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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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