# Trigonometric Identities dilemma

If $\cos^2 x + \sin^2(x) =1$

Does $\cos2x = \cos^2(x) - \sin^2(x) = 1$ too? meaning $\cos^2(x) - \sin^2(x) = 1$ and $\cos^2 x + \sin^2(x) =1$

How so? It doesn't make sense to me.

• The first is an identity. The second statement is possible for a few specific choices of $x$. Jun 5 '15 at 7:17

The first statement is true for every real number $x$. If you replace $x$ for every real number you can imagine the equality will hold. So for example $\sin^2 1+\sin^2 1=1$, $\sin^2 1000+\cos^2 1000=1$, $\sin^2 \pi+\cos^2 \pi=1$, etc. You can prove this easily using the pythagoras theorem in the trigonometric circle.

The second equality is true just for a few choices of $x$. For example, $\cos (2\cdot0)=\cos^2 0-\sin^2 0=1$, but $\cos(2\cdot\pi/2)=\cos^2 (\pi/2)-\sin^2 (\pi/2)=-1$. For the general case $\cos (2x)=\cos^2x-\sin^2x=1$ if and only if $x=k\pi$ for some $k\in \mathbb Z$.

Any questions I'm glad to help.

• Your penultimate sentence should read "For the general case $\cos(2x) = \color{red}{\cos^2x - \sin^2x} = 1$ if and only if $x = k\pi$ for every $k \in \mathbb{Z}$." Jun 5 '15 at 8:48
• @N.F.Taussig I'm sorry it was a typo. Thank you. Jun 5 '15 at 8:52
• @N.F.Taussig in fact it should read "… for some $k \in \mathbb Z$"! Jun 12 '16 at 22:04

$\cos^2(x)-\sin^2(x) = 1$ is just an expression. It is valid for only some particular values of $x$.

That is it is true for $x=\pi \space n$ and $n \in \mathbb{Z}$

As evident from:

• The evidence may be helped by adding a Ticks -> {{-3Pi, -2Pi, -Pi, 0, Pi, 2Pi, 3Pi}, {-1, 1}}. :-) Jun 12 '16 at 22:06