Evaluate arcsin(1/2) + arccos(1/2)

Question

Evaluate $\sin^{-1}\!\left(\frac{1}{2}\right) + \cos^{-1}\!\left(\frac{1}{2}\right)$ .

Solution — Step by Step

Recall the principal value ranges:

$\sin^{-1}(x)$ has range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\cos^{-1}(x)$ has range $[0, \pi]$

$\sin^{-1}\!\left(\frac{1}{2}\right) = \frac{\pi}{6}$ since $\sin\frac{\pi}{6} = \frac{1}{2}$ and $\frac{\pi}{6}$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ✓

$\cos^{-1}\!\left(\frac{1}{2}\right) = \frac{\pi}{3}$ since $\cos\frac{\pi}{3} = \frac{1}{2}$ and $\frac{\pi}{3}$ lies in $[0, \pi]$ ✓

\sin^{-1}\!\left(\frac{1}{2}\right) + \cos^{-1}\!\left(\frac{1}{2}\right) = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}

There is a beautiful identity:

\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \quad \text{for } x \in [-1, 1]

Applying this with $x = \frac{1}{2}$ :

\sin^{-1}\!\left(\frac{1}{2}\right) + \cos^{-1}\!\left(\frac{1}{2}\right) = \frac{\pi}{2} \quad \checkmark

Why This Works

The identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ comes from the complementary relationship between sine and cosine.

If $\sin^{-1}(x) = \theta$ , then $\sin\theta = x$ . Since $\sin\theta = \cos\!\left(\frac{\pi}{2} - \theta\right)$ , we get $\cos\!\left(\frac{\pi}{2} - \theta\right) = x$ , which means $\cos^{-1}(x) = \frac{\pi}{2} - \theta$ .

Therefore: $\sin^{-1}(x) + \cos^{-1}(x) = \theta + \left(\frac{\pi}{2} - \theta\right) = \frac{\pi}{2}$ .

This identity holds for all $x \in [-1, 1]$ , not just $x = \frac{1}{2}$ .

Common Mistake

Some students confuse the principal value ranges. For $\sin^{-1}$ , the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (symmetric about 0), while for $\cos^{-1}$ , the range is $[0, \pi]$ (only non-negative values). A common error: writing $\cos^{-1}(-\frac{1}{2})$ as $-\frac{\pi}{3}$ — but this is outside the range of $\cos^{-1}$ ! The correct value is $\frac{2\pi}{3}$ (the second quadrant angle where cosine is $-\frac{1}{2}$ ).

For JEE, the complementary identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ is tested regularly in simplification problems. Similarly, $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$ and $\sec^{-1}(x) + \csc^{-1}(x) = \frac{\pi}{2}$ (for appropriate domains). Memorise all three pairs.

Question

Solution — Step by Step

Why This Works

Common Mistake

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