Example 16 - Chapter 3 Class 11 Trigonometric Functions

Last updated at Dec. 16, 2024 by

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Example 16 Prove that ๐‘๐‘œ๐‘ โกใ€–7๐‘ฅ + ๐‘๐‘œ๐‘ โก5๐‘ฅ ใ€—/๐‘ ๐‘–๐‘›โกใ€–7๐‘ฅ โˆ’ ๐‘ ๐‘–๐‘›โก5๐‘ฅ ใ€— = cot x Solving L.H.S. We solve cos 7x + cos 5x & sin 7x โ€“ sin 5x separately cos 7x + cos 5x = 2 cos ((๐Ÿ•๐’™ + ๐Ÿ“๐’™)/๐Ÿ) cos ((๐Ÿ•๐’™ โˆ’ ๐Ÿ“๐’™)/๐Ÿ) = 2 cos (12๐‘ฅ/2) cos (2๐‘ฅ/2) = 2 cos 6x cos x sin 7x โ€“ sin 5x = 2 cos ((๐Ÿ•๐’™ + ๐Ÿ“๐’™)/๐Ÿ) sin((๐Ÿ•๐’™ โˆ’ ๐Ÿ“๐’™)/๐Ÿ) = 2 cos (12๐‘ฅ/2) sin (2๐‘ฅ/2) = 2 cos 6x sin x Now ใ€–๐‘๐‘œ๐‘  ใ€—โกใ€–7๐‘ฅ + ๐‘๐‘œ๐‘ โก5๐‘ฅ ใ€—/๐‘ ๐‘–๐‘›โกใ€–7๐‘ฅ โˆ’ ๐‘ ๐‘–๐‘›โก5๐‘ฅ ใ€— = (๐Ÿ ใ€– ๐’„๐’๐’” ใ€—โกใ€–๐Ÿ”๐’™ ๐’„๐’๐’”โก๐’™ ใ€—)/(๐Ÿ ๐’„๐’๐’”โกใ€– ๐Ÿ”๐’™ ๐’”๐’Š๐’โก๐’™ ใ€— ) = ๐‘๐‘œ๐‘ โกใ€– ๐‘ฅใ€—/๐‘ ๐‘–๐‘›โกใ€– ๐‘ฅใ€— = cot x = 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo