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prove that sin 20*sin40+sin60*sin80=3/16 how does it comes
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
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
Since sin 60= root3/2 => root3/2 [ sin 20 . Sin 40 . Sin 80]=> root3/2 [ sin 20 . Sin (60-20) . Sin (60+ 20) ]We know that sin x . Sin (60-x) . Sin (60+ x) = ¼ sin 3xTherefore, root3/2 [ ¼ sin60]Root 3/2 . ¼ . root 3/2 => 3/16.
Hello student,The problem posted by you needs a small correction, it should besin20 * sin40 * sin60 * sin80 = 3/16I hope the attached solution image will solve your doubts.Thank YouAll the best for your exams.
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