Example 17 - Chapter 3 Class 11 Trigonometric Functions (Term 2)

Transcript

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