Question

Evaluate the exact numerical value of the given inverse trigonometric expression: $\sin\left[\cos^{-1} \cos\left(\frac{7\pi}{6}\right)\right]$.

Hint:

Always remember the principal value ranges: - $\sin^{-1}(x) \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ - $\cos^{-1}(x) \in [0, \pi]$ Never blindly cancel an inverse function with its regular function without verifying the angle range first!

Answer 1

Concept: To accurately evaluate composite functions involving trigonometric and inverse trigonometric operations, we must strictly respect the principal value branches. The principal value branch of the inverse cosine function, $y = \cos^{-1}(x)$, is strictly restricted to the closed domain interval $[0, \pi]$. Therefore, the simplification identity $\cos^{-1}(\cos \theta) = \theta$ holds true if and only if $\theta$ lies within $[0, \pi]$. If the angle $\theta$ lies outside this range, we must first utilize trigonometric reduction identities to map the angle into the principal interval without changing its fundamental cosine evaluation.

Step 1: Analyzing and reducing the inner angle $\frac{7\pi}{6}$ to fit within the principal value branch.
The given internal angle is $\theta = \frac{7\pi}{6}$. Let us check if this angle falls inside the principal value range $[0, \pi]$: \[ \frac{7\pi}{6} = 1.166\pi > \pi \] Since $\frac{7\pi}{6} \notin [0, \pi]$, we cannot directly simplify $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{7\pi}{6}$. We must rewrite the angle using the periodic and symmetric properties of the cosine function. We know that $\cos(2\pi - \alpha) = \cos(\alpha)$. Let us apply this identity: \[ \cos\left(\frac{7\pi}{6}\right) = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos\left(\frac{12\pi - 7\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) \] Now, let us verify if the new angle $\frac{5\pi}{6}$ lies inside our principal branch interval $[0, \pi]$: \[ 0 \leq \frac{5\pi}{6} \leq \pi \] This is true. Thus, $\frac{5\pi}{6}$ is the correct principal angle representation.

Step 2: Evaluating the inverse cosine component.
Now we can substitute our simplified, valid principal angle expression back into the composition: \[ \cos^{-1}\left[\cos\left(\frac{7\pi}{6}\right)\right] = \cos^{-1}\left[\cos\left(\frac{5\pi}{6}\right)\right] \] Applying the direct inversion property $\cos^{-1}(\cos \alpha) = \alpha$ since $\alpha = \frac{5\pi}{6} \in [0, \pi]$: \[ \cos^{-1}\left[\cos\left(\frac{5\pi}{6}\right)\right] = \frac{5\pi}{6} \]

Step 3: Calculating the final outermost sine evaluation.
The final expression requires us to take the sine of the value calculated in
Step 2: \[ \text{Value} = \sin\left(\frac{5\pi}{6}\right) \] To compute this without a calculator, we expand the angle using the standard second-quadrant identity $\sin(\pi - \alpha) = \sin(\alpha)$: \[ \sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) \] We know from standard exact trigonometric values that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Thus, the final evaluated result of the entire expression is exactly $\frac{1}{2}$.