Question

tan^-1 (tan 6pi/7) = - pi/7

Can someone explain to me why this is the answer for the inverse function?

Answer 1

The range of the inverse tangent function is angles in the first and fourth quadrants.\(\dfrac{6\pi}{7}\) lies in the second quadrant.

Answer 2

What is confusing me is how would I know that lies within the second quadrant since I know the the radians in the second quadrant are 2pi/3, 3pi/4, and 5pi/6. I understand how it goes negative now because it needs to stay between [-pi/2, pi/2] but, How by looking at 6pi/7 do you know it lies at the second quadrant?

Answer 3

Consider the unit circle, which (within one revolution) contains all angles from \(0\) to \(2\pi\). If you have some angle \(k\pi\), with \(0<k<1\), then the angle lies between \(0\pi=0\) and \(1\pi=1\), which comprise the first two quadrants.

These two quadrants are split at the midpoint, \(\dfrac{\pi}{2}\). So if your angle \(k\pi\) has \(0<k<\dfrac{1}{2}\), so that \(0<k\pi<\dfrac{\pi}{2}\), then the angle lies in Q1.

Otherwise, if \(\dfrac{1}{2}<k<1\), then \(\dfrac{\pi}{2}<k\pi<\pi\), so \(k\pi\) lies in Q2.

Does that make sense?

Answer 4

For reference, if the angle \(0\le k\pi\le2\pi\),

Answer 5

\[\frac{ \frac{7}{2} \pi}{7}<\frac{6 \pi}{7} < \frac{7\pi}{7} \\ \text{ or in other words } \\\frac{\pi}{2} < \frac{6\pi}{7} < \pi \\ \]

Answer 6

3.5<6<7