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प्रश्न
Prove that:
sin A sin(60° + A) sin(60° – A) = `1/4` sin 3A
बेरीज
उत्तर
LHS = sin A sin(60° + A) sin(60° – A)
= sin A [sin2 60° - sin2A]
[∵ sin (A + B) sin (A - B) = sin2A - sin2B]
= sin A `[(sqrt3/2)^2 - sin^2"A"]`
= sin A `[3/4 - sin^2"A"]`
= sin A `[(3 - 4 sin^2 "A")/4]`
`= 1/4` [3 sin A - 4 sin3A]
`= 1/4` sin 3A [∵ sin 3A = 3 sin A - 4 sin3A]
= RHS.
Hence proved.
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