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