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Prove that if 1/2 ≀ x ≀ 1, then cos βˆ’1 x + cos βˆ’1 [x/2+√(3 - 3x 2 )/2] = Ο€/3

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

You can download the question paper hereΒ  https://www.teachoo.com/cbse/sample-papers/


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Question 7 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 7 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3

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Transcript

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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.