Sin20°sin40° sin60°. Sin80° =3/10

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Solution

sin20. sin 40. sin60. sin80

=> sin60[sin20.sin40.sin80]

=>√3/2[sin20.sin(60-20).sin(60+20)]

=>√3/2[sin 3(20)/4]

=>√3/2[sin 60/4]

=>√3/2[√3/24]

=>√3/2√3/8

=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

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Similar questions

$s i n 20^{\circ} s i n 40^{\circ} s i n 60^{\circ} s i n 80^{\circ}$ =

sin 20 \cdot sin 40 \cdot sin 60. s i n 80

sin 20 \times sin 40 \times sin 60 \times sin 80

sin 20^{\circ} sin 40^{\circ} sin 60^{\circ} sin 80^{\circ} = \frac{3}{16}

sin 20

sin 40

sin 60

sin 80 = \frac{3}{16}

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