Trigonometric Functions Alt

  • The integrals which we have obtained are not only general expressions which satisfy the differential equation, they represent in the most distinct manner the natural effect which is the object of the phenomenon…when this condition is fulfilled, the integral is, properly speaking, the equation of the phenomenon; it expresses clearly the character and progress of it, in the same manner as the finite equation of a line or curved surface makes known all the properties of those forms. (Fourier)
  • God does not care about our mathematical difficulties. He integrates empirically. (Einstein)

Defining Differential Equation

The trigonometric functions \sin(t) and \cos(t) are elementary functions defined by the solution (x(t),v(t)) of the IVP of a harmonic oscillator (see also alternative presentation):

  • \frac{d}{dt}x(t) =v(t), \quad \frac{d}{dt}v(t) =-x(t)\quad\mbox{for }t>0 (dx =v*dt, dv=-x*dt)
  • x(0)=0, \quad v(0)=1,

as \sin(t) = x(t) and \cos(t)= v(t). These functions can be extended to t<0 as solutions to the same differential equations for t<0. Watch and play with code.

                                           The trigonometric functions \sin(t) and \cos(t).

Properties of Trigonometric Functions

By definition, we have

  • \frac{d}{dt}\sin(t)=\cos(t),\quad\frac{d}{dt}\cos(t)=-\sin(t).

Further, we have with x(t)=\sin(t) and v(t)=\cos(t),

  • \frac{d}{dt}(x^2+v^2)=2(x\dot x+v\dot v) = 2(xv -xv)=0,

showing that x^2(t)+v^2(t) is constant in time, and since x^2(0)+v^2(0)=1, we have for all t

  • \sin^2(t)+\cos^2(t)=1.   (1)

Geometric Interpretation

We shall now give an interpretation of \sin(t) and \cos(t) in the plane with the usual Cartesian coordinate system x=(x_1,x_2).

If we write x_1(t)=\cos(t) and x_2(t)=\sin(t), then the defining differential equation is written

  • \dot x_1(t)=-x_2(t),\quad \dot x_2(t)=x_1(t) ,

from which follows with \dot x(t) =(\dot x_1(t),\dot x_2(t)) that

  • \dot x(t)\cdot x(t) =-x_2(t)x_1(t)+x_1(t)x_2(t)= 0 ,

which means that the velocity vector \dot x(t) is perpendicular to the vector x connecting the origin with the point $x$.

Recalling from (1) that

  • x_1^2(t)+x_2^2(t)= \dot x_2^1(t)+\dot x_1^2(t)=1  (2)

it follows that as t varies, the point (x_1(t),x_2(t))=(\cos(t),\sin(t)) moves along a unit circle centered at the origin with unit speed, as illustrated in the following picture:

We can thus choose time t to be a measure of the angle, from the horisontal.

Pythagoras Theorem

Note that we can interpret  (1) and (2) as Pythagoras Theorem. We have thus given a proof of Pythagoras theorem, which is different from standard proofs based on similarity.

Measuring Angles in Radians

We have seen that (\cos (t),\sin(t)) are the coordinates of a point moving counterclockwise on the unit circle with unit velocity starting at (1,0) for t=0. Let us denote by \frac{\pi}{2} the smallest t for which \cos(t))=0 and \sin(t)=1.

By periodicity it follows that for t=2\pi the point will be back to (1,0) and thus the length of the circumference of a unit circle is equal to 2\pi. If we agree to measure the angle formed by the line from the origin to the point (\cos(t),\sin(t) by t, which represents the length of the circle arc from (1,0) to (\cos(t),\sin(t), then we measure the angle in the unit of radians. One revolution will then correspond to 2\pi radians. In other words 360^o = 360\quad\mbox{degrees}=2\pi radians, or

  • \mbox{one degree} = 1^o = \frac{\pi}{180}\quad\mbox {radians} .

We shall see below that the choice of y=(0,1) in fact covers the general case (since a general vector y of length 1 can be rotated to (0,1) by an orthogonal transformation which does not change the scalar product).

Angle vs Scalar Product

Let x=(x_1,x_2) and y=(1,0) be two points in the plane with corresponding vectors (or arrows) from the origin also denoted by x=(x_1,x_2) and y=(1,0). Since

  • x\cdot y=x_1=\vert x\vert \cos(t) ,

where t is the angle in radians between x and y. This formula extends to any two vectors x and y:

  • x\cdot y=\vert x\vert\vert y\vert\cos(t) ,

where t is the angle between the vectors.

Table of Values

To Think About

  • How are trigonometric functions defined in standard Calculus texts?
                                                                 Triangulations by Olle Baertling

Archimedes computed the value of \pi by inscribing and circumscribing polygons (octagons) toa circle. What value did he obtain?

Complex Numbers

We define \exp(it) = \cos(t)+i\sin(t) with i the imaginary unit satisfying i^2=-1. We compute

  • \frac{d}{dt}\exp (it) = i\exp(it),

and thus find that \exp(it) is the solution to the IVP

  • \dot u (t)=iu(t) for all real t with u(0)=1.

We have thus defined the exponential function \exp(it) for all imaginary numbers it. The rules for the usual exponential \exp(t) extend to \exp(it):

  • \exp(ia+ib)=\exp(ia)\exp(ib),   \exp(c(ia+ib))=\exp(ia+ib)^c

for real numbers a,b and c.


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