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प्रश्न
Evaluate `cos[pi/6 + cos^-1 (- sqrt(3)/2)]`
उत्तर
Let `cos^-1 ((-sqrt(3))/2)` = y
∴ cos y = `(-sqrt(3))/2`
= `- cos (pi/6)`
= `cos (pi - pi/6)`
= `cos (5pi)/6`
The principal value branch of cos−1 is [0, π] and `0 ≤ (5pi)/6 ≤ pi`.
∴ y = `(5pi)/6`
∴ `cos^-1 ((-sqrt(3))/2) = (5pi)/6`
∴ `pi/6 + cos^-1 ((-sqrt(3))/3)`
= `pi/6 + (5pi)/6`
= π
∴ `cos[pi/6 + cos^-1 (- sqrt(3)/2)]` = cos π = −1
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