# Why is $\cos(i)>1$?

I always thought that cosine only ranges from $$-1$$ to $$1$$. But, I found out that $$\cos(i)=\frac12\left(e+\frac{1}{e}\right)$$ which is certainly greater than $$1$$. Why is that?

• $-1\le\cos{x}\le1$ if $x$ is real. $\cos{z}$ is unbounded if $z$ is complex. Jan 5, 2019 at 17:41
• Complex numbers typically break rules that were established within the confines of the real numbers.
– Blue
Jan 5, 2019 at 17:41
• Cosine has a geometric interpretation as a function on $\mathbb{R}$, but it is also useful to extend it to a function on $\mathbb{C}$. There are all kinds of extensions one could define, but it's very nice to preserve analyticity. This is easily accomplished by using the series expansion $\cos(z)=\sum_{n=0}^\infty\frac{(-1)^nz^{2n}}{(2n)!}$. In fact, someone correct me if I'm wrong, but I'm pretty sure that's the unique analytic extension to $\mathbb{C}$. From this falls out all kinds of nice identities like $\cos(z)=(e^{iz}+e^{-iz})/2$. Jan 5, 2019 at 17:53
• If $\cos(z)$ where bounded, it would be constant. Jan 5, 2019 at 17:55
• Not a bad question, but what do you actually know about complex trigonometric functions? What do you know about complex functions in general? Do you know definitions of exponential and trigonometric functions by their power series? By differential equations? Jan 5, 2019 at 18:39

It's true that the cosine of a real number must be between $$-1$$ and $$1$$, but this is not true for the cosine of a complex number. In fact, complex-differentiable functions can never be bounded (unless they are constant).

Here is an analogy, if you like. Let $$f(x) = x^2$$. Then we learn some rule that $$f(x) \geq 0$$ for all $$x$$. But wait a second, $$f(i)$$ is negative. There's no scandal, since $$i$$ is not a real number.

• +1 for "There's no scandal". Jan 5, 2019 at 17:46
• To be precise: non-constant complex-differentiable functions in $\Bbb C$ can not be bounded. Jan 5, 2019 at 17:46

The general definition of $$\cos(z)$$ is $$\cos(z)=\frac{{e^{iz}}+e^{-iz}}{2}$$ When you plug in complex numbers into $$\cos(z$$), you can get values greater than $$1$$ or less than $$-1$$

• Can you prove this? Jan 5, 2019 at 17:47
• Sure, do you want me to add the proof as part of the answer? Jan 5, 2019 at 17:48
• @MathLover: If you want to know where the value of $\cos(i)$ comes from, then you should say so in your question. As it is, your question seems to be simply about why the value is unexpectedly large.
– Blue
Jan 5, 2019 at 17:58
• @Math Lover and anyone else reading, also you can come to this formula by trying to calculate cos(i) from Euler's formula, or just come to it visually. I (think) it's valid because it's an analytic continuation. At z=ix you get the average of the graph $e^x$ and the same graph flipped over the y axis, which isn't bounded. Just try some sort of large values or set its absolute value greater than one and solve algebraically. Jan 5, 2019 at 18:20
• @Math Lover to look deeper into this topic, look into hyperbolic trig. cos(ix)=cosh(x). We define a trig function as circular because it parametrizes a circle. Now, just because you put an imaginary input into a trig function doesn't mean it's not a trig function, but then it parametrizes the shape of a hyperbola and is a hyperbolic trig function. sin(ix)=i sinh(x). Putting in an imaginary angle ix gives the hyperbolic version of the trigonometric function and tanh, sech, etc satisfy the same relations to the other hyperbolic functions. Hyperbolic angles are better called imaginary and on cont Jan 5, 2019 at 18:29

The function $$\cos z$$ belongs in the interval $$\left[-1,1 \right]$$ when $$z$$ is a real number, not necessarily when $$z$$ is a complex number. An example of this is $$\cos i > 1$$, as you correctly pointed out.

Hint

Use the identity$$\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$$

Here's what happening visually: cosh is essentially cosine with domain of imaginary numbers, as I said in the comments. Cosine and sine trace the unit circle, and COSH (the specific one you asked about) and sinh for the unit hyperbola. The circle parametrization functions (sin and cos) have range between $$-1$$ and $$1$$, whereas the unit hyperbola ($$x^2-y^2=1$$) is unbounded. You could look up these graphs to see this or even deeper see their shape through their relation to conic sections. So that's some visual intuition if want a reason whilst skipping all the knarly mindless algebra.

*If you continue the graph of the unit circle, putting in outputs outside the unit circles range and getting out imaginary numbers, it traces the unit hyperbola. And it sort of makes sense that if a real angle gives the circle, a hyperbola might come from something imaginary. Anyways, there are also other reasons this particular hyperbola is considered "unit" too. cosh and sinh parametrize the unit hyperbola because subbing the cos and sin for a function of their hyperbolic counterparts in the identity $$cos^2(x)+sin^2(x)=1$$ you get $$cosh^2(x)-sinh^2(x)=1$$, which is why the hyperbolic sine and cosine parametrize the unit hyperbola, which again is $$x^2-y^2=1$$, letting x=cosh(t) and y=sinh(t).