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प्रश्न
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
योग
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उत्तर
L.H.S. = cos 3A = cos 90° = 0
R.H.S. = 4 cos3 A – 3 cos A
= 4 cos3 30° – 3 cos 30°
= `4 (sqrt3/2)^3 - 3(sqrt3/2)`
= `(4 xx 3sqrt(3))/8 - (3sqrt(3))/2`
= `(3sqrt(3))/2 - (3sqrt(3))/2 = 0`
L.H.S. = R.H.S.
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