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.