Asked by Julia
Prove the identity of the following equation:
(cos 2x)/(1/(cos x)) * (sin(pi + x))/(tan x) = (sec(x) - csc(x)) * (csc(x))/(sec^2 x) * (1 - cos^2 x)
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(cos 2x)/(1/(cos x)) * (sin(pi + x))/(tan x) = (sec(x) - csc(x)) * (csc(x))/(sec^2 x) * (1 - cos^2 x)
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Answers
Answered by
Reiny
take apart the LS
(cos 2x)/(1/(cos x)) is just (cos 2x)(cosx)
(sin(pi + x))/(tan x) = -sinx(cosx)/sinx = -cosx
so LS = (cos 2x)(cosx)*(-cosx)
= -(cos 2x)(cos^2 x)
RS = (sec(x) - csc(x)) * (csc(x))/(sec^2 x) * (1 - cos^2 x)
= (1/cosx - 1/sinx)*(cos^2 x)/sinx * (sin^2 x)
= (sinx - cosx)/(sinxcosx)*(cos^2 x)sinx
= (sinx - cosx)(cosx)
testing for x = 30°
LS = -.375
RS = -.316...
The equation is not an identity the way you typed it
(cos 2x)/(1/(cos x)) is just (cos 2x)(cosx)
(sin(pi + x))/(tan x) = -sinx(cosx)/sinx = -cosx
so LS = (cos 2x)(cosx)*(-cosx)
= -(cos 2x)(cos^2 x)
RS = (sec(x) - csc(x)) * (csc(x))/(sec^2 x) * (1 - cos^2 x)
= (1/cosx - 1/sinx)*(cos^2 x)/sinx * (sin^2 x)
= (sinx - cosx)/(sinxcosx)*(cos^2 x)sinx
= (sinx - cosx)(cosx)
testing for x = 30°
LS = -.375
RS = -.316...
The equation is not an identity the way you typed it
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