Trigonometric functions from their differential equations

[lpetrich]

Michael Penn runs a nice YouTube channel: (3) Michael Penn - YouTube

I recall him making a video about deriving trigonometric functions' properties from their differential equations, and I will redo that reasoning here.

First, a simpler version of that reasoning, with the exponential function:

y' = y

For independent variable x, call the solution e(x). Since the integration constant multiplies it, let us take e(0) = 1 for definiteness.

It is easy to show that a shifted solution e(x+y) is also a solution, and that it must differ by a constant:

e(x+y) = A*e(x) = B*e(y) = C*e(x)*e(y)

From e(0) = 1, e(x+y) = e(x)*e(y) -- the exponential function

Let's consider its inverse function, ie: e(ie(x)) = ie(e(x)) = x. Take the derivative:

e'(ie(x)) * ie'(x) = 1 -- e(ie(x) * ie'(x) = 1
ie'(e(x)) * e'(x) = 1 -- ie'(e(x)) * e(x) = 1

giving ie'(x) = 1/x -- ie(x) is the natural-logarithm function of x

 

[lpetrich]

Let's extend this reasoning further with y'' = y

It has a linear combination of two solutions: c(x) and s(x) which can be fixed with c(0) = s'(0) = 1 and c'(0) = s(0) = 0.

Consider c(x+y) and s(x+y) again. In general, they are A*c(x)*c(y) + B*c(x)*s(y) + C*s(x)*c(y) + D*s(x)*s(y)

y = 0 gives us c(x), s(x) = A*c(x) + C*s(x) and x = 0 gives us c(y), s(y) = A*c(y) + B*s(y)
and thus c(x+y) = c(x)*c(y) + D*s(x)*s(y) and s(x+y) = c(x)*s(y) + s(x)*c(y) + D*s(x)*s(y)

Taking d/dy and then y = 0 gives us c'(x) = Dc*s(x) and s'(x) = c(x) + Ds*s(x)
Combining them and then using the defining differential equation,
c''(x) = Dc*(c(x) + Ds*s(x)) = c(x)

Thus,
c(x+y) = c(x)*c(y) + s(x)*s(y)
s(x+y) = c(x)*s(y) + s(x)*c(y)
c'(x) = s(x)
s'(x) = c(x)

Set Q(x) = c(x)^2 - s(x)^2
We find that Q'(x) = 0 and thus Q is a constant. Evaluating it at 0 gives Q = 1
Thus, c(x)^2 - s(x)^2 = 1

With y = -x,
1 = c(x)*c(-x) + s(x)*s(-x)
0 = c(x)*s(-x) + s(x)*c(-x)

Solving for c(-x) and s(-x) gives us
c(-x) = c(x)
s(-x) = - s(x)

Taking combinations
cs1(x) = c(x) + s(x) and cs2(x) = c(x) - s(x)
we find
cs1(x+y) = cs1(x)*cs1(x) and cs2(x+y) = cs2(x)*cs2(y)
cs1'(x) = cs1(x) and cs2'(x) = - cs2(x)

Thus,
cs1(x) = e(x) and cs2(x) = e(-x)
and
c(x) = (1/2)*(e(x) + e(-x))
s(x) = (1/2)*(e(x) - e(-x))

So I've rediscovered hyperbolic functions.

 

[lpetrich]

I now turn to y'' = - y and proceed like with hyperbolic functions.

Use constraints c(0) = 1, s(0) = 0, c'(0) = 0, s'(0) = 1

With c(x), s(x) = A*c(x)*c(y) + B*c(x)*s(y) + C*s(x)*c(y) + D*s(x)*s(y)
we get
c(x+y) = c(x)*c(y) + Dc*s(x)*s(y) and s(x+y) = c(x)*s(y) + s(x)*c(y) + Ds*s(x)*s(y)

Taking d/dy and then y = 0,
c'(x) = Dc*s(x) and s'(x) = c(x) + Ds*s(x)

Combining and using the defining differential equation:
c''(x) = Dc*(c(x) + Ds*s(x)) = - c(x)

Giving us
c(x+y) = c(x)*c(y) - s(x)*s(y)
s(x+y) = c(x)*s(y) + s(x)*c(y)
c'(x) = - s(x)
s'(x) = c(x)

This time, Q(x) = c(x)^2 + s(x)^2. Here also, Q'(x) = 0 and Q(0) = 1. Thus,
c(x)^2 + s(x)^2 = 1

With y = -x,
1 = c(x)*c(-x) + s(x)*s(-x)
0 = c(x)*s(-x) + s(x)*c(-x)
giving us
c(-x) = -c(x)
s(-x) = -s(x)

Taking combinations
cs1(x) = c(x) + i*s(x) and cs2(x) = c(x) - i*s(x)
we find
cs1(x+y) = cs1(x)*cs2(x) and cs2(x+y) = cs2(x)*cs2(y)
cs1'(x) = cs1(x) and cs2'(x) = - cs2(x)

Thus,
cs1(x) = e(i*x) and cs2(x) = e(-i*x)
and
c(x) = (1/2)*(e(i*x) + e(-i*x))
s(x) = (1/(2i))*(e(i*x) - e(-i*x))

 

[lpetrich]

Repeating what we did with the exponential function, and using the identities that relate c(x) and s(x), we find:

The inverse functions of hyperbolic c(x), s(x):
ic'(x) = 1/sqrt(x^2 - 1)
is'(x) = 1/sqrt(x^2 + 1)

The inverse functions of trigonometric c(x), s(x):
ic'(x) = - 1/sqrt(1 - x^2)
is'(x) = 1/sqrt(1 - x^2)

The trig functions have some additional features, related to how they increase or decrease with increasing arg.

The exponential function increases for increasing arg, since its derivative is positive, from it itself being positive.

For arg > 0, the hyperbolic functions are positive with positive derivs, making them increasing for increasing arg.

But trigonometric functions are a different story.

Start at x = 0 and increase. c(x) is initially > 0 and it makes s(x) increase. But c(x) is decreasing as a result of s(x) being > 0, and there will be a certain x value where it is 0. Call that value q.

c(q) = 0, s(q) = 1 from the sum-of-squares identity and s(x) increasing for increasing x.

From the addition identities,
c(x+q) = -s(x) and s(x+q) = c(x)
c(x+2q) = -c(x) and s(x+2q) = -s(x)
c(x+3q) = s(x) and s(x+3q) = -c(x)
c(x+4q) = c(x) and s(x+4q) = s(x)

Thus, these two functions are periodic with period 4q. One can find q with their inverse functions:

q = integral over x from 0 to 1 of 1/sqrt(1 - x^2)

From the definition of pi, q = pi/2.