How do you verify (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x(sinx)(tanxcosx−cotxcosx)=1−2cos2x?

2 Answers
Ho Kae Boon
May 1, 2018

Pretty sure the question is (sinx)(tanxcosx-cotxcos x)=1-2cos^2x(sinx)(tanxcosx−cotxcosx)=1−2cos2x ,or else it will be not provable.

Explanation:

Some basic knowledge to begin with:
1. sin^2x+cos^2x=1sin2x+cos2x=1
2. tanx=sinx/cosxtanx=sinxcosx
3. cotx=cosx/sinxcotx=cosxsinx

Let's start from the left hand side
(sinx)(tanxcosx-cotxcos x)(sinx)(tanxcosx−cotxcosx)

=sinxtanxcosx-sinxcotxcosx=sinxtanxcosx−sinxcotxcosx

=sinx(sinx/cosx)cosx-sinx(cosx/sinx)cosx=sinx(sinxcosx)cosx−sinx(cosxsinx)cosx

=sin^2x-cos^2x=sin2x−cos2x

=sin^2x+cos^2x-2cos^2x=sin2x+cos2x−2cos2x

=1-2cos^2x=1−2cos2x

Jim G.
May 1, 2018

"see explanation"see explanation

Explanation:

"using the "color(blue)"trigonometric identities"using the trigonometric identities

•color(white)(x)tanx=sinx/cosx" and "cotx=cosx/sinx∙xtanx=sinxcosx and cotx=cosxsinx

•color(white)(x)sin^2x+cos^2x=1∙xsin2x+cos2x=1

rArrsin^2x=1-cos^2x⇒sin2x=1−cos2x

"consider the left side"consider the left side

"distributing the parenthesis"distributing the parenthesis

sinx xxsinx/cancel(cosx)xxcancel(cosx)-cancel(sinx)xxcosx/cancel(sinx)xxcosx

=sin^2x-cos^2x

=(1-cos^2x)-cos^2x

=1-2cos^2x=" right side "rArr"verified"

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