# Intuitive understanding of why trigonometric functions relate to both triangles and the complex exponential

The trigonometric functions sine and cosine can be defined in terms of the complex exponential:

$$\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$$

$$\cos(x) = \frac{1}{2}(e^{ix} + e^{ix})$$

And can also be defined in terms of the lengths of the sides of a triangle where $$x$$ now represents the angle between the hypotenuse and the adjacent sides $$H$$, $$A$$, and $$O$$ represent the lengths of the hypotenuse, adjacent, and opposite sides respectively:

$$\sin(x) = \frac{O}{H}$$

$$\cos(x) = \frac{A}{H}$$

I find it utterly bizarre that something as seemingly unrelated as the dimensions of a triangle relates to exponentiation and the root of a negative.

Mathematically, how is this possible?

Intuitively, why is this possible?

• You can define sine and cosine just by drawing pictures of triangles or of the unit circle, and you can discover power series for $\sin(x)$ and $\cos(x)$ and $e^x$ just by doing calculus with real numbers and never thinking about complex numbers. It does seem to be shocking to me that if you plug $x = i \theta$ into the power series for $e^x$ the result is $\cos(\theta) + i \sin(\theta)$. I think that doing that calculation, with no expectation about what the result should be, must be one of the most shocking moments in math. Commented Aug 15, 2022 at 19:08
• From whichever definition that you have for $e^x$, I suppose you have proven that $e^{x+y}=e^xe^y$, using this and your first definition of the trigs you can check that $\cos^2(x)+\sin^2(x)=1$. Pythagoras gives that a triangle with sides $\cos(x),\sin(x),1$ has a right angle between the first two sides. So, the point $(\cos(x),\sin(x))$ in Cartesian coordinates is on the unit circle. The geometric sine and cosines are the sides of the triangle.
– plop
Commented Aug 15, 2022 at 19:10
• If you consider the unit circle in the complex plane $e^{i x}=\cos(x)+i \sin(x)$ for $0<x<2 \pi$, then the real part $\cos(x)$ is the x-axis (adjacent side), the imaginary part $\sin(x)$ is the y-axis (opposite side), and the radial is the hypotenuse. Commented Aug 15, 2022 at 19:19
• The geometric interpretation of complex number multiplication (“add the angles and multiply the lengths”) appears to me to be a basic miracle, in some sense equivalent to the miracle that $e^{i \theta} = \cos \theta + i \sin \theta$. If we accept that the function $h(\theta) = \cos \theta + i \sin \theta$ satisfies $h(x+y) = h(x) h(y)$, then that formula suggests a connection to the exponential function. But then we have just shifted the miracle elsewhere. Commented Aug 15, 2022 at 19:48
• Your equation for cosine is wrong: $$\cos(x) = \frac{1}{2}(e^{ix} + e^{ix})$$ One of the exponents should be $-ix$. Commented Aug 18, 2022 at 5:06

I would say the relationship is purely derivative-related. You have that sine and cosine cycle back to $$(-1)^n$$ times themselves on differentiating $$2n$$ times. $$e^x$$ or more generally $$e^{cx}$$ for a constant $$c$$, is of course related to cyclic differentiation.

So you would expect some connection.

A good approach is to write the differential equation both sine and cosine satify: $$f''(x)=-f(x) \tag{\star}\label{1}$$ We know we will get some exponential solution(s); let's solve the ODE to know the exponentials that satisfy it. Substitute $$f(x)=e^{\lambda x}$$: $$\lambda^2=-1\\ \lambda=\pm i\\ f(x)=c_1e^{ix}+c_2e^{-ix}$$ So, we can write sine as: $$\sin x=c_1e^{ix}+c_2e^{-ix}$$ To solve for the $$2$$ constants we need $$2$$ equations, we already have one, to get the other, just differentiate: $$\cos x=ic_1e^{ix}-ic_2e^{-ix}$$ Now, plug in $$0$$ for $$x$$ in both equations to solve for the constants and recall $$\sin(0)=0$$ and $$\cos(0)=1$$: $$0=c_1+c_2\\ 1=ic_1-ic_2\\ i=-c_1+c_2\\ c_2=\frac{i}{2}\\ c_1=-\frac{i}{2}\\ \sin x=-i\frac{e^{ix}-e^{-ix}}{2}$$ Multiply by $$\frac{i}{i}$$: $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ Plug the value for the constants in the equation for cosine to get: $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ So we don't actually need taylor series to derive this formula! And we actually get to understand its origin and cause too!

For an intuitive explanation, you have to take a trip down memory lane, from (presumably millenium or older) trigonometry, to Real Analysis to Complex Analysis.

