2
$\begingroup$

By Eulers identity $e^{i \theta}=\cos(\theta)+i \sin(\theta)$ sine and cosine can be written in exponential form as

$\sin(\theta)=\frac {e^{i \theta}-e^{-i \theta}}{2i}$ and $\cos(\theta)=\frac {e^{i \theta}+e^{-i \theta}}{2}$.

Could you calculate specific values of the trigonometric function with these formulae? My guess is that the complex exponential can only be calculated using Euler's identity so you have to know the values of sine / cosine to begin with. Is there any method to calculate the value of sine / cosine using the identity above? Is there any reason why it isn't (is?) possible?

$\endgroup$
  • 1
    $\begingroup$ You could use powers of $e^{i\theta}$ for instance. $\left(e^{i\theta}\right)^n = e^{in\theta}$ are used to compute $\cos(n\theta)$ and $\sin(n\theta)$ $\endgroup$ – Guy May 28 '17 at 13:28
  • $\begingroup$ The development by Euler was to observe that the complex function $f(z)=\sum_{j=0}^{\infty}z^j/j!$ satisfies $f(z_1)f(z_2)$ for all complex $z_1,z_2,$ and also $f(x)=e^x$ for all $x\in \mathbb R.$ So it is reasonable to call this function $e^z.$ And then we obtain $e^{\pm ix}=\cos x \pm i\sin x$ for real $x.$ So calculating $e^{\pm ix}$ for real $x $ is the same thing as calculating its real and imaginary parts $\cos x$ and $\pm \sin x.$ $\endgroup$ – DanielWainfleet May 28 '17 at 15:19
6
$\begingroup$

You can evaluate it at imaginary values:

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

$$\cos(i)=\frac{e^{-1}+e}2$$

But not much else.


Good for deriving some formulas though:

$$\cos^2(x)=\left[\frac{e^{ix}+e^{-ix}}2\right]^2=\frac{e^{2ix}+2+e^{-2ix}}4=\frac{\cos(2x)+1}2$$

$\endgroup$
  • 2
    $\begingroup$ Don't forget that the imaginary values lead to the hyperbolic functions, as well. $\endgroup$ – Cye Waldman May 28 '17 at 19:33
  • $\begingroup$ @CyeWaldman Of course they do :P $\endgroup$ – Simply Beautiful Art May 28 '17 at 20:23
1
$\begingroup$

For me, the best use of Euler's formula arises in the rotation of coordinates, that is,

$$ x'=x\cos\theta+y\sin\theta\\ y'=-x\sin\theta+y\cos\theta $$

In complex coordinates this transform and it inverse are simply

$$z'=z\,e^{i\theta}\text{ and }z=z'\,e^{-i\theta}$$

Basically, anything you can do in Cartesian coordinates can be done in the complex plane, and frequently much simpler (in my opinion) .

$\endgroup$
0
$\begingroup$

Technically, you can use the Maclaurin series of the exponential function to evaluate sine and cosine at whatever value of $\theta$ you want. But you will find that when you simplify the resulting expressions, they turn out to be exactly the same expressions as the Maclaurin series of sine and cosine themselves. So you have not made the situation any easier.

$\endgroup$
0
$\begingroup$

There are many many things you can use the imaginary exponential for without knowing much if anything about sine and cosine of a particular value. Perhaps the most well-known example is this: $$ 1 = e^0 = e^{i\theta}e^{-i\theta} = (\cos \theta + i \sin \theta)(\cos \theta - i \sin \theta) = \cos^2 \theta + \sin^2 \theta. $$ Furthermore, the imaginary exponential is no more useless than sine and cosine themselves in as much as numerics are concerned. We forget, but, sine and cosine don't come for free. However, the algebraic structure of the imaginary exponential is far more lucid than that of sine and cosine. Unfortunate it is that teach them more we do not.

$\endgroup$

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.