Can anybody show me how to prove that $\cos^2(z)+\sin^2(z)=1$ for $z\in \mathbb{C}$? I can prove it for the case where $z$ is real, but can't seem to find a way to prove it for complex numbers in general.
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My professor insists that it is true. I can't find a way to prove him wrong. If you can provide a proof of why it is not true for complex numbers in general, that would be awesome. Thank you! – Jeff M. Oct 27 '16 at 00:36
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Note that $\cos^2(z)=\frac12(1+\cos(2z))$ and $\sin^2(z)=\frac12 (1-\cos(2z))$. – Mark Viola Oct 27 '16 at 00:37
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7@user8960: Your comment is wrong. – P Vanchinathan Oct 27 '16 at 00:38
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2@user8960 "That's because it is false for complex numbers in general" What? It's an entire function that equals $1$ on the real axis, therefore it's $1$ everywhere by the identity principle. – zhw. Oct 27 '16 at 01:02
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1@user8960 ... Perhaps user has confused this with the assertion that $|\cos z|^2 + |\sin z|^2 = 1$, which is, indeed, false. – GEdgar Oct 27 '16 at 01:18
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1Some older related posts: Is $\sin^2(z) + \cos^2(z)=1$ still true for $z \in \Bbb{C}$? Is it true that $ |\sin^2z+\cos^2z|=1, \forall z \in\Bbb C$? How to prove that $\cos^2(z)+\sin^2(z)=1$, where $z$ is a complex variable (if it is true)?. Found using Approach0 - you can probably find others by refining the search query or browsing through the search results. – Martin Sleziak Oct 27 '16 at 02:18
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Apologies. The individual $\sin(z), \cos(z)$ are not bounded in any half-plane, but the identity does hold, and the simplest proof I found coincides with that from Rene Schipperus. – user8960 Oct 27 '16 at 16:46
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Ok so here is a conceptual proof: the function $f(z)=\cos^2 z+\sin^2 z-1$ is analytic and is zero on a set with a limit point (ie $\mathbb{R}$) and thus $f(z)=0$ identically. The comment above is incorrect. A more computational proof can be given from $$\cos z=\frac{e^{iz}+e^{-iz}}{2}$$ $$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$
Rene Schipperus
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The very same proof that works for real values works for complex values: both $\sin z$ and $\cos z$ admit convergent powerseries on the whole complex plane, which you can differentiate term by term, giving that $$(\cos z) ' = -\sin z$$ $$(\sin z)' = \cos z$$
If you consider $f(z) = \cos^2 z +\sin^2 z-1$ the above gives $f'(z) =0$:
$$f'(z) = -2\cos z\sin z+2\sin z\cos z=0$$ Thus $f$ must be constant, and because $f(0)=0$, it is constantly zero.
Pedro
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For real trigonometric functions have interpretation in terms of ratios in right-angled triangle. Then we have calculus (in one variable) that helps us calculate the Taylor series for $\sin x$ and $\cos x$. For complex values we define trigonometric function by this infinite series and so function values will also be complex numbers.
These series happen to converge throughout the complex plane and define an entire (analytic everywhere) function.
Now consider the complex function $f(z)$ defined by $f(z)=\cos^2 z +sin ^2 z-1$. This function is also entire. And is identically zero on the real line (by the identity known for highschool trig functions).
Now invoke the Principle of Permanence of Identities in complex analysis. This essentially means the set of points where an analytic function takes the value zero (or any other complex number for that matter) is an isolated set: i.e., a set with no limit point, unless the function is a constant function.
Here our $f(z)$ takes value zero on the real line, a set with limit points. SO it should be constantly zero. SO the identity is valid for complex values of $z$ too.
P Vanchinathan
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Did you mean to write $f(z)=cos^2(z)-sin^2(z)$? I'm confused why you defined your $f(z)$ the way you did. – Jeff M. Oct 27 '16 at 00:46
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Let $x,y\in\mathbb{R}$ a real number, so : $$ \cos(y+ix)=\frac{e^{i(y+ix)}+e^{-i(y+ix)}}{2}=\frac{e^{iy}e^{-x}+e^{-iy}e^{x}}{2} $$ and the same for $\sin$ : $$ \sin(y+ix)=\frac{e^{i(y+ix)}-e^{-i(y+ix)}}{2i}=\frac{e^{iy}e^{-x}-e^{-iy}e^{x}}{2i} $$ if we put $z=y+ix$ : \begin{eqnarray} 4(\cos^2(z)+\sin^2(z))&=&(e^{iy}e^{-x}+e^{-iy}e^{x})^2 -(e^{iy}e^{-x}-e^{-iy}e^{x})^2\\ &=& e^{-2x+2iy}+e^{2x-2iy}+2e^{-x+iy+x-iy}-(e^{-2x+2iy}+e^{2x-2iy}-2e^{-x+iy+x-iy})\\ &=&4 \end{eqnarray} so : $$ \cos(z)^2+\sin(z)^2=1 \qquad \forall z \in \mathbb{C} $$
Hamza
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