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Transcript

Ex 3.3, 15 Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x) Solving L.H.S. cot 4x (sin 5x + sin 3x) = cot 4x Γ— [ 2 sin ((5π‘₯ + 3π‘₯)/2) cos ((5π‘₯ βˆ’ 3π‘₯)/2) ] = cot 4x Γ— [ 2sin (8π‘₯/2) cos (2π‘₯/2)] = 2 cot 4x sin 4x cos x = 2 π‘π‘œπ‘ β‘4π‘₯/𝑠𝑖𝑛⁑4π‘₯ Γ— sin 4x Γ— cos x = 2 cos 4x cos x Solving R.H.S. cot x (sin 5x – sin 3x) = cot x ( 2 cos (5π‘₯ + 3π‘₯)/2 sin (5π‘₯ βˆ’ 3π‘₯)/2 ) = cot x ( 2 cos (8π‘₯/2) sin (2π‘₯/2)] = cot x ( 2 cos 4x sin x) = 2 cos 4x sin x cot x = 2 cos 4x sin x Γ— π‘π‘œπ‘ β‘π‘₯/sin⁑π‘₯ = 2 cos 4x cos x = L.H.S Hence L.H.S. = R.H.S. Hence Proved

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Davneet Singh

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.