$$dfrac2sin (x)cos(x)1-2cos(x)-1 = dfrac2sin(x)cos(x)-2cos^2(x) = dfrac2sin(x)-2cos(x) = - an (x)$$

I think I"ve gone wrong somewhere.

You are watching: Simplify cos^2x-sin^2x


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We have

$$cos(2x) = cos^2x-sin^2x=2cos^colorred2(x)-1=1-2sin^2 x$$

and then

$$dfracsin(2x)1-cos (2x)=dfrac2sin x cos x1-1+2sin^2 x=cot x$$


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You gained one the the indications mixed up.

$$fracsin(2x)1-cos (2x) = dfracsin(2x)1-(2cos^2x-1) = fracsin(2x)1-2cos^2x+1$$

Instead, girlfriend accidentally used

$$fracsin(2x)1-cos (2x) colorred eq fracsin(2x)1-2cos^2x-1$$

As for the leveling itself,

$$fracsin(2x)1-cos (2x) = frac2sin xcos x1-(1-2sin^2x) = frac2sin xcos x2sin^2 x = fraccos xsin x = cot x$$

Also, to allude out, you offered $cos(2x) = 2cos^2x-1$, which no actually acquire you anywhere since you acquire two $+1$’s. Using $cos^2x = 1-2sin^2x$ is the way to go.


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edited Nov 23 "18 in ~ 20:45
reply Nov 23 "18 in ~ 20:40
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KM101KM101
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$$frac2sin xcos x1-(1-2sin^2x)=fraccos xsin x.$$


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edited Nov 23 "18 in ~ 21:05
answer Nov 23 "18 in ~ 20:50
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