Q.Find the value of the following: $\cos^{-1} \left( \frac{1}{2} \right) + 2 \sin^{-1} \left( \frac{1}{2} \right)$

Concept: Inverse Trigonometric Graphs — recall the principal value ranges: $\cos^{-1}x \in [0,\pi]$ and $\sin^{-1}x \in [-\frac{\pi}{2},\frac{\pi}{2}]$.

Step 1: Evaluate $\cos^{-1}\left(\frac{1}{2}\right)$.

Since $\cos\frac{\pi}{3} = \frac{1}{2}$ and $\frac{\pi}{3} \in [0,\pi]$, we get $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$.

Step 2: Evaluate $\sin^{-1}\left(\frac{1}{2}\right)$. …

The key to this problem is knowing the principal values of inverse trigonometric functions. These are the specific angles (within a defined range) that the functions return. For $\cos^{-1}$, the principal value range is $[0, \pi]$, and for $\sin^{-1}$, it is $[-\pi/2, \pi/2]$. Once you recall the standard angles, the arithmetic is trivial.

Let’s work through it.

1. Evaluate $\cos^{-1}(1/2)$.

   We need an angle $\theta$ in $[0, \pi]$ such that $\cos \theta = 1/2$.

   From the unit circle, $\cos(\pi/3) = 1/2$, and $\pi/3$ lies in $[0, \pi]$.

   So $\cos^{-1}(1/2) = \pi/3$.

2. Evaluate $\sin^{-1}(1/2)$.

   We need an angle $\phi$ in $[-\pi/2, \pi/2]$ such that $\sin \phi = 1/2$.

   The standard angle is $\pi/6$, since $\sin(\pi/6) = 1/2$ and $\pi/6$ is within the principal range.

   So $\sin^{-1}(1/2) = \pi/6$.

3. Plug into the expression.

   The given expression is $\cos^{-1}(1/2) + 2 \sin^{-1}(1/2)$.

   Substituting: $\pi/3 + 2 \cdot (\pi/6) = \pi/3 + \pi/3 = 2\pi/3$. …

Method: Combining inverse trig values with coefficients

Apply this when an expression multiplies an inverse trig term by a constant before adding.

Steps

Step 1: Evaluate each inverse trig value first

Find each principal value on its own — e.g. $\cos^{-1}$ in $[0,\pi]$ and $\sin^{-1}$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ — using standard angles.

Step 2: Apply the coefficient to the angle, not the argument …

Common Mistakes

Mistake 1: Pulling the coefficient inside as $\sin^{-1}\left(2 \cdot \frac{1}{2}\right) = \sin^{-1}(1)$

Why it's wrong: the $2$ multiplies the resulting angle, not the argument; $2\sin^{-1}\frac{1}{2} = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3}$, whereas $\sin^{-1}(1) = \frac{\pi}{2}$. Correct approach: evaluate the inverse function first, then multiply.

Mistake 2: Confusing the standard values of $\cos^{-1}\frac{1}{2}$ and $\sin^{-1}\frac{1}{2}$ …