Question

How do you evaluate $\tan \left( \dfrac{11\pi }{8} \right)$ ?
(a) Using trigonometric angle identities
(b) Using linear formulas
(c) a and b both
(d) none of the above

Answer 1

Hint: In this problem we are to find the value of $\tan \left( \dfrac{11\pi }{8} \right)$. We will try to use the trigonometric double angle identities of $\tan 2\theta $ to find and simplify the value of our needed problem. We can start with the fact that $\tan \left( \pi +\theta \right)=\tan \theta $ and consider $\theta =\dfrac{3\pi }{4}$ to get ahead with the problem and evaluate the value.

Complete step by step solution:
According to the question, we start with,
$\tan \left( \dfrac{11\pi }{8} \right)$
We can also write $\dfrac{11\pi }{8}=\pi +\dfrac{3\pi }{8}$ ,
$\Rightarrow \tan \left( \pi +\dfrac{3\pi }{8} \right)$
As, $\tan (\pi +\theta )=\tan \theta $ ,because it is in the 3rd quadrant, we get,
$\Rightarrow \tan \left( \dfrac{3\pi }{8} \right)$
Now, say, $\theta =\dfrac{3\pi }{8}$ , then,$2\theta =\dfrac{3\pi }{4}$ ,
Then we go on with the known trigonometric quantities,
$\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
So, we can write, $\tan \left( \dfrac{3\pi }{4} \right)=\dfrac{2t}{1-{{t}^{2}}}$ , where \[t=\tan \theta =\tan \left( \dfrac{3\pi }{8} \right)\].
As, $\tan \left( \dfrac{3\pi }{4} \right)=\tan \left( \pi -\dfrac{\pi }{4} \right)=-\tan \left( \dfrac{\pi }{4} \right)=-1$, it is in the second quadrant where the value of the tangent function is always negative.
$\Rightarrow -1=\dfrac{2t}{1-{{t}^{2}}}$
Multiplying both sides with $1-{{t}^{2}}$, we are getting,
$\Rightarrow {{t}^{2}}-1=2t$
Again, simplifying,
$\Rightarrow {{t}^{2}}-2t-1=0$
Using the Sridharacharya formula, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for an quadratic equation, $a{{x}^{2}}+bx+c=0$ ,
We get,
$t=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4.1.\left( -1 \right)}}{2.1}$
Simplifying,
$t=\dfrac{2\pm \sqrt{4+4}}{2}$
$\Rightarrow t=\dfrac{2\pm \sqrt{8}}{2}$
So, after more simplifying,
$\Rightarrow t=\dfrac{2\pm 2\sqrt{2}}{2}$
Dividing numerator and denominator by 2, we will get,
Then, $t=1\pm \sqrt{2}$ .
Now, $t=1-\sqrt{2}=\tan \left( \dfrac{3\pi }{8} \right)$ which is not possible as, $\dfrac{3\pi }{8}$ lies in the first quadrant where the value of tan is always positive.
So, $t=1+\sqrt{2}=\tan \left( \dfrac{3\pi }{8} \right)$,
Then, we can conclude,
$\tan \left( \dfrac{11\pi }{8} \right)=\tan \left( \dfrac{3\pi }{8} \right)=\sqrt{2}+1$

So, the correct answer is “Option a”.

Note: To understand how the values of trigonometric ratios like $\tan \left( \dfrac{3\pi }{8} \right)$ change in different quadrants, first we have to understand ASTC rule. The ASTC rule is nothing but the "all sin tan cos" rule in trigonometry. The angles which lie between 0° and 90° are said to lie in the first quadrant. The angles between 90° and 180° are in the second quadrant, angles between 180° and 270° are in the third quadrant and angles between 270° and 360° are in the fourth quadrant. In the first quadrant, the values for sin, cos and tan are positive. In the second quadrant, the values for sin are positive only. In the third quadrant, the values for tan are positive only. In the fourth quadrant, the values for cos are positive only.