I am finishing a proof. It seems like I can use $\cos^2 + \sin^2 = 1$ to figure this out, but I just can't see how it works. So I've got two questions.
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$?
And if it does, then how?
$\endgroup$ 24 Answers
$\begingroup$Observe that $$ \begin{align*} \sin^2(x)-\cos^2(x)&=\sin^2(x)+\bigl(\cos^2(x)-\cos^2(x)\bigr)-\cos^2(x)\\ &= (\sin^2(x)+\cos^2(x))-2\cos^2(x)\\ &= 1-2\cos^2(x). \end{align*} $$ More easily, just subtract $2\cos^2(x)$ from both sides of $\sin^2(x)+\cos^2(x)=1$ to get the result.
$\endgroup$ $\begingroup$Not really different but just another way of looking at it. Start with $\sin^2x+\cos^2x=1$ and subtract $2\cos^2x$ from both sides. Done!
$\endgroup$ $\begingroup$Here is my favorite way to verify trigonometric identities:
First note that the equation of a circle gives us the rational parameterizations
$$\sin\theta=\frac{2t}{1+t^2}\qquad\cos\theta=\frac{1-t^2}{1+t^2}.$$
Substitute these expressions in. Now the equation we want to verify is $$\left(\frac{2t}{1+t^2}\right)^2-\left(\frac{1-t^2}{1+t^2}\right)^2\overset{?}{=}1-2\left(\frac{1-t^2}{1+t^2}\right)^2.$$ Now just find a common denominator and compare numerators, so we want to know $$(2t)^2-(1-t^2)^2\overset{?}{=}(1+t^2)^2-2(1-t^2)^2.$$
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.
$\endgroup$ $\begingroup$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. do algebraic manipulations to make the left hand side of the equation $\sin^2 x - \cos^2 x$.
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