Using principal value, evaluate: sin−1(sin53π).Answer: 52πSolution53π∈(2π,π), which lies outside the range [−2π,2π] of sin−1.Use sin53π=sin(π−53π)=sin52π, and 52π∈[−2π,2π].sin−1(sin53π)=sin−1(sin52π)=52π52πCBSE Class 12 Board · 2009 · Subjective