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Question
The value of `sin^-1 [cos((33pi)/5)]` is ______.
Options
`(3pi)/5`
`(-7pi)/5`
`pi/10`
`(-pi)/10`
Solution
The value of `sin^-1 [cos((33pi)/5)]` is `(-pi)/10`.
Explanation:
`sin^-1 [cos((33pi)/5)] = sin^-1[cos(6pi + (3pi)/5)]`
= `sin^-1[cos (3pi)/5]` .......[∵ cos(2nπ + x) = cos x]
= `sin^-1 [cos(pi/2 + pi/10)]`
= `sin^-1[-sin (pi/10)]` ......`[because cos(pi/2 + theta) = - sin theta]`
= `sin^-1[sin((-pi)/10)]`
= `(-pi)/10`
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