Question

How do you find the value of $\tan 6\pi ?$.

Answer 1

Hint: We will use the circular function of the trigonometry to find the value of $\tan 6\pi $. We will use the formula $\tan \left( 2\pi +\theta \right)=\tan \theta $ and we will first find the value of $\theta $ as $\theta $ must lies in the range of $\left[ 0,\dfrac{\pi }{2} \right]$ because we generally do not know the value of trigonometric function for angle greater than $\dfrac{\pi }{2}$.

Complete step-by-step solution:
We will first recall the concept of the circular function of the trigonometric. All trigonometric functions can also be seen on the unit circle. The unit circle is divided into quadrants by the Cartesian coordinates, so the signs of each circular function can be determined by the value of t. If the value of t has a positive value for both x and y in Quadrant I, then the trigonometric function will have positive value. In Quadrant II, only the value of y is positive, and x has a negative value, so sin t and cosecant t will be positive. All other functions will be negative. In Quadrant III, both coordinates for x and y are negative. This means that tangents and their reciprocal cotangents will be positive. In Quadrant 4, x is positive and y is negative, so cosines and their reciprocal, secants, will be positive.
Since, we know that the whole angle of the circle is equal to $360{}^\circ $or $2\pi $ and when we move in anticlockwise direction the angle is counted as positive and when we move in clockwise direction angle is counted as negative.
So, to represent $6\pi $ we have to move a whole circle three times in anticlockwise direction, so the angle will lie on the x-axis so we can say that it is equivalent to $0{}^\circ $ and we know that tangent of any angle in first quadrant is positive and negative in the fourth quadrant.
So, we can write $\tan 6\pi =\tan 0{}^\circ $.
And, since from standard trigonometric table we know that $\tan 0{}^\circ =0$
$\Rightarrow \tan 6\pi =\tan 0{}^\circ =0$
This is our required solution.

Note: We can also solve the above question alternatively by using the trigonometric formula \[\tan \left( 2\pi +\theta \right)=\tan \theta \]. When we put $\theta =4\pi $, we will get:
$\Rightarrow \tan \left( 2\pi +4\pi \right)=\tan 4\pi $
Now, again we will use the formula and write $4\pi =2\pi +2\pi $ as $\tan 4\pi =\tan \left( 2\pi +2\pi \right)$.
$\Rightarrow \tan 4\pi =\tan \left( 2\pi +2\pi \right)$
$\Rightarrow \tan 4\pi =\tan \left( 2\pi \right)$
$\Rightarrow \tan \left( 2\pi \right)=0$