Trigonometric Functions

The trigonometric functions $x=cos(t)$ and $y=sin(t)$ appear as solutions to the initial value problem for $t>0$ : (see alternative presentation)

$\frac{dx}{dt} = -y$ (1)
$\frac{dy}{dt} = x$ (2)

or in time stepping code form

$x=x-y*dt$ (as $dx = -y*dt$ )
$y=y+x*dt$ (as $dy = x*dt$ )

with the following initial values for $t=0$ :

$x=\cos(0)=1$ and $y=\sin(0)=0$ .

Watch and play with the code.

Understand that by construction with $D=\frac{d}{dt}$

$D\cos(t) = -\sin(t)$
$D\sin(t) = \cos(t)$ .

Conclude that both $cos(t)$ and $sin (t)$ solves the second-order differential equation (initial value problem) in a function $u(t)$

$D^2u + u \equiv \frac{du^2}{dt^2}+u=0$ for $t > 0$ ,

with the following initial values:

$u(t)=cos(t): \quad u(0)=1,\quad Du(0)=0$ ,
$u(t)=sin(t): \quad u(0)=0,\quad Du(0)=1$ ,

Observe that the differential equations (1)-(2) express that the velocity vector $(vx,vy)=(\frac{dx}{dt},\frac{dy}{dt})$ and the position vector $(x,y)$ satisfies

$vx* x + vy*y = -y*x + x*y =0$ , (velocity is orthogonal to position)

which means that the point $(x,y)$ moves around a circle with radius 1, which give $\cos(t)$ and $\sin(t)$ a geometric meaning with t appearing as an angle. Find out the details of the geometry.

Check out this more detailed presentation.

To do:

Compute Pi (3.14….) by computing the first value of t > 0 such that sin(t)=0 by time stepping (1)-(2).
Show that by construction $\sin(s+t) =\sin(s)\cos(t)+\cos(s)\sin(t)$ . Hint: Note that $u(t)=sin(s+t)$ solves $D^2u + u = 0$ with initial conditions $u(0)=sin(s)$ and $Du(0)=cos(s)$ and so is the linear combination $sin(s)cost(t)+cos(s)sin(t)$ .
Similarly show other properties of $\sin(t)$ and $\cos(t)$ .
See that (1)-(2) models a harmonic oscillator of a body connected to a fixed point with a spring modeled by dx = – v*dt and dv = x*dt, where the acceleration dv/dt is balanced by a spring force scaling with position x measuring the extension of the spring.

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