find exact value of $\cos^{-1}(\cos(9\pi/7))$
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.
$\endgroup$3 Answers
$\begingroup$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 $\cos^{-1} (\cos \phi)=\phi$
Just like the tan problem you posted before, inverses cancel. Therefore, you should get $9\pi/7$
$\endgroup$ 1 $\begingroup$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}
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