Ex 3.3, 20 - Chapter 3 Class 11 Trigonometric Functions

Transcript

Ex 3.3, 20 Prove that 〖sin x − 〗⁡sin⁡3x /(sin2⁡x − cos2⁡x ) = 2 sin x Solving L.H.S. 〖sin x −〗⁡sin⁡3x /(sin2⁡x − cos2⁡x ) We solve sin x – sin 3x & sin2 x – cos2 x separately sin x – sin 3x = 2 cos ((x + 3x)/2) sin((x − 3x)/2) = 2 cos (4𝑥/2) sin ((−2𝑥)/2) = 2 cos 2x sin (–x) sin2 x – cos2 x = – cos 2x We know that cos 2x = cos2 x – sin2 x cos 2x = – ( sin2 x – cos2 x) Thus, −( sin2 x – cos2 x) = cos 2x − ( sin2 x – cos2 x) = – cos 2x Now, sin⁡〖𝑥 − sin⁡3𝑥 〗/sin2⁡〖𝑥 − cos2⁡𝑥 〗 = 〖𝟐 𝒄𝒐𝒔 〗⁡〖𝟐𝒙 〖 𝒔𝒊𝒏〗⁡〖(−𝒙)〗 〗/〖−𝒄𝒐𝒔〗⁡𝟐𝒙 = 〖2 𝑐𝑜𝑠〗⁡〖2𝑥 〖(− sin〗⁡𝑥)〗/〖−𝑐𝑜𝑠〗⁡2𝑥 = 2sin x = R.H.S. Hence L.H.S. = R.H.S. Hence proved