Question

If 4 cos² A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin³ A

Answer 1

4 cos² A − 3 = 0

`cos A = sqrt(3)/2`

We know `cos 30^circ = sqrt(3)/2`

So, A = 30°

L.H.S. = sin 3A = sin 90° = 1

R.H.S. = 3 sin A – 4 sin³ A

= 3 sin 30° – 4 sin³ 30°

= `3 xx 1/2 - 4 xx (1/2)^3`   ...{∵ sin 30° = `1/2`}

= `3/2 - 4 xx 1/8`

= `3 /2 - 1/2`

= `2/2`

= 1

L.H.S. = R.H.S.