0
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ The first is an identity. The second statement is possible for a few specific choices of $x$. $\endgroup$
    – Macavity
    Jun 5 '15 at 7:17
3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $k \in \mathbb{Z}$." $\endgroup$ Jun 5 '15 at 8:48
  • $\begingroup$ @N.F.Taussig I'm sorry it was a typo. Thank you. $\endgroup$
    – user42912
    Jun 5 '15 at 8:52
  • $\begingroup$ @N.F.Taussig in fact it should read "… for some $k \in \mathbb Z$"! $\endgroup$
    – LSpice
    Jun 12 '16 at 22:04
2
$\begingroup$

$\cos^2(x)-\sin^2(x) = 1$ 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:

cos2x=1

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.