In Analytical Geometry (in days of yore), the domain of the sine and cosine functions were angles, and the range of the sine and cosine functions were (real valued) ratios, between (for example) the side of one of the legs of a right triangle, and the hypotenuse of the right triangle.

In days of yore, Mathematicians knew that if two right triangles had the same angles (i.e. similar triangles), then the corresponding ratios would be equivalent. So, the Mathematicians were able to regard the domain of the sine and cosine functions as angles (in a right triangle) independent of how large the (right) triangle was.

This is a perfectly sensible definition.

Then (to oversimplify), during the birth of Real Analysis (about 300-400 years ago), Mathematicians got tired of overdosing on Reimann Sum rectangles to determine (for example) $$~\displaystyle \int_0^1 x^2 dx.$$

So, differential and integral Calculus evolved, motivated by the desire to find easier ways of attacking specific problems, than the consideration of an infinite number of infinitely narrow rectangles.

Then, Mathematicians found other problems that they wanted to solve. One such classic problem is

$$\int_0^1 \frac{dt}{1 + t^2}. \tag1$$

They discovered that you could facilitate attacking (1) above if you redefined the sine and cosine function in a way that (at first glance) seems nonsensical.

Instead of having the domain of such numbers be angles, the domain of such numbers would be real numbers. Excerpting from "Calculus Volume 1, 2nd Edition" (Tom Apostol):

• The sine and cosine functions are defined for all real numbers.

• $$\cos(0) = \sin(\pi/2) = 1, ~\cos(\pi) = -1.$$

• $$\cos(y - x) = \cos(y)\cos(x) + \sin(y)\sin(x).$$

• For $$0 < x < (\pi/2)~$$ :
$$\displaystyle 0 < \cos(x) < \frac{\sin(x)}{x} < \frac{1}{cos(x)}.$$

Mathematicians found that all of the (modern) notions of trig formulas, including trig derivatives/integrals can be proven using the above $$(4)$$ axioms. Mathematicians also discovered a strong relationship between this new and bizarre definition of the sine and cosine functions, and the old sensible Analytical Geometry definitions.

They discovered that if you confined your attention to the unit circle (i.e. of radius $$1$$, centered at $$(x,y) = (0,0),$$ then the old angle-based definition of the sine and cosine functions was equivalent to the new definition, as long as you focused on the arc length.

So, under the new bizarre definition, the domain of the sine and cosine functions were the arc length formed by the unit circle sector formed by the corresponding angle.

For example, under the Analytical Geometry definition, $$\sin(45^\circ) = \dfrac{1}{\sqrt{2}}.$$

Under the new definition, $$\sin(t) = \dfrac{1}{\sqrt{2}},$$ when $$~\displaystyle t = \frac{1}{8} \times$$ the circumference of the unit circle (i.e. $$~\displaystyle \frac{1}{8} \times 2\pi$$).

One example of the benefit of this bizarre redefinition is (1) above, which turns out to equal

$$\arctan(1) - \arctan(0) = \frac{\pi}{4}.$$

Another advantage is that Taylor Series approximations for the sine and cosine functions were possible. That is,

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots - ...$$

and

$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots - ...$$

Mathematicians also knew that $$~e^x ~: ~x \in \Bbb{R}~$$ had the Taylor series

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.$$

Naturally, the above approaches provided good job security to renaissance Mathematicians, because you had to be legally insane to comprehend all of this. However, some of these Mathematicians were crazier than others. The really crazy Mathematicians found that if they created $$(i) = \sqrt{-1}$$ and invented Complex numbers as $$~z = x + iy~ : ~x,y \in \Bbb{R}, ~z \in \Bbb{C}$$ that this would facilitate easier solutions to other problems.

They found that if $$z \in \Bbb{C}$$ and $$z \neq 0$$, then (for example) $$z$$ could be uniquely expressed as

$$z = r[\cos(\theta) + i\sin(\theta)] ~: ~r \in \Bbb{R^+}, ~\theta \in \Bbb{R}, ~-\pi < \theta \leq \pi.$$

The following simple example shows a classic problem that is facilitated by Complex Analysis.

Consider the equation
$$0 = f(x) = (x^3 - 1) = (x-1)(x^2 + x + 1).$$

The old (harder) approach was that the roots are
$$(x = 1)~$$ and $$~\displaystyle x = \frac{1}{2} \left[-1 \pm \sqrt{-3}\right].$$

So, one immediate development was that the three roots were the elements in

$$\left\{ ~1, ~\frac{-1 + i\sqrt{3}}{2}, ~\frac{-1 - i\sqrt{3}}{2} ~\right\}. \tag2$$

Then, Mathematicians discovered that

$$\left\{r_1[\cos(\theta_1) + i\sin(\theta_1)]\right\} \times \left\{r_2[\cos(\theta_2) + i\sin(\theta_2)]\right\}$$

$$= \left(r_1 \times r_2\right) \times \left[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right]. \tag3$$

One of the consequences of this is that if $$z,w$$ were Complex numbers on the unit circle (i.e. |z| = 1 = |w|), with corresponding angles $$~\theta_z, \theta_w~$$ then you could multiply $$z \times w$$ merely by adding $$\theta_z + \theta_w$$.

