Question

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

Answer 1

Here, in this question, we need to find the exact value of $\tan 165^\circ$ using the half angle identity. In order to use the half angle identity, we will first rewrite the given value and then simplify in such a way that we get to use half angle identity and find the answer.

Complete step-by-step solution:
The given value is $\tan 165^\circ$ , we need to find its exact value by using the half angle identity.
First, let $\tan 165^\circ$ be $t$ …………….... (1)
Now, we will use half angle on $\tan 165^\circ$ ,
\Rightarrow$$$\tan 2\left( {165^\circ } \right) = \tan 330^\circ $$ Now, $$\tan 330^\circ $$ can also written using $$360^\circ $$ by \Rightarrow\tan 330^\circ = \tan (360^\circ - 30^\circ )$$ We clearly know a property of $\tan $, i.e., $\tan (\pi - \theta ) = \tan ( - \theta )$ , it becomes $\Rightarrow\tan 330^\circ = \tan (360^\circ - 30^\circ ) = \tan ( - 30^\circ ) There is another property that is $\tan ( - \theta ) = - \tan \theta $ , therefore,\tan ( - 30^\circ )becomes, $\Rightarrow$$$\tan ( - 30^\circ ) = - \tan 30^\circ
We know that $\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$ , therefore we get
\Rightarrow$$$ - \tan 30^\circ = - \dfrac{1}{{\sqrt 3 }}$$ Hence $$\tan 330^\circ = - \dfrac{1}{{\sqrt 3 }}$$ ……………... (2) Now, using the identity of trigonometric functions, \Rightarrow\tan 2t = \dfrac{{2\tan t}}{{1 - {{\tan }^2}t}}$$ , And now from $(1)$ we have $$\tan 165^\circ = t$$ This implies that $\tan 2t = \tan 330^\circ $ , $\Rightarrow\tan 2t = \tan 2(165^\circ ) = \dfrac{{2t}}{{1 - {t^2}}} From $(2)$ we can say that $\tan 2t = \dfrac{1}{{\sqrt 3 }}$ . Now, putting all this together we get, $\Rightarrow$ $\tan 2(165^\circ ) = \tan 2t = \dfrac{{2t}}{{1 - {\operatorname{t} ^2}}} = \dfrac{1}{{\sqrt 3 }}$ This implies, $\Rightarrow$ $\dfrac{{2t}}{{1 - {\operatorname{t} ^2}}} = \dfrac{1}{{\sqrt 3 }}$ Cross multiplying this, we get, $\Rightarrow$ $2t(\sqrt {3)} = 1 - {t^2}$ Removing the brackets, $\Rightarrow$ $2\sqrt 3 t = 1 - {t^2}$ Forming a quadratic equation by transferring the terms to one side we get, $\Rightarrow$$${t^2} - 2\surd 3t - 1 = 0
After forming an equation, we can solve that quadratic equation by using the quadratic equation formula:
${\text{x = }}\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$
The equation is ${t^2} - 2\surd 3t - 1 = 0$
Here $a = 1,b = - 2\sqrt 3 ,c = - 1$
Applying in the formula,
$\Rightarrow$ $x = \dfrac{{ - ( - 2\sqrt 3 ) \pm \sqrt {{{( - 2\sqrt 3 )}^2} - 4( - 1)(1} )}}{{2(1)}}$
\Rightarrow$$$x = \dfrac{{2\sqrt 3 \pm \sqrt {12 + 4} }}{2}$$ \Rightarrowx = \dfrac{{2\sqrt 3 \pm \sqrt {16} }}{2}$$ $\Rightarrowx = \dfrac{{2\sqrt 3 \pm 4}}{2} Now we can split it into two, as the expression involves $ \pm $ in it, one can be plus the other can be minus. $\Rightarrow$$$x = \dfrac{{2\sqrt 3 + 4}}{2},\dfrac{{2\sqrt 3 - 4}}{2}
\Rightarrow$$$x = \dfrac{{2\left( {\sqrt 3 + 2} \right)}}{2},\dfrac{{2\left( {\sqrt 3 - 2} \right)}}{2}$$ \Rightarrow$$$x = \sqrt 3 + 2,\sqrt 3 - 2$$
$\Rightarrow$ $x = \sqrt 3 + 2,x = \sqrt 3 - 2$
Therefore, the values of $x$ are $\sqrt 3 + 2,\sqrt 3 - 2$ .

**Since $\tan 165^\circ$ , and $\tan 165{\text{ }} < {\text{ }}0$ , and the answer $\sqrt 3 - 2$ is accepted. **

Note: Alternative method:
If the question is silent about the method that should be used, we can solve this $\tan 165^\circ$ in another easy method.
By using the property of trigonometric functions,
$\tan (a - b) = \dfrac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}$
So here we have$\tan 165^\circ$ , which can be written as,
$\tan 165^\circ = \tan (180 - 15)^\circ$
Appling in the formula,
$\tan 165^\circ = \tan (180 - 15)^\circ = \dfrac{{\tan 180^\circ - \tan 15^\circ }}{{1 + \tan 180^\circ \tan 15^\circ }}$
$\tan 165^\circ = \dfrac{{0 - 2 - \sqrt 3 }}{{1 + 0.2 - \sqrt 3 }}$
$\tan 165^\circ = \dfrac{{ - 2 + \sqrt 3 }}{1}$
$\begin{gathered} \tan 165^\circ = - 2 + \sqrt 3 \\\ \tan 165^\circ = \sqrt 3 - 2 \\\ \end{gathered}$
Therefore, the answer is $\sqrt 3 - 2$.