prove that
sin 20 sin 40 sin 80 =under root 3/8...pls answer asap
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1/2 sin20 (2sin40 sin80)
2sinA sinB=cos (A-B)-cos (A+B)
1/2sin20 [cos(80-40)-cos (80+40)]
1/2sin20 [cos40-cos120]
1/2sin20 [cos40-cos (90+30)]
1/2sin20 [cos40+sin30]
1/2sin20cos40+1/2sin20sin30
1/4 [sin60-sin20]+1/4sin20
1/4sin60-1/4sin20+1/4sin20
1/4×root3/2
root3/8
hence proved
2sinA sinB=cos (A-B)-cos (A+B)
1/2sin20 [cos(80-40)-cos (80+40)]
1/2sin20 [cos40-cos120]
1/2sin20 [cos40-cos (90+30)]
1/2sin20 [cos40+sin30]
1/2sin20cos40+1/2sin20sin30
1/4 [sin60-sin20]+1/4sin20
1/4sin60-1/4sin20+1/4sin20
1/4×root3/2
root3/8
hence proved
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