Question

sin20.sin40.sin60.sin80 is??

Answer 1

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