# List of trigonometric identities

I am finishing a proof. It seems lượt thích I can use \$cos^2 + sin^2 = 1\$ to lớn figure this out, but I just can"t see how it works. So I"ve got two questions.

Does \$sin^2 x - cos^2 x = 1-2cos^2 x\$?

And if it does, then how?  Observe that\$\$eginalign*sin^2(x)-cos^2(x)&=sin^2(x)+igl(cos^2(x)-cos^2(x)igr)-cos^2(x)\&= (sin^2(x)+cos^2(x))-2cos^2(x)\&= 1-2cos^2(x).endalign*\$\$More easily, just subtract \$2cos^2(x)\$ from both sides of \$sin^2(x)+cos^2(x)=1\$ khổng lồ get the result.

Bạn đang xem: List of trigonometric identities Not really different but just another way of looking at it. Start with \$sin^2x+cos^2x=1\$ & subtract \$2cos^2x\$ from both sides. Done! Here is my favorite way khổng lồ verify trigonometric identities:

First note that the equation of a circle gives us the rational parameterizations

\$\$sin heta=frac2t1+t^2qquadcos heta=frac1-t^21+t^2.\$\$

Substitute these expressions in. Now the equation we want lớn verify is\$\$left(frac2t1+t^2 ight)^2-left(frac1-t^21+t^2 ight)^2overset?=1-2left(frac1-t^21+t^2 ight)^2.\$\$Now just find a common denominator và compare numerators, so we want to lớn know\$\$(2t)^2-(1-t^2)^2overset?=(1+t^2)^2-2(1-t^2)^2.\$\$

But this is true because\$\$(1+t^2)^2-(1-t^2)^2=(1+2t^2+t^4)-(1-2t^2+t^4)=4t^2=(2t)^2,\$\$thus the identity is true.

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answered Feb 12 "13 at 6:53 Chris BrooksChris Brooks
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To go with your idea, you could try solving \$cos^2 x + sin^2 x = 1\$ for \$sin^2 x - cos^2 x\$. i.e. vị algebraic manipulations to lớn make the left h& side of the equation \$sin^2 x - cos^2 x\$.

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answered Feb 12 "13 at 4:52
user14972user14972
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