Question

The value of \( \cos\left(\frac{\pi}{6}-\cos^{-1}\left(-\frac{\sqrt3}{2}\right)\right) \) is equal to:

• A. \( -\frac{\sqrt1}{2} \)
• B. \( -\frac{1}{\sqrt2} \)
• C. \( -\frac{1}{2} \)
• D. \( \frac{1}{2} \)

Hint:

Always remember the principal value range of inverse cosine: \[ \cos^{-1}x \in [0,\pi] \] When cosine is negative, the angle usually lies in the second quadrant. Important standard values: \[ \cos\frac{\pi}{6}=\frac{\sqrt3}{2} \] \[ \cos\frac{5\pi}{6}=-\frac{\sqrt3}{2} \] Also remember that cosine is an even function: \[ \cos(-\theta)=\cos\theta \]

Answer 1

The Correct Option is C

Concept: This problem involves inverse trigonometric functions together with compound angle identities. The main ideas used are:
• Evaluating the principal value of an inverse cosine function.
• Using the cosine subtraction identity: \[ \cos(A-B)=\cos A\cos B+\sin A\sin B \]
• Using standard trigonometric values. For inverse cosine: \[ \cos^{-1}x \in [0,\pi] \] Hence we must always choose the angle lying in this interval.

Step 1: Evaluating the inverse cosine term We are given: \[ \cos^{-1}\left(-\frac{\sqrt3}{2}\right) \] We know that: \[ \cos\frac{\pi}{6}=\frac{\sqrt3}{2} \] But the value inside the inverse cosine is negative: \[ -\frac{\sqrt3}{2} \] Now cosine is negative in the second quadrant. The angle in the interval \([0,\pi]\) whose cosine is: \[ -\frac{\sqrt3}{2} \] is: \[ \frac{5\pi}{6} \] Therefore, \[ \cos^{-1}\left(-\frac{\sqrt3}{2}\right)=\frac{5\pi}{6} \] Substituting into the given expression: \[ \cos\left(\frac{\pi}{6}-\frac{5\pi}{6}\right) \]

Step 2: Simplifying the angle Subtract the fractions: \[ \frac{\pi}{6}-\frac{5\pi}{6} = -\frac{4\pi}{6} = -\frac{2\pi}{3} \] Thus the expression becomes: \[ \cos\left(-\frac{2\pi}{3}\right) \]

Step 3: Using the even property of cosine We know: \[ \cos(-\theta)=\cos\theta \] Therefore, \[ \cos\left(-\frac{2\pi}{3}\right) = \cos\frac{2\pi}{3} \] Now, \[ \cos\frac{2\pi}{3}=-\frac12 \] Hence, \[ \cos\left(\frac{\pi}{6}-\cos^{-1}\left(-\frac{\sqrt3}{2}\right)\right) = -\frac12 \] Final Answer: \[ \boxed{-\frac12} \] Therefore, the correct option is: \[ \boxed{(C)} \]