# 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  ?

Feb 6, 2018

${c}^{2} + {s}^{2} = 1$

#### Explanation:

We have

$s ' = c \Rightarrow s \cdot s ' = s \cdot c$ and
$s = - c ' \Rightarrow c \cdot s = - c \cdot c '$ then

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

${c}^{2} + {s}^{2} = {C}_{0}$

now putting the initial conditions

${c}^{2} \left(0\right) + {s}^{2} \left(0\right) = 0 + 1 = {C}_{0} \Rightarrow {C}_{0} = 1$

and finally

${c}^{2} + {s}^{2} = 1$

Feb 6, 2018

${\left[s \left(x\right)\right]}^{2} + {\left[c \left(x\right)\right]}^{2} = 1$

#### Explanation:

We are given that:

$s ' \left(x\right) = c \left(x\right) \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ ..... [A]
$c ' \left(x\right) = - s \left(x\right) \setminus \setminus \setminus \setminus \setminus \setminus$ ..... [B]

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

$c ' ' \left(x\right) = - s ' \left(x\right)$

And then incorporating the first equation [A]:

$- c ' ' \left(x\right) = c \left(x\right)$

Or:

$c ' ' \left(x\right) + c \left(x\right) = 0$

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

${m}^{2} + 1 = 0 \implies m = \pm i$

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

$c \left(x\right) = A \cos x + B \sin x$

And then using [B] we have:

$s \left(x\right) = - c ' \left(x\right)$
$\therefore s \left(x\right) = - \left\{- A \sin x + B \cos x\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = A \sin x - B \cos x$

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

$A + 0 = 1 \implies A = 1$
$0 - B = 0 \implies B = 0$

Thus we have:

$c \left(x\right) = \cos x$
$s \left(x\right) = \sin x$

And so we infer that:

${\left[s \left(x\right)\right]}^{2} + {\left[c \left(x\right)\right]}^{2} = {\sin}^{2} x + {\cos}^{2} = 1$

Apr 29, 2018

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

#### Explanation:

We have:

$\left(\begin{matrix}s ' \\ c '\end{matrix}\right) = \left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) \left(\begin{matrix}s \\ c\end{matrix}\right)$

$\implies m a t h b f s ' = M m a t h b f s$

That solves trivially as:

• $m a t h b f s = {e}^{x M} m a t h b f {s}_{o}$

Now, "what can we say":

${s}^{2} + {c}^{2} = m a t h b f {s}^{T} m a t h b f s$

$= {\left({e}^{x M} m a t h b f {s}_{o}\right)}^{T} \left({e}^{x M} m a t h b f {s}_{o}\right)$

$= m a t h b f {s}_{o}^{T} {\left({e}^{x M}\right)}^{T} \cdot {e}^{x M} m a t h b f {s}_{o}$

$= m a t h b f {s}_{o}^{T} {e}^{x {M}^{T}} \cdot {e}^{x M} m a t h b f {s}_{o}$

Looking the matrices:

$M {M}^{T} = \left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right) = m a t h \boldsymbol{I} = {M}^{T} M$

Symmetry....and Commutation

$\implies {e}^{x {M}^{T}} \cdot {e}^{x M} = {e}^{x \left({M}^{T} + M\right)}$

$= m a t h b f {s}_{o}^{T} {e}^{x \left(\left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) + \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right)\right)} m a t h b f {s}_{o}$

$= m a t h b f {s}_{o}^{T} {e}^{x \left(0\right)} m a t h b f {s}_{o} = {s}_{o}^{2} = 1$

So:

${s}^{2} + {c}^{2} = 1$