrac{{\sin x}}{{1 + \cos x}}$ ….. (1)
Now, we need to find the value of $\tan 165^\circ$. So, we have
$\Rightarrow \dfrac{x}{2} = 165^\circ$
Multiply both sides by 2,
$\Rightarrow x = 330^\circ$
Let us substitute the value in equation (1)
$\Rightarrow \tan 165^\circ = \dfrac{{\sin 330^\circ }}{{1 + \cos 330^\circ }}$
As we know $\cos \left( {2\pi - \theta } \right) = \cos \theta$ and $\sin \left( {2\pi - \theta } \right) = - \sin \theta$. Then,
$\Rightarrow \tan 165^\circ = \dfrac{{\sin \left( {360^\circ - 30^\circ } \right)}}{{1 + \cos \left( {360^\circ - 30^\circ } \right)}}$
Simplify the terms,
$\Rightarrow \tan 165^\circ = \dfrac{{ - \sin 30^\circ }}{{1 + \cos 30^\circ }}$
Now substitute the values,
$\Rightarrow \Rightarrow \tan 165^\circ = \dfrac{{ - \dfrac{1}{2}}}{{1 + \dfrac{{\sqrt 3 }}{2}}}$
Take LCM and cancel out the common factor,
$\Rightarrow \tan 165^\circ = \dfrac{{ - 1}}{{2 + \sqrt 3 }}$
Now, rationalize the denominator by multiplying by its conjugate,
$\Rightarrow \tan 165^\circ = \dfrac{{ - 1}}{{2 + \sqrt 3 }} \times \dfrac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }}$
Simplify the terms,
$\Rightarrow \tan 165^\circ = \dfrac{{ - 2 + \sqrt 3 }}{{4 - 3}}$
Simplify the terms,
$\Rightarrow \tan 165^\circ = \sqrt 3 - 2$

Hence, the value of $\tan 165^\circ$ is $\sqrt 3 - 2$

Note:
We should perform each step carefully in order to avoid calculation mistakes and confusion. We should keep in mind the nature of the values of trigonometric functions in different quadrants while solving this type of problem. Similarly, we can expect the formulas to find the value of $\sin 165^\circ$ and $\cos 165^\circ$ using the formula of $\cos 2\theta$.