Ex 3.3

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Ex 3.3, 13 Prove that cos2 2π₯ β cos2 6π₯ = sinβ‘4π₯ sinβ‘8π₯ Solving L.H.S. cos2 2x β cos2 6x = (cos 2x + cos 6x) (cos 2x β cos 6x) Lets calculate (cos 2x + cos 6x) and (cos 2x β cos 6x) separately cos 2x + cos 6x = 2 cos ((2x+6x)/2) cos ((2xβ6x)/2) = 2 cos (8π₯/2) cos ((β4π₯)/2) = 2 cos 4x cos (-2x) cos 2x β cos 6x = β 2 sin ((2x+6x)/2) sin((2xβ6x)/2) = β 2 sin (8π₯/2) sin ((β4π₯)/2) = β 2 sin 4x sin (β2x) Hence ππππ ππ β ππππ ππ = (cosβ‘2π₯ + cosβ‘6π₯) (cosβ‘2π₯ β 6π₯) = (2 cosβ‘γ4π₯ πππβ‘γ(βππ)γ γ ) (β2 sinβ‘4π₯ (πππβ‘γ(βππ)γ )) = (2 cosβ‘γ4π₯ πππβ‘γ(ππ)γ γ ) (β2 sinβ‘4π₯ (γβπππγβ‘γ(ππ)γ )) = (2 cosβ‘γ4π₯ cosβ‘γ(2π₯)γ γ ) (2 sinβ‘4π₯ sinβ‘γ(2π₯)γ ) = (π πππβ‘ππ πππβ‘ππ) (π πππβ‘ππ πππβ‘ππ) We know that sin 2x = 2 sin x cos x Putting 4x instead of x And putting 2x instead of x = sinβ‘8π₯ sinβ‘4π₯ = R.H.S. Hence, L.H.S. = R.H.S. Hence proved