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
Explanation:
We have
now putting the initial conditions
and finally
[ 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
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,
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
If you're going to submit an inefficient answer, it may as well be interesting !
Explanation:
We have:
That solves trivially as:
mathbf s = e^(x M) mathbf s_o
Now, "what can we say":
Looking the matrices:
Symmetry....and Commutation
So: