prove that : sin10°sin50°sin70°=1/8
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sin10sin50sin70
= 1/2(2sin10sin50sin70)
And 2sinAsinB = cos (A-B) - cos (A+B)
So 1/2(2sin10sin50sin70) = 1/2(sin10{cos20 - cos120})
And cos 120 = cos (90 + 30) = 1/2
So 1/2 {sin10(cos20 +1/2)}
= 1/2sin10cos20 +1/4sin10
= 1/4(2sin10cos20) + 1/4sin10
As 2sinAcosB = sin(A+B)-sin(A-B)
= 1/4 (sin 30 -sin10) +1/4sin10
= 1/4sin30
= 1/8 = RHS
= 1/2(2sin10sin50sin70)
And 2sinAsinB = cos (A-B) - cos (A+B)
So 1/2(2sin10sin50sin70) = 1/2(sin10{cos20 - cos120})
And cos 120 = cos (90 + 30) = 1/2
So 1/2 {sin10(cos20 +1/2)}
= 1/2sin10cos20 +1/4sin10
= 1/4(2sin10cos20) + 1/4sin10
As 2sinAcosB = sin(A+B)-sin(A-B)
= 1/4 (sin 30 -sin10) +1/4sin10
= 1/4sin30
= 1/8 = RHS
Asif11111:
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