Sin20.sin40.sin60.sin80=1/16
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Posted by Kshitiz Chaudhary 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
since,sin60=√ 3/2
= √ 3/2( sin20sin40sin80)
=√ 3/2( sin20sin80sin40)
=√ 3/4 [(2sin20sin40)sin80]
on applying [cos(A-B)-cos(A+B) = 2sinAsinB]
we get,
= √ 3/4 (cos20-cos60)sin80 [since,cos(-a)=cosa]
= √ 3/4(cos20sin80-cos60sin80)
= √ 3/8(2sin80cos20-sin80)
= √ 3/8(sin100+sin60-sin80)
= √ 3/8( √ 3/2+sin100-sin80 )
= √ 3/8( √ 3/2+sin(180-80)-sin80 )
= √ 3/8( √ 3/2+sin80-sin80 ) [since,sin(180-a)=sina]
= √ 3/8( √ 3/2)
= 3/16
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