Question

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

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

Hint:

Logic Tip: Always map arguments back into the appropriate principal range (Quadrants 1 \& 4 for sine, Quadrants 1 \& 2 for cosine) using ASTC rules before canceling the function with its inverse.

Answer 1

Answer: (E)

Concept:
Inverse trigonometric functions $f^{-1}(f(x))$ only equal $x$ if $x$ falls within their principal value branches. For $y = \cos^{-1}(\cos x)$, the range must be $[0, \pi]$. For $y = \sin^{-1}(\sin x)$, the range must be $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Step 1: Evaluate the inverse cosine term.
The argument is $\frac{2\pi}{3}$. We check if it falls within the principal range $[0, \pi]$. Since $0 \le \frac{2\pi}{3} \le \pi$, the identity holds perfectly: $$\cos^{-1}\left(\cos\frac{2\pi}{3}\right) = \frac{2\pi}{3}$$
Step 2: Evaluate the inverse sine term.
The argument is $\frac{2\pi}{3}$. We check if it falls within the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $\frac{2\pi}{3}>\frac{\pi}{2}$, we cannot simply output $\frac{2\pi}{3}$. We must find a reference angle in the principal range that has the same sine value. In the second quadrant, $\sin(\pi - \theta) = \sin \theta$. $$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)$$ Now, apply the inverse sine to this equivalent expression. Since $\frac{\pi}{3}$ is within $[-\frac{\pi}{2}, \frac{\pi}{2}]$: $$\sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3}$$
Step 3: Add the two evaluated terms together.
$$\text{Sum} = \frac{2\pi}{3} + \frac{\pi}{3} = \frac{3\pi}{3} = \pi$$