This implied that a much easier approach to reaching the result in (2) above, is that you are looking for all unit circle elements $$z$$, with angle $$\theta_z$$, such that $$3 \times \theta_z$$ is a multiple of $$(2\pi).$$

This is because (for example) $$1 = \cos(2\pi) + i\sin(2\pi).$$

Turns out that there was this lunatic named Euler, that looked at the Taylor Series for the sine and cosine functions, as well as the Taylor Series for $$e^x$$, and thought:

Suppose that I pretend that the Taylor Series around $$e^x ~: ~x \in \Bbb{R}$$ can be extended to a Taylor Series around $$e^z ~: ~z \in \Bbb{C}.$$

This will produce syntactic sugar. One of the results is that
for $$x,y \in \Bbb{R},$$
you can define $$e^{x + iy} = e^x \times [\cos(y) + i\sin(y)].$$

Euler found that then, the Real Analysis exponent law (i.e. $$e^{x_1} \times e^{x_2} = e^{x_1 + x_2}$$) carried over into Complex Analysis.

But (lunatic or not), Euler was crafty. He decided that instead of referring to this definition of $$e^z = e^{x + iy}$$ as syntactic sugar, he would refer to it as a profound result.

All the other lunatics cheered, because this was (if you'll pardon the pun) Real (Analysis) job security. So, they rewarded Euler by naming the $$2.71828...$$ to be the constant $$e$$.

So, instead of having to express a non-zero Complex number $$z$$ as $$r[\cos(\theta) + i\sin(\theta)]$$, you could express it as $$z = re^{i\theta}.$$

$$e^{ix}=\cos x + i \sin x$$ is intuitively a circle of radius $$|e^{ix}|=1$$ in the complex plane with the radius, the real part, and the complex part forming a right-angled triangle.

Every number $$w$$ (whether real or complex) has positive magnitude $$r = |w|$$ and an angle. (It's just that all positive real number have an angle of $$0$$ and all negative real numbers all have an angle of $$180$$ and there are no other options).

Now if we multiply a number $$w$$ by a scalar $$a$$, all $$aw$$ means is we've stretched $$w$$ by the scale value $$a$$.

When we multiply a number $$w$$ by real number $$v$$ the magnitudes multiply and stretch accordingly as $$|wv| = |w|\cdot |v|$$ but what of there $$\pm$$ or $$0, 180$$ angles? Well. They just rotate each other. If $$\angle 0 \times whatever = whatever$$ and $$\angle 180 \times whatever = -whatever$$.

This becomes more apparent when we notice that multipling a number by $$i$$ does what? Well of $$w = a+bi=$$ the point graphed by $$(a,b)$$ then multiplying by $$i$$ will $$a+bi \to (a+bi)i =ai + bi^2 \to -b + ai$$ or the point graphed by $$(-b, a)$$. In other words multiplying by $$i: (a,b) \to (-b, a)$$. THAT'S JUST ROTATING BY 90 DEGREES!!

.....

So that's all multiplication is! If $$w$$ has magnitude $$r$$ and and angle $$\theta$$ and $$v$$ has magnitude $$s$$ and angle $$\eta$$ then $$w\cdot v$$ is going to have magnitude $$rs$$ and angle $$\theta + \eta$$.

Now here's the crazy knock you down with a feather what's the world all about crazy part.

If we consider what rotating $$\theta +\eta$$ does. $$\sin(\theta + \eta) = \sin\theta \cos \theta +\cos \theta \sin \eta$$. While $$\cos(\theta + \eta) = \cos \theta \cos \eta - \sin\theta \sin \eta$$

and compare it to multiplying complex numbers

$$(S + Ci)(s + ci) = Ss + Sci +Cis +Cici = (Ss -Cc) + (Sc + Cs)i$$

we notice those are the exact same thing! If $$S=\sin \theta; s=\sin\eta;C=\cos \theta; c=\cos \eta$$, then simple multplication gives us $$(S+Ci)(s+ci) = \cos(\theta + \eta) + \sin (\theta + \eta)i$$!

Hence... multiplying complex number results in 1) multiplying their magnitudes and 2) adding their angles and the rules for getting the real, imaginary values of adding the angles is EXACTLY the same as the rules for getting the sines and cosines of added angles.

....

ANd that's the intuition.