CBSE Class 12 Math 2009 Solved Paper

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Question : 1
Total: 29
Using principal value, evaluate the following: sin1(sin
3π
5
)

Solution:  
As sin1 (sin θ) = θ so sin1(sin(
3π
5
)
)
=
3π
5

But
3π
5
[
π
2
,
π
2
]

So,
sin1(sin(
3π
5
)
)
= sin1(sin(π
2π
5
)
)

= sin1(sin
2π
5
)

=
2π
5
[
π
2
,
π
2
]

∴ Principal value is
2π
5
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