prove that : sin20 sin40 sin60 sin80 = 3/16
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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
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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