Question 2: Find the principal value of cos¯¹(√3/2)
Class 12 - Math - Inverse Trigonometric Functions
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let cos-¹(-√3/2) = ∅
cos∅ = -√3/2
since the range of the principal value branch of cos-¹(x) is [0,π] ,therefore 0≤∅≤π
Now cos∅ = -√3/2 = -cosπ/6 = cos(π-π/6)
=> cos5π/6
therefore ∅ = 5π/6
cos∅ = -√3/2
since the range of the principal value branch of cos-¹(x) is [0,π] ,therefore 0≤∅≤π
Now cos∅ = -√3/2 = -cosπ/6 = cos(π-π/6)
=> cos5π/6
therefore ∅ = 5π/6
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