Name says it all. The two seem so unrelated? What's more, if I'm not mistaken the exponential version contains an imaginary part. I'm kind of ignorant about imaginary numbers, but does this mean that sin extends info the imaginary plane? If so, how is that possible if sin is a continuous function?

  • $\begingroup$ The ratio of two sides of a triangle has nothing to do with the exponential function. Its the ${\bf angle}$ that does. $\endgroup$ – Christian Blatter Nov 19 '15 at 14:39

Let's look at exponential arithmetic first. We have $a^x\cdot a^y=a^{(x+y)}$ which becomes the functional equation for a generalised exponential function $$f(x)\cdot f(y)=f(x+y)$$The only well-behaved real valued functions which satisfy this equation are exponential functions with $a$ being some positive real number.

Suppose you have a number $i$ which has the proper $i^2=-1$ and consider the function $f(x)=\cos x+i\sin x$. We have that $$f(x)\cdot f(y)=\cos x\cos y-\sin x\sin y+i(\cos x\sin y+\sin x\cos y)=\cos (x+y)+i\sin (x+y) $$


Note: I have used the well-known formulae for $\cos (x+y), \sin (x+y)$ which can be given an elementary geometrical proof (amongst other proofs).

So this particular combination of $\sin x$ and $\cos x$ with a square root of $-1$ satisfies the functional equation for the exponential function. It turns out that this works as an exponential function based on the particular real number we call $e$.

It is hard to go very much further without getting much more into complex numbers, but it turns out not to be a coincidence that this works, but rather a delicate and beautiful set of mathematical connections.

Further analysis leads to defining the sine function as an infinite sum, which converges both for real and complex numbers, namely

$$\sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}-\frac {x^7}{7!}+\dots$$

Where the terms are successive odd powers of $x$ with alternating signs and divided by the factorial of the exponent. Compare the exponential function $$e^x=\frac {x^0}{0!}+\frac {x^1}{1!}+\frac {x^2}{2!}+\frac {x^3}{3!}+\dots$$where you have all the nonnegative powers of $x$ and the signs are the same. If you plug $x=iy$ into this formula and use $i^2=-1$ you should start to see some connections - and find a related series for $\cos x$.

Note that for these formulae to work the angle $x$ is measured in radians.


The two seem so unrelated.

Appearances can be so deceiving $\ldots$ ;-$)$ A triangle is formed by three non-collinear points. But, at the same time, three non-collinear points uniquely determine a circle. What is the equation of a circle centered at the origin ? $x^2+y^2=r^2$. And if in the latter we were to replace the sum by a difference, we would get a hyperbola $\ldots$ which, when rotated by an angle of $45^\circ$, and scaled by a factor of $\sqrt2$, becomes $xy=r^2\iff y=\dfrac{r^2}x.~$ Now, what happens when we integrate the latter, and what is the relation of its anti-derivative to the exponential function ? :-$)$

  • $\begingroup$ Brilliantly put. Gives a very palpable understanding of the idea, and just cryptic enough to force me to use my head. But unfortunately I'm a bit dense so let me make sure I follow. So when you integrate x you get lnx so does this man that the integration of a the delta of two sides of a circle is like the inverse of the exponential function $\endgroup$ – user3256725 Nov 17 '15 at 18:33
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    $\begingroup$ @user3256725: It should comes as no surprise that, given the similarity between their algebraic equations, $x^2\pm y^2=r^2$, both trigonometric and hyperbolic functions are constructed from the same “bricks”. These “bricks” are all of the form $\dfrac{x^n}{n!}$ . What differs, however, is the mode of their construction. In the case of a circle, they alternate, each new term canceling the power and magnitude of the previous one; whereas, in the case of a hyperbola, this does not happen, but rather, with each new term being added, the sum becomes ever greater, causing it to grow exponentially. $\endgroup$ – Lucian Nov 17 '15 at 19:25

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