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Question
Find the value of the following:
`cos^(-1) (cos (13pi)/6)`
Solution
We know that cos−1 (cos x) = x if `x in [0,pi]`, which is the principal value branch of cos −1x.
Here, `(13pi)/6 !in [0 ,pi]`
Now `cos^(-1) (cos (13pi)/6)` can be written as
`cos^(-1) (cos (13pi)/6) `
`= cos^(-1) [cos(2pi + pi/6)]`
` = cos^(-1) [cos(pi/6)], " where " pi/6 in [0, pi]`
`:. cos^(-1) (cos (13pi)/6) `
`= cos^(-1)[cos (pi/6)] `
`= pi/6`
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