Question

The value of $\cos^{-1}\left[\cos \left(\frac{\pi}{2}\right)\right]+\cos^{-1}\left[\sin \left(\frac{2\pi}{3}\right)\right]$ is

• A. $\frac{2\pi}{3}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\pi$

Answer 1

Answer: (A)

$\frac{2\pi}{3}$

Answer 2

1. Evaluate $\cos^{-1}(\cos(\pi/2))$.

   Since $\pi/2$ lies in the principal range $[0, \pi]$ for $\cos^{-1}$, we have:

   $$\cos^{-1}(\cos(\pi/2)) = \pi/2.$$

2. Evaluate $\cos^{-1}(\sin(2\pi/3))$.

   Note that:

   $$\sin(2\pi/3) = \sin\left(\pi - \pi/3\right) = \sin(\pi/3) = \sqrt{3}/2.$$

   Since $\cos(\pi/6) = \sqrt{3}/2$ and $\pi/6$ is in $[0, \pi]$, we get:

   $$\cos^{-1}(\sqrt{3}/2) = \pi/6.$$

3. Sum the results:

   $$\pi/2 + \pi/6 = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}.$$