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CBSE Class 12 Math 2009 Solved Paper

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Question : 1
Total: 29
Using principal value, evaluate the following: sin−1(sin
3π
5
)
Solution:  
As sin−1 (sin θ) = θ so sin−1(sin(
3π
5
)) =
3π
5
But
3π
5
 ∉ [−
π
2
,
π
2
]
So,
sin−1(sin(
3π
5
)) = sin−1(sin(π−
2π
5
))
= sin−1(sin
2π
5
)
=
2π
5
 ∊ [−
π
2
,
π
2
]
∴ Principal value is
2π
5
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