prove thatsin A sin (60 - A) sin (60 + A) = 1/4 sin 3A
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lhs = sinAsin(60-A)sin(60+A)
=sinA[sin60cosA-sinAcos60][sin60cosA+sinAcos60}
=sinA[(sin60cosA)²-(sinAcos60)²]
=sinA[sin²60cos²A- sin²Acos²60]
=sinA[3/4cos²A-sin²A*1/4]
=sinA*1/4[3cos²A-sin²A]
=sinA/4[3(1-sin²A)-sin²A]
=sinA/4[3-3sin²A-sin²A]
=1/4*sinA[3-4sin²A]
=1/4[3sinA-4sin³A]
=1/4*sin 3A
=rhs
=sinA[sin60cosA-sinAcos60][sin60cosA+sinAcos60}
=sinA[(sin60cosA)²-(sinAcos60)²]
=sinA[sin²60cos²A- sin²Acos²60]
=sinA[3/4cos²A-sin²A*1/4]
=sinA*1/4[3cos²A-sin²A]
=sinA/4[3(1-sin²A)-sin²A]
=sinA/4[3-3sin²A-sin²A]
=1/4*sinA[3-4sin²A]
=1/4[3sinA-4sin³A]
=1/4*sin 3A
=rhs
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