Question

Evaluate \(\sin\left[\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)\right]\).

Hint:

Since 7pi/6 is outside [0, pi], first write cos(7pi/6) = cos(5pi/6), then the inverse cosine gives 5pi/6 and sin(5pi/6) = 1/2.

Answer 1

Step 1: Recall the principal value branch:
The identity \(\cos^{-1}(\cos\theta)=\theta\) holds only when \(\theta\) lies in \([0,\pi]\).
Here \(\theta=\dfrac{7\pi}{6}\), which is bigger than \(\pi\). So we cannot cancel the two functions directly.

Step 2: Bring the angle into the principal range:
Cosine is even and has period \(2\pi\), so \(\cos\theta=\cos(2\pi-\theta)\).
\[ \cos\dfrac{7\pi}{6}=\cos\left(2\pi-\dfrac{7\pi}{6}\right)=\cos\dfrac{5\pi}{6} \]
The angle \(\dfrac{5\pi}{6}\) lies in \([0,\pi]\).

Step 3: Apply the inverse cosine:
\[ \cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)=\cos^{-1}\left(\cos\dfrac{5\pi}{6}\right)=\dfrac{5\pi}{6} \]

Step 4: Find the sine:
\[ \sin\dfrac{5\pi}{6}=\sin\left(\pi-\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{6}=\dfrac{1}{2} \]

Final Answer:
The value of the expression is one half. \[ \boxed{\dfrac{1}{2}} \]