Question

Prove that 2sin x/ sin 3x = 1 - tan x * cot 3x

Answer 1

#.Prove that 2sin x/ sin 3x = 1 - tan x * cot 3x

2sin x/ sin 3x = 1 - tan x * cot 3x
or, 2sin x/ sin 3x + tan x * cot 3x=1
if we can prove that 2sin x/ sin 3x + tan x * cot 3x = 1 then its prove
so,


L.H.S
2sin(x)/sin(3x) + tan(x)cot(3x)
=2sin(x)/sin(3x) + (sin(x)/cos(x)) * (cos(3x)/sin(3x))
=2sin(x)/sin(3x) + (sin(x)cos(3x))/(sin(3x)cos(x))
=(sin(x)/sin(3x)) * (2 + cos(3x)/cos(x))
=(sin(x)/sin(3x)) * (2cos(x) + cos(3x)) / cos(x)
=[sin(x) * (2cos(x) + cos(3x))] / (sin(3x)cos(x))
=[sin(x) * (2cos(2x - x) + cos(2x + x)) ]/ (sin(3x)cos(x))
=[sin(x) * (2cos(2x)cos(x) + 2sin(2x)sin(x) + cos(2x)cos(x) - sin(2x)sin(x)) ]/ (sin(3x)cos(x))
=[sin(x) * (3cos(2x)cos(x) + sin(2x)sin(x)) ]/ (sin(3x)cos(x))
=[sin(x) * (3cos(x) * (cos(x)^2 - sin(x)^2) + 2sin(x)^2 * cos(x)) ]/ (sin(3x) * cos(x))
=[sin(x) * (3cos(x)^3 - 3sin(x)^2 * cos(x) + 2sin(x)^2 * cos(x)) ]/ (sin(3x) * cos(x))
=[sin(x) * cos(x) * (3cos(x)^2 - sin(x)^2) ]/ (sin(3x)cos(x))
=[sin(x) * (3 - 4sin(x)^2)] / sin(3x)
=[sin(x) * (3 - 4sin(x)^2)] / (sin(2x)cos(x) + sin(x)cos(2x))
=[sin(x) * (3 - 4sin(x)^2)] / [2sin(x)cos(x)^2 + sin(x)cos(x)^2 - sin(x)^3]
=(3 - 4sin(x)^2) / (2cos(x)^2 + cos(x)^2 - sin(x)^2)
=(3 - 4sin(x)^2) / (3cos(x)^2 - sin(x)^2)
=(3 - 4sin(x)^2) / (3 - 3sin(x)^2 - sin(x)^2)
=(3 - 4sin(x)^2) / (3 - 4sin(x)^2)
=1 R.H.S