Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Ex 3.3, 19 Prove that γ€–sin x +〗⁑sin⁑3x /(π‘π‘œπ‘ β‘x + π‘π‘œπ‘ β‘3x ) = tan 2x Solving L.H.S. γ€–sin x +〗⁑sin⁑3x /(π‘π‘œπ‘ β‘x + π‘π‘œπ‘ β‘3x ) We solve sin x + sin 3x & cos x + cos 3x seperately sin x + sin 3x = 2 sin ((x+3x)/2) cos ((xβˆ’3x)/2) = 2 sin (4π‘₯/2) cos ((βˆ’2π‘₯)/2) = 2 sin 2x cos (–x) cos x + cos 3x = 2 cos ((x+3x)/2) cos ((5xβˆ’3x)/2) = 2 cos (4π‘₯/2) cos ((βˆ’2π‘₯)/2) = 2 cos 2x cos (–x) Now 𝑠𝑖𝑛⁑〖π‘₯ + 𝑠𝑖𝑛⁑3π‘₯ γ€—/π‘π‘œπ‘ β‘γ€–π‘₯ + π‘π‘œπ‘ β‘3π‘₯ γ€— = (𝟐 γ€– π’”π’Šπ’ γ€—β‘γ€–πŸπ’™ 𝒄𝒐𝒔⁑〖(βˆ’π’™)γ€— γ€—)/(𝟐 𝒄𝒐𝒔⁑〖 πŸπ’™ 𝒄𝒐𝒔⁑〖(βˆ’π’™)γ€— γ€— ) = 𝑠𝑖𝑛⁑〖 2xγ€—/cos⁑〖 2xγ€— = tan 2x = R.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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.