Suppose s(x) and c(x) are 2 functions where: 1) s'(x) = c(x) and c'(x) = -s(x); 2) s(0) = 0 and c(0) = 1. What can you say about the quantity: \qquad [ s(x) ]^2 + [ c(x) ]^2 ?

3 Answers
Feb 6, 2018

c^2+s^2=1

Explanation:

We have

s'=c rArr s cdot s' = s cdot c and
s=-c' rArr c cdot s = -c cdot c' then

s cdot s'+c cdot c' = 0 or

c^2+s^2=C_0

now putting the initial conditions

c^2(0)+s^2(0) = 0+1=C_0 rArr C_0 = 1

and finally

c^2+s^2=1

Feb 6, 2018

[ s(x) ]^2 + [ c(x) ]^2 =1

Explanation:

We are given that:

s'(x) = c(x) \ \ \ \ \ \ \ \ \ \ \ ..... [A]
c'(x) = -s(x) \ \ \ \ \ \ ..... [B]

Differentiating the second equation [B] wrt x we get:

c''(x) = -s'(x)

And then incorporating the first equation [A]:

-c''(x) = c(x)

Or:

c''(x) + c(x) = 0

Which is a Second Order ODE with constant coefficients, so we consider the associated Auxiliary equation:

m^2 + 1 = 0 => m= +- i

So as we have two pure imaginary roots, the solution is of the

c(x) =Acosx + Bsinx

And then using [B] we have:

s(x) = -c'(x)
:. s(x) = -{-Asinx+Bcosx}
\ \ \ \ \ \ \ \ \ \ \ = Asinx-Bcosx

Using the given condition, c(0)=1 and s(0)=0 we have:

A + 0 = 1 => A = 1
0-B = 0 => B=0

Thus we have:

c(x) = cosx
s(x) = sinx

And so we infer that:

[ s(x) ]^2 + [ c(x) ]^2= sin^2x+cos^2 =1

Apr 29, 2018

If you're going to submit an inefficient answer, it may as well be interesting !

Explanation:

We have:

((s'),(c')) = ((0, 1),(-1,0)) ((s),(c))

implies mathbf s' = M mathbf s

That solves trivially as:

  • mathbf s = e^(x M) mathbf s_o

Now, "what can we say":

s^2 + c^2 = mathbf s^T mathbf s

= (e^(x M) mathbf s_o)^T (e^(x M) mathbf s_o)

= mathbf s_o ^T (e^(x M))^T * e^(x M) mathbf s_o

= mathbf s_o ^T e^(x M^T) * e^(x M) mathbf s_o

Looking the matrices:

M M^T = ((0, 1),(-1,0)) ((0, -1),(1,0)) = mathbb I = M^T M

Symmetry....and Commutation

implies e^(x M^T) * e^(x M)= e^(x (M^T+ M))

= mathbf s_o ^T e^(x (((0, 1),(-1,0)) + ((0, -1),(1,0)))) mathbf s_o

= mathbf s_o ^T e^(x( 0)) mathbf s_o = s_o^2 = 1

So:

s^2 + c^2 = 1