find exact value of $\cos^{-1}(\cos(9\pi/7))$

Question

anybody know how to find the exact value of $\cos^{-1}(\cos(\frac{9\pi}7))$ using inverse functions without the use of a calculator, i have no idea how to use $\frac{9\pi}7$ or how to get it to a term on the unit circle. thanks.

Answer 1

Hints:

• $\dfrac{9\pi}{7}$ is slightly more than half a circle
• $\cos\left(\pi+\theta\right)=\cos\left(\pi - \theta\right)$
• There is some $\phi$ for which $\cos^{-1} (\cos \phi)=\phi$

Answer 2

Just like the tan problem you posted before, inverses cancel. Therefore, you should get $9\pi/7$

Answer 3

Let $P\left(\dfrac{2\pi}{7}\right) = (u,v)$ so that $u = \cos\left(\dfrac{2\pi}{7}\right)$ and $v = \sin\left(\dfrac{2\pi}{7}\right)$

Note that $\pi - \dfrac{2\pi}{7} = \dfrac{5\pi}{7}$, so $P\left(\dfrac{5\pi}{7}\right)=(-u,v)$ and is in the second quadrant.

It follows that $\cos^{-1}\left(\cos\left(\frac{9\pi}{7}\right)\right) = \cos^{-1}(-u) = \dfrac{5\pi}{7}$

You can verify this by showing that

\begin{align} \cos\left(\dfrac{5\pi}{7}\right) & = \cos\left(\pi - \dfrac{2\pi}{7}\right) \\ & = \cos(\pi) \cos\left(\dfrac{2\pi}{7}\right) -\sin(\pi) \sin\left(\dfrac{2\pi}{7}\right)\\ &=-\cos\left(\dfrac{2\pi}{7}\right)\\ &= -u \end{align}