Q.Evaluate $\cos\left[\cos^{-1}\left(\frac{-\sqrt3}{2}\right)+\frac{\pi}{6}\right]$.

Concept: Inverse trigonometric functions and angle addition formulas.

First, evaluate the inner inverse cosine.

$\cos^{-1}\left(\frac{-\sqrt3}{2}\right)$ is the angle in $[0, \pi]$ whose cosine is $-\frac{\sqrt3}{2}$. That angle is $\frac{5\pi}{6}$.

Now the expression becomes:

$\cos\left(\frac{5\pi}{6} + \frac{\pi}{6}\right) = \cos\left(\frac{6\pi}{6}\right) = \cos(\pi)$.

We know $\cos(\pi) = -1$.

Concept and Intuition

This problem tests your understanding of inverse trigonometric functions — specifically, the range of $\cos^{-1}$ and how to handle a negative input inside it. Many students rush to say $\cos^{-1}(-\sqrt3/2) = 5\pi/6$ or $7\pi/6$, but only one of these lies in the principal value range of $\cos^{-1}$, which is $[0, \pi]$. Once we correctly identify that angle, the rest is just a straightforward cosine of a sum.

The trap here is forgetting that $\cos^{-1}$ does not return an angle in the second or fourth quadrant arbitrarily — it always returns an angle between $0$ and $\pi$ (inclusive). For a negative cosine value, that angle must be in the second quadrant (between $\pi/2$ and $\pi$).

Step-by-step solution

1. Identify the principal value of $\cos^{-1}(-\sqrt3/2)$ We know $\cos(\pi/6) = \sqrt3/2$, so $\cos(5\pi/6) = -\sqrt3/2$ because cosine is negative in the second quadrant. Since $5\pi/6$ lies in $[0, \pi]$, it is the correct principal value. Therefore:

2. Add $\pi/6$ to this angle

3. Evaluate the cosine of the sum

Method: Evaluating a trig function of (inverse-trig $+$ angle)

For any expression of the form $\cos\!\big[\cos^{-1}(a)+\beta\big]$ (or with $\sin$, $\tan$), work strictly inside-out: resolve the inverse to a genuine angle first, then do the arithmetic, then apply the outer function.

Steps

Step 1: Evaluate the inner inverse to a principal-value angle.

Find the unique angle in the inverse function's range whose value is the given argument. For $\cos^{-1}$ that angle lives in $[0,\pi]$, so a negative argument gives a second-quadrant angle.

Step 2: Combine with the added angle.

Do the exact fraction addition — keep everything as multiples of $\pi$ so the sum stays exact (e.g. $\tfrac{5\pi}{6}+\tfrac{\pi}{6}=\pi$).

Step 3: Apply the outer trig function to that single angle.

Now it is an ordinary evaluation, e.g. $\cos\pi=-1$. Because the inner step returned an exact standard angle, the final value is exact — no decimals or estimation needed.

Common Mistakes

Mistake 1: Taking $\cos^{-1}\!\big(-\tfrac{\sqrt3}{2}\big)=\tfrac{7\pi}{6}$.

Why it's wrong: $\tfrac{7\pi}{6}>\pi$, so it is outside the arccos range $[0,\pi]$. Correct approach: for a negative cosine the principal angle is in the second quadrant, giving $\tfrac{5\pi}{6}$.

Mistake 2: Treating $\cos^{-1}\!\big(-\tfrac{\sqrt3}{2}\big)$ as $-\tfrac{\pi}{6}$ by pulling the sign out.

Why it's wrong: unlike $\sin^{-1}$, $\cos^{-1}$ is not odd, so $\cos^{-1}(-a)\ne-\cos^{-1}(a)$. Correct approach: use $\cos^{-1}(-a)=\pi-\cos^{-1}(a)=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}$.

Mistake 3: Applying the outer cosine before resolving the inner inverse.

Why it's wrong: $\cos[\cos^{-1}(\cdot)+\tfrac{\pi}{6}]$ is not $-\tfrac{\sqrt3}{2}$ plus something — the $+\tfrac{\pi}{6}$ is an angle addition. Correct approach: first get $\tfrac{5\pi}{6}$, add $\tfrac{\pi}{6}$ to reach $\pi$, then take $\cos\pi=-1$.