Question
Evaluate sin−1(sin67π).
Solution — Step by Step
The function sin−1(x) has a restricted domain: it only returns values in [−2π,2π].
Our angle 67π lies between π and 23π — clearly outside this range. So we cannot directly write sin−1(sin67π)=67π.
67π is in the third quadrant (between π and 23π), where sine is negative.
sin67π=sin(π+6π)=−sin6π=−21
The problem reduces to:
sin−1(sin67π)=sin−1(−21)
We know sin6π=21, so:
sin−1(−21)=−6π
This is valid because −6π∈[−2π,2π].
sin−1(sin67π)=−6π
Why This Works
The inverse trig function sin−1 is not the true inverse of sin — it is the inverse of sin restricted to [−2π,2π]. This is the principal value branch, chosen because sine is one-one on this interval.
The identity sin−1(sinθ)=θ holds only when θ∈[−2π,2π]. When θ is outside this range, we must first reduce sinθ to its numerical value, then find the unique angle in the principal branch that gives the same sine.
Think of it this way: many angles share the same sine value (e.g., 6π, 65π, −611π all have sin=21). The function sin−1 always picks exactly one answer — the one in [−2π,2π].
There’s a direct formula for angles in the third quadrant. When θ∈[π,23π]:
sin−1(sinθ)=−(θ−π)⋅... wait, let’s derive it cleanly.
For θ∈[π,23π], write θ=π+α where α∈[0,2π].
Then sinθ=−sinα, so sin−1(sinθ)=−α=−(θ−π).
Applying to our problem: θ=67π, so α=67π−π=6π.
sin−1(sin67π)=−6π✓
Memorise these two range-reduction formulas for board and JEE Main:
- θ∈[2π,π]: sin−1(sinθ)=π−θ
- θ∈[π,23π]: sin−1(sinθ)=−(θ−π)
They save 30–40 seconds per question in a timed paper.
Common Mistake
The most common error — seen in JEE Main 2023 and countless board papers — is writing the answer directly as 67π.
Students apply the “cancel rule” sin−1(sinθ)=θ without checking whether θ is in the principal branch. Since 67π∈/[−2π,2π], this is flat-out wrong.
The answer 67π is not even in the range of sin−1, which only outputs values between −2π and 2π. If your answer is greater than 2π, stop and recheck.