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Ex 3.3, 13 - Prove that cos2 2x - cos2 6x = sin⁡4x sin⁡8x - (x + y) formula

Ex 3.3, 13 - Chapter 3 Class 11 Trigonometric Functions - Part 2

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Ex 3.3, 13 Prove that cos2 2𝑥 – cos2 6𝑥 = sin⁡4𝑥 sin⁡8𝑥 Solving L.H.S. cos2 2x – cos2 6x Using a2 – b2 = (a + b) (a - b) = (cos 2x + cos 6x) (cos 2x – cos 6x) Lets calculate (cos 2x + cos 6x) and (cos 2x + cos 6x) separately Hence cos2 2𝑥 – cos2 6𝑥 = (cos⁡2𝑥 + cos⁡6𝑥) (cos⁡2𝑥 – 6𝑥) = (2 cos⁡〖4𝑥 cos⁡〖(−2𝑥)〗 〗 ) (−2 sin⁡4𝑥 (sin⁡〖(−2𝑥)〗 )) = (2 cos⁡〖4𝑥 cos⁡〖(2𝑥)〗 〗 ) (−2 sin⁡4𝑥 (〖−sin〗⁡〖(2𝑥)〗 )) = (2 cos⁡〖4𝑥 cos⁡〖(2𝑥)〗 〗 ) (2 sin⁡4𝑥 sin⁡〖(2𝑥)〗 ) = (2 sin⁡4𝑥 cos⁡4𝑥) (2 sin⁡2𝑥 cos⁡2𝑥) = sin⁡8𝑥 sin⁡4𝑥 = R.H.S. Hence, L.H.S. = R.H.S. Hence proved

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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.