\$\$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  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\$\$ 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\$\$

\$\$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 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
user65203user65203
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