How do you verify #(1-cos2x)/ tanx = sin2x#?

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Answer 1 · 2015-08-06T13:45:14

#\frac{1-cos 2x}{tan x} = \frac{2sin^2 x}{tan x} = 2sin x cos x = sin 2x #

#cos2x = cos^2 x-sin^2 x = 1 - 2sin^2 x#
# sin 2x = 2sin x cos x #
#tan x = \frac{sin x}{cos x} #

#\frac{1-cos 2x}{tan x} = \frac{2sin^2 x}{tan x} = 2sin x cos x = sin 2x #

Answer 2 · 2015-08-07T16:12:37

Starting with #(1-cos2x)/(tanx) = sin2x# and using the identity #(1-cos2x)/2 = sin^2x#:

#= (2sin^2x)/(tanx) = sin2x#

Of course, #tanx = (sinx)/(cosx)#:

#= (2cosxsin^cancel(2)x)/(cancel(sinx)) = sin2x#

and:

#2sinxcosx = sinxcosx + cosxsinx#
# = sin(x+x) = sin2x#:

#=> color(blue)(sin2x = sin2x)#