Prove that if 1/2 ≤ x ≤ 1, then cos −1 x + cos −1 [x/2+√(3 - 3x 2 )/2] = π/3

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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Question 7 Prove that if 1/2 ≀ x ≀ 1 then cosβˆ’1 x + cosβˆ’1 [π‘₯/2+√(3 βˆ’ 3π‘₯^2 )/2] = πœ‹/3 Solving LHS cosβˆ’1 x + cosβˆ’1 [π‘₯/2+√(3 βˆ’ 3π‘₯^2 )/2] = cosβˆ’1 x + cosβˆ’1 [π‘₯/2+√(3(1 βˆ’ π‘₯^2))/2] Let x = cos πœƒ i.e. cosβˆ’1 x = πœƒ Theory Since it is √(1βˆ’π‘₯^2 ) we can use x = sin πœƒ or x = cos πœƒ Since it is given cos-1 x, we use x = cos πœƒ = πœƒ + cosβˆ’1 [cosβ‘πœƒ/2+√(3(1βˆ’cos^2β‘γ€–πœƒ)γ€— )/2] Using sin ΞΈ = 1 – cos2 ΞΈ = πœƒ + cosβˆ’1 [cosβ‘πœƒ/2+√(3 sin^2β‘πœƒ )/2] = πœƒ + cosβˆ’1 [cosβ‘πœƒ/2+γ€–βˆš3 sinγ€—β‘πœƒ/2] = πœƒ + cosβˆ’1 [1/2 Γ—cosβ‘πœƒ+√3/2 Γ—sinβ‘πœƒ ] Theory We convert 1/2 Γ—cosβ‘πœƒ+√3/2 Γ—sinβ‘πœƒ into cos (x – y) Since cos (x – y) = cos x cos y + sin x sin y We use x = 60Β° and y = πœƒ = πœƒ + cosβˆ’1 [cos⁑〖60Β°γ€— Γ—cosβ‘πœƒ+sin⁑〖60Β°γ€—Γ—sinβ‘πœƒ ] = πœƒ + cosβˆ’1 [cosβ‘γ€–πœ‹/3γ€— Γ—cosβ‘πœƒ+sinβ‘γ€–πœ‹/3γ€—Γ—sinβ‘πœƒ ] Using cos (x – y) = cos x cos y + sin x sin y = πœƒ + cosβˆ’1 [cos⁑(πœ‹/3βˆ’πœƒ) ] = πœƒ + cosβˆ’1 [π‘π‘œπ‘ (πœ‹/3βˆ’πœƒ)] = πœƒ + πœ‹/3 βˆ’ πœƒ = πœ‹/3 = RHS Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.