How do you verify #-cot(x/2) = (sin2x + sinx) / (cos2x - cosx)#?

1 Answer
Nghi N.
Nov 2, 2015

Verify trig expression
#-cot (x/2) = (sin 2x + sin x)/(cos 2x - cos x)#

Explanation:

Use the trig identities to transform the right side:
sin 2x = sin x.cos x
#cos 2x = 2cos^2 x - 1#.
Transform the right side:
RS = #(sin 2x + sin x)/(cos 2x - cos x) = (sin x(2cos x + 1))/(2cos^2 x - cos x - 1# =

Factor the trinomial #(2cos^2 x - cos x - 1).#
Since (a + b + c = 0), use the Shortcut, the 2 factors are (cos x - 1) and (2cos x + 1). Finally,

#RS = (sin x(2cos x + 1))/((cos x - 1)(2cos x + 1)) = (sin x)/(cos x - 1)#
Since:
#sin x = 2sin (x/2).cos (x/2) and#
#(cos x - 1) = -2sin^2 (x/2)#, therefor
#RS = (2sin (x/2)(cos x/2))/(-2sin^2 (x/2)) = #
#= - cos (x/2)/(sin (x/2)) = -cot x/2#

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