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Question
`"cos"^-1 ("cos" (7pi)/6)` = _________.
Options
`(7pi)/6`
`(5pi)/6`
`pi/6`
`(3pi)/2`
Solution
`"cos"^-1 ("cos" (7pi)/6) = (5pi)/6`.
Explanation:
You can write `(7pi)/6 as (pi + pi/6)`
Thus, we can clearly see the angle falls in the third quadrant, and the cosine value in this quadrant is always negative.
Hence, `cos(pi + pi/6) = -cos(pi/6)`
coming back to the question
`cos^(-1)[cos((7pi)/6)] = cos^(-1)[-cos(pi/6)]`
= `pi - cos^(-1)[cos(pi/6)]`
= `pi - pi/6`
= `(5pi)/6`
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