# sin 20 sin 40 sin 60 sin 80=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 hope you get it regards

• 144

sin(20) sin(40) sin (60) sin (80)substitute sin(60) = √3 /2√3/2 [ sin(20) sin(40) sin(80) ]= (√3/2) sin(20) [ sin(40) sin(80) ]use the formula sin A sin B = (1/2) [ cos(A - B) - cos(A + B) ]= √3/2 sin(20) (1/2)[ cos(40) - cos(120) ]= √3/4 sin(20) [ cos(40) + cos(60) ]= √3/4 sin(20) [ cos(40) + 1/2 ]= √3/4 sin(20)cos(40) + (√3/8) sin(20)use the formula sin A cos B = 1/2 [ sin(A + B) + sin(A - B) ]= (√3/4)(1/2) [ sin(60) + sin(-20) ]+ (√3/8)sin(20)= (√3/8) [ (√3 / 2) - sin(20) ]+ (√3/8)sin(20)= 3/16 - (√3/8)sin(20) + (√3/8)sin(20)= 3/16

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OR BETTER USE THIS WAY :

=sin(20) sin(40) sin (60) sin (80)=sin(60) sin(20) sin(40) sin (80)=√3/2 [sin(20) sin(40) sin(80)] -------------- (1) {since sin(60) = √3 /2}Taking sin(20) sin(40) sin(80)= sin (20) [ sin(40) sin (80) ]= sin (20) [sin (60-20) sin (60+20)]= sin (20) [sin^2 (60) - sin^2 (20)]= sin (20) [(√3/2)^2 - sin^2 (20)]= sin (20) [3/4 - sin^2 (20)]= 3/4 sin (20) - sin^3 (20)Taking LCM= 3sin (20) - 4sin^3 (20)/4= 1/4 [3sin (20) - 4sin^3 (20)]= 1/4 sin 3(20)= 1/4 sin (60)= 1/4 (√3/2)= √3/8 --------------------- (2)substitute equation (2) in equation (1)= (√3/2)(√3/8)= 3/16 ------------------------ RHS

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