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Question
Prove the following:
`sin^-1(-1/2) + cos^-1(-sqrt(3)/2) = cos^-1(-1/2)`
Solution
Let `sin^-1(-1/2) = α, "where" - pi/(2) ≤ α ≤ pi/(2)`
∴ sin α = `-1/2 = -sin pi/(6)`
∴ sin α = `sin(-pi/6)` ...[∵ sin(– θ) = – sin θ]
∴ α = `- pi/(6) ...[∵ - pi/(2) ≤ - pi/(6) ≤ pi/(2)]`
∴ `sin^-1(-1/2) = - pi/(6)` ...(1)
Let `cos^-1(- sqrt(3)/2)` = β, where 0 ≤ β ≤ π
∴ cos β = `- sqrt(3)/(2) = - cos pi/(6)`
∴ cos β = `cos(pi - pi/6)` ...[∵ cos(π – θ) = – cos θ]
∴ cos β = `cos (5pi)/(6)`
∴ β = `(5pi)/(6) ...[∵ 0 ≤ (5pi)/(6) ≤ pi]`
∴ `cos^-1(- sqrt(3)/2) = (5pi)/(6)` ...(2)
Let `cos^-1(- 1/2)` = ϒ, where 0 ≤ ϒ ≤ π
∴ cos ϒ = `-(1)/(2) = - cos pi/(3)`
∴ cos ϒ = `cos(pi - pi/3)` ...[∵ cos(π – θ) = – cos θ]
∴ cos ϒ = `cos (2pi)/(3)`
∴ ϒ = `(2pi)/(3) ...[∵ 0 ≤ (2pi)/(3) ≤ pi]`
∴ `cos^-1(- 1/2) = (2pi)/(3)` ...(3)
L.H.S. = `sin^-1(- 1/2) + cos^-1(- sqrt(3)/2)`
= `- pi/(6) + (5pi)/(6)` ...[By (1) and (2)]
= `(4pi)/(6) = (2pi)/(3)`
= `cos^-1(- 1/2)` ...[By (3)]
= R.H.S.
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