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


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Ex 3.3, 16 Prove that π‘π‘œπ‘ β‘γ€–9π‘₯ βˆ’γ€– π‘π‘œπ‘ γ€—β‘5π‘₯ γ€—/(𝑠𝑖𝑛 17π‘₯ βˆ’ 𝑠𝑖𝑛⁑3π‘₯ ) =βˆ’π‘ π‘–π‘›β‘γ€–2π‘₯ γ€—/π‘π‘œπ‘ β‘10π‘₯ Solving L.H.S π‘π‘œπ‘ β‘γ€–9π‘₯ βˆ’γ€– π‘π‘œπ‘ γ€—β‘5π‘₯ γ€—/(𝑠𝑖𝑛 17π‘₯ βˆ’ 𝑠𝑖𝑛⁑3π‘₯ ) We solve cos 9x – cos 5x & sin 17x – sin 3x seperately cos 9x – cos 5x = – 2 sin ((9x+5x)/2) sin((9xβˆ’5x)/2) = – 2 sin (14π‘₯/2) sin (4π‘₯/2) = – 2 sin 7x sin (2x) sin 17x – sin 3x = 2 cos ((17x+3x)/2) sin((17xβˆ’3x)/2) = 2 cos (20π‘₯/2) sin (14π‘₯/2) = 2 cos 10x sin 7x Now, π‘π‘œπ‘ β‘γ€–9π‘₯ βˆ’γ€– π‘π‘œπ‘ γ€—β‘5π‘₯ γ€—/(𝑠𝑖𝑛 17π‘₯ βˆ’ 𝑠𝑖𝑛⁑3π‘₯ ) = (βˆ’πŸ 〖𝐬𝐒𝐧 〗⁑〖(πŸ•π±)〖𝐬𝐒𝐧 〗⁑〖(𝟐𝐱)γ€— γ€—)/(𝟐 𝒄𝒐𝒔⁑〖(𝟏𝟎𝐱)𝐬𝐒𝐧⁑〖 (πŸ•π±)γ€— γ€— ) = γ€–βˆ’sin〗⁑〖(2x)γ€—/π‘π‘œπ‘ β‘γ€–(10x)γ€— = R.H.S So, 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.