10/09/2026
π Evaluating Inverse Sine Functions: Stay Within the Boundaries!
π CBSE | Class 12 | Maths
Chapter: Inverse Trigonometric Functions
Think you can always cancel (\sin^{-1}) and (\sin)? π
Not so fast!
The rule
[\sin^{-1}(\sin\theta)=\theta]
works directly only when (\theta) lies within the principal value range of (\sin^{-1}x):
[-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}]
If (\theta) lies outside this range, you need to find an equivalent angle within the principal range.
π§ Remember it like this:
Step 1 β Check the angle
β¬οΈ
Step 2 β Is it inside (\left[-\frac{\pi}{2},\frac{\pi}{2}\right])?
β¬οΈ
YES β Keep the angle
NO β Find an equivalent angle inside the range
For example:
[\sin^{-1}\left(\sin\frac{5\pi}{6}\right)] is not simply (\frac{5\pi}{6}), because (\frac{5\pi}{6}) is outside the principal range.
Using[\sin\frac{5\pi}{6}=\sin\frac{\pi}{6}] we get: [\sin^{-1}\left(\sin\frac{5\pi}{6}\right)=\frac{\pi}{6}]
β‘ Key Rule:
Inverse trig functions return values only within their principal value ranges.
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