Question 7
Prove that if 1/2 ≤ x ≤ 1 then cos−1 x + cos−1 [𝑥/2+√(3 − 3𝑥^2 )/2] = 𝜋/3
Solving LHS
cos−1 x + cos−1 [𝑥/2+√(3 − 3𝑥^2 )/2]
= cos−1 x + cos−1 [𝑥/2+√(3(1 − 𝑥^2))/2]
Let x = cos 𝜃
i.e. cos−1 x = 𝜃
Theory
Since it is √(1−𝑥^2 )
we can use x = sin 𝜃 or x = cos 𝜃
Since it is given cos-1 x,
we use x = cos 𝜃
= 𝜃 + cos−1 [cos𝜃/2+√(3(1−cos^2〖𝜃)〗 )/2]
Using sin θ = 1 – cos2 θ
= 𝜃 + cos−1 [cos𝜃/2+√(3 sin^2𝜃 )/2]
= 𝜃 + cos−1 [cos𝜃/2+〖√3 sin〗𝜃/2]
= 𝜃 + cos−1 [1/2 ×cos𝜃+√3/2 ×sin𝜃 ]
Theory
We convert 1/2 ×cos𝜃+√3/2 ×sin𝜃
into cos (x – y)
Since cos (x – y) = cos x cos y + sin x sin y
We use x = 60° and y = 𝜃
= 𝜃 + cos−1 [cos〖60°〗 ×cos𝜃+sin〖60°〗×sin𝜃 ]
= 𝜃 + cos−1 [cos〖𝜋/3〗 ×cos𝜃+sin〖𝜋/3〗×sin𝜃 ]
Using cos (x – y) = cos x cos y + sin x sin y
= 𝜃 + cos−1 [cos(𝜋/3−𝜃) ]
= 𝜃 + cos−1 [𝑐𝑜𝑠(𝜋/3−𝜃)]
= 𝜃 + 𝜋/3 − 𝜃
= 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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