# Trigonometry functions

How do I verify the following identity:

$$\frac{1-(\sin x - \cos x)^2}{\sin x} = 2\cos x$$

Tony

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are you sure your brackets are in the right place? – oks Mar 7 '13 at 18:19
Multiple across by $\sin x$ and use a basic relationship involving $\sin^2 x + \cos^2 x$. – copper.hat Mar 7 '13 at 18:22
What you've written isn't an identity, it's an equation. An equation asks "For which values of $x$ are the left- and right-hand sides equal." For an identity, you need to use $\equiv$. For example $\cos^2x+\sin^2 \equiv 1$. This means that the left- and right-hand sides are identical. So writing $\cos^2x + \sin^2x \equiv 1$ is the same as saying that $\sin^x+\cos^2=1$ for all values of $x$. – Fly by Night Mar 7 '13 at 18:25
@FlybyNight: An identity (according to every trig book I've ever seen) is is an equation that holds whenever both sides of the equation are defined. This is an identity in that sense. – Cameron Buie Mar 7 '13 at 20:53
@CameronBuie I agree 100%. If $\sin^2x + \cos^2x = 1$ for all $x$ then $\sin^2x + \cos^2x \equiv 1$. (Which is of course true.) However, an equation, in general, is not an identity. – Fly by Night Mar 7 '13 at 21:07

$$\frac{1-(\sin x -\cos x)^2}{\sin x}=\frac{1-(\sin^2x+\cos^2x)+2\sin x\cos x}{\sin x}=\frac{2\cos x\sin x}{\sin x}=2\cos x$$
Here I used the identity: $\sin^2x+\cos^2x=1$