Ex 3.3, 20 - Chapter 3 Class 11 Trigonometric Functions

Last updated at August 8, 2026 by

Ex 3.3, 20 - Prove sin x - sin 3x / sin2 x - cos2 x = 2 sin x - Ex 3.3

part 2 - Ex 3.3, 20 - Ex 3.3 - Serial order wise - Chapter 3 Class 11 Trigonometric Functions

Take a fresh quiz. Then take another.
Every attempt is a new AI-adaptive Teachoo quiz with 10 questions, selected from your answers, mistakes, and progress.
Remove Ads

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.