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Question
Find the value of the following:
`cos^(-1) (cos (13pi)/6)`
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Solution
`cos^(-1) (cos (13pi)/6) ≠ (13 pi)/6` as the range of the principal value branch of cos−1 is [0, π].
So, `cos^(-1) (cos (13pi)/6)`
= `cos^(-1) [cos(2pi + pi/6)]`
= `cos^(-1) [cos(pi/6)]`
= `pi/6`
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