Question 7
Prove that if 1/2 β€ x β€ 1 then cosβ1 x + cosβ1 [π₯/2+β(3 β 3π₯^2 )/2] = π/3
Solving LHS
cosβ1 x + cosβ1 [π₯/2+β(3 β 3π₯^2 )/2]
= cosβ1 x + cosβ1 [π₯/2+β(3(1 β π₯^2))/2]
Let x = cos π
i.e. cosβ1 x = π
Theory
Since it is β(1βπ₯^2 )
we can use x = sin π or x = cos π
Since it is given cos-1 x,
we use x = cos π
= π + cosβ1 [cosβ‘π/2+β(3(1βcos^2β‘γπ)γ )/2]
Using sin ΞΈ = 1 β cos2 ΞΈ
= π + cosβ1 [cosβ‘π/2+β(3 sin^2β‘π )/2]
= π + cosβ1 [cosβ‘π/2+γβ3 sinγβ‘π/2]
= π + cosβ1 [1/2 Γcosβ‘π+β3/2 Γsinβ‘π ]
Theory
We convert 1/2 Γcosβ‘π+β3/2 Γsinβ‘π
into cos (x β y)
Since cos (x β y) = cos x cos y + sin x sin y
We use x = 60Β° and y = π
= π + cosβ1 [cosβ‘γ60Β°γ Γcosβ‘π+sinβ‘γ60Β°γΓsinβ‘π ]
= π + cosβ1 [cosβ‘γπ/3γ Γcosβ‘π+sinβ‘γπ/3γΓsinβ‘π ]
Using cos (x β y) = cos x cos y + sin x sin y
= π + cosβ1 [cosβ‘(π/3βπ) ]
= π + cosβ1 [πππ (π/3βπ)]
= π + π/3 β π
= π/3
= RHS
Hence proved

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.