Example 17 - Chapter 3 Class 11 Trigonometric Functions

Transcript

Example 17 Prove that sin⁡〖5x − 〖2sin 3x +〗⁡sin⁡x 〗/𝑐𝑜𝑠⁡〖5x − 𝑐𝑜𝑠⁡x 〗 = tan x Solving 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 cos 5x – cos x = – 2 sin ((𝟓𝒙 + 𝒙)/𝟐) sin((𝟓𝒙 − 𝒙)/𝟐) = – 2 sin (6𝑥/2) sin (4𝑥/2) = – 2 sin 3x sin 2x Solving L.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 = (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