Ex 3.3, 13 - Chapter 3 Class 11 Trigonometric Functions

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Ex 3.3, 13 Prove that cos2 2𝑥 – cos2 6𝑥 = sin⁡4𝑥 sin⁡8𝑥 Solving L.H.S. cos2 2x – cos2 6x = (cos 2x + cos 6x) (cos 2x – cos 6x) Lets calculate (cos 2x + cos 6x) and (cos 2x – cos 6x) separately cos 2x + cos 6x = 2 cos ((2x+6x)/2) cos ((2x−6x)/2) = 2 cos (8𝑥/2) cos ((−4𝑥)/2) = 2 cos 4x cos (-2x) cos 2x – cos 6x = – 2 sin ((2x+6x)/2) sin((2x−6x)/2) = – 2 sin (8𝑥/2) sin ((−4𝑥)/2) = – 2 sin 4x sin (–2x) Hence 𝒄𝒐𝒔𝟐 𝟐𝒙 – 𝒄𝒐𝒔𝟐 𝟔𝒙 = (cos⁡2𝑥 + cos⁡6𝑥) (cos⁡2𝑥 – 6𝑥) = (2 cos⁡〖4𝑥 𝒄𝒐𝒔⁡〖(−𝟐𝒙)〗 〗 ) (−2 sin⁡4𝑥 (𝒔𝒊𝒏⁡〖(−𝟐𝒙)〗 )) = (2 cos⁡〖4𝑥 𝒄𝒐𝒔⁡〖(𝟐𝒙)〗 〗 ) (−2 sin⁡4𝑥 (〖−𝒔𝒊𝒏〗⁡〖(𝟐𝒙)〗 )) = (2 cos⁡〖4𝑥 cos⁡〖(2𝑥)〗 〗 ) (2 sin⁡4𝑥 sin⁡〖(2𝑥)〗 ) = (𝟐 𝒔𝒊𝒏⁡𝟒𝒙 𝒄𝒐𝒔⁡𝟒𝒙) (𝟐 𝒔𝒊𝒏⁡𝟐𝒙 𝒄𝒐𝒔⁡𝟐𝒙) We know that sin 2x = 2 sin x cos x Putting 4x instead of x And putting 2x instead of x = sin⁡8𝑥 sin⁡4𝑥 = R.H.S. Hence, L.H.S. = R.H.S. Hence proved