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Example 17 - Prove sin 5x - 2 sin 3x + sin x / cos 5x - cos x

Example 17 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Example 17 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Example 17 - Chapter 3 Class 11 Trigonometric Functions - Part 4

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Example 17 Prove that sin⁡〖5x − 〖2sin 3x +〗⁡sin⁡x 〗/𝑐𝑜𝑠⁡〖5x − 𝑐𝑜𝑠⁡x 〗 = tan x Taking L.H.S. sin⁡〖5x + 〖sin x − 〗⁡2sin⁡3x 〗/𝑐𝑜𝑠⁡〖5x − 𝑐𝑜𝑠⁡x 〗 = 〖(sin〗⁡〖5x + 〖sin x) − 〗⁡〖2 sin〗⁡3x 〗/𝑐𝑜𝑠⁡〖5x − 𝑐𝑜𝑠⁡x 〗 Solving numerator and denominator separately sin 5x + sin x = 2 sin ((5𝑥 + 𝑥)/2) cos ((5𝑥 − 𝑥)/2) = 2 sin (6𝑥/2) cos (4𝑥/2) = 2 sin 3x cos 2x sin 5x + sin x = 2 sin ((5𝑥 + 𝑥)/2) cos ((5𝑥 − 𝑥)/2) = 2 sin (6𝑥/2) cos (4𝑥/2) = 2 sin 3x cos 2x sin x + sin y = 2 sin (𝑥 + 𝑦)/2 cos (𝑥 − 𝑦)/2 Putting x = 5x & y = x sin x + sin y = 2 sin (𝑥 + 𝑦)/2 cos (𝑥 − 𝑦)/2 Putting x = 5x & y = x cos x – cos y = –2 sin (𝑥 + 𝑦)/2 sin (𝑥 − 𝑦)/2 Putting x = 5x & y = x Solving R.H.S 𝐬𝐢𝐧⁡〖𝟓𝐱 + 〖𝐬𝐢𝐧 𝐱 − 〗⁡2sin⁡3x 〗/𝒄𝒐𝒔⁡〖𝟓𝐱 − 𝒄𝒐𝒔⁡𝐱 〗 Putting values = (2 sin⁡3𝑥 cos⁡2𝑥 − 2 sin⁡3𝑥)/(−2 sin⁡〖3𝑥 sin⁡2𝑥 〗 ) = (2 sin⁡3𝑥 (cos⁡〖2𝑥 − 1)〗)/(−2 sin⁡〖3𝑥 sin⁡2𝑥 〗 ) = ( (cos⁡〖2𝑥 − 1)〗)/(−sin⁡2𝑥 ) = ( −(cos⁡〖2𝑥 −1) 〗)/sin⁡2𝑥 = (〖1 − 𝐜𝐨𝐬〗⁡𝟐𝒙 )/𝒔𝒊𝒏⁡𝟐𝒙 "Using cos 2x = 1 – 2sin2 x" "& sin 2x = 2 cos x sin x" = (1 − (𝟏 − 𝟐 𝐬𝐢𝐧𝟐⁡𝒙 ) )/(𝟐 𝒄𝒐𝒔⁡𝒙 𝒔𝒊𝒏⁡𝒙 ) = (1 − 1 + 2 sin2⁡𝑥)/(2 cos⁡〖𝑥 〗 sin⁡𝑥 ) = (0 + 2 sin2⁡𝑥)/(2 cos⁡〖𝑥 〗 sin⁡𝑥 ) = (2 sin2⁡𝑥)/(2 cos⁡〖𝑥 〗 sin⁡𝑥 ) = sin⁡〖𝑥 〗/cos⁡〖𝑥 〗 = tan 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.