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Example 16 - Prove cos 7x + cos 5x / sin 7x - sin 5x = cot x

Example 16 - Chapter 3 Class 11 Trigonometric Functions - Part 2

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Example 16 Prove that π‘π‘œπ‘ β‘γ€–7π‘₯ + π‘π‘œπ‘ β‘5π‘₯ γ€—/𝑠𝑖𝑛⁑〖7π‘₯ βˆ’ 𝑠𝑖𝑛⁑5π‘₯ γ€— = cot x Taking L.H.S. We solve cos 7x + cos 5x & sin 7x – sin 5x separately cos x + cos y = 2 cos (π‘₯ + 𝑦)/2 cos (π‘₯ βˆ’ 𝑦)/2 Putting x = 7x & y = 5x cos 7x + cos 5x = 2 cos ((7π‘₯ + 5π‘₯)/2) cos ((7π‘₯ βˆ’ 5π‘₯)/2) = 2 cos (12π‘₯/2) cos (2π‘₯/2) = 2 cos 6x cos x sin x – sin y = 2 cos (π‘₯ + 𝑦)/2 sin (π‘₯ βˆ’ 𝑦)/2 Putting x = 7x & y = 5x sin 7x – sin 5x = 2 cos ((7π‘₯ + 5π‘₯)/2) sin((7π‘₯ βˆ’ 5π‘₯)/2) = 2 cos (12π‘₯/2) sin (2π‘₯/2) = 2 cos 6x sin x Now γ€–π‘π‘œπ‘  〗⁑〖7π‘₯ + π‘π‘œπ‘ β‘5π‘₯ γ€—/𝑠𝑖𝑛⁑〖7π‘₯ βˆ’ 𝑠𝑖𝑛⁑5π‘₯ γ€— = (2 γ€– π‘π‘œπ‘  〗⁑〖6x π‘π‘œπ‘ β‘π‘₯ γ€—)/(2 π‘π‘œπ‘ β‘γ€– 6π‘₯ si𝑛⁑π‘₯ γ€— ) = π‘π‘œπ‘ β‘γ€– π‘₯γ€—/𝑠𝑖𝑛⁑〖 π‘₯γ€— = cot x = R.H.S. Hence L.H.S. = R.H.S. Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.