5
$\begingroup$

Is $$\sin^2(z) + \cos^2(z)=1$$ still true for all $z \in \Bbb{C}$? I've tried rewriting it using complex definitions of $\sin$ and $\cos$, and I don't see why it wouldn't, but the text I'm reading asks this question as if it shouldn't hold?

$\endgroup$
6

2 Answers 2

6
$\begingroup$

You can derive this identity from nothing more than the fundamental differential identities that define $\sin$ and $\cos$. For suppose you have functions $s(x)$ and $c(x)$ (that play the obvious roles) with $s'(x) = c(x)$ and $c'(x) = -s(x)$, where $s(0) = c'(0) = 0$ and $c(0)=s'(0)=1$. (If you are only given the more common requirement that $s'' = -s$ and $c''=-c$ then you usually obtain the above derivatives as a result of knowing that certain ODEs with specified initial conditions have unique solutions; I don't want to invoke this here.)

This is just because $(s^2+c^2)' = 2ss' + 2cc' = 2sc-2sc = 0$, and so it's constant; but you also know that $s^2(0)+c^2(0)=1$ by the definition of $s$ and $c$. In this proof I never actually used that these are real functions; just the formulas their derivatives satisfy.

There are other ways to conclude, too. One standard way is Euler's formula; another is the fact that, because $\sin$ and $\cos$ are both holomorphic, $\sin^2+\cos^2-1$ is holomorphic, and we know it's zero on the real line. The identity theorem says that any holomorphic function that vanishes on a set with a limit point (which the real line, of course, is) vanishes everywhere, so $\sin^2(z)+\cos^2(z)=1$ everywhere. It's not hard to prove the identity theorem, but the identity $\sin^2+\cos^2=1$ is even more fundamental than it is.

$\endgroup$
0
$\begingroup$

You can also prove this with inversions of Euler's identity:$$ \begin{align} \operatorname{e}^{iz} &= \cos z + i \sin z \Rightarrow \\ \cos z & = \frac{\operatorname{e}^{iz} + \operatorname{e}^{-iz}}{2},\ \mathrm{and} \\ \sin z & = \frac{\operatorname{e}^{iz} - \operatorname{e}^{-iz}}{2i}. \end{align}$$ These lead to: $$\begin{align} \cos^2 z & = \frac{\operatorname{e}^{2iz} + \operatorname{e}^{-2iz} + 2}{4},\ \mathrm{and} \\ \sin^2 z & = -\frac{\operatorname{e}^{2iz} + \operatorname{e}^{-2iz} - 2}{4}. \end{align}$$ Add these together, and you get $1$. You can make the same proof with the power series definitions, trig substitutions with the power reduction formulae, etc.

$\endgroup$

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.