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.

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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.

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 $sin^2 x - cos^2 x$.

answered Feb 12 "13 at 4:52
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