Question

let c and s both be continuous function
if \[s'(x)=c(x)~~,~~ c'(x)=-s(x) ~~and~~ s(0)=0,~~c(0)=1\]
prove that
\[[s(x)]^2+[c(x)]^2=1\]

Answer 1

i know this sin and cos
i'm stuck oh how to show they are

Answer 2

the conditions satisfy sin and cos functions

Answer 3

i was thinking of adding the two derivative

Answer 4

\[\begin{cases}s'(x)=c(x)&(1)\\c'(x)=-s(x)&(2)\end{cases}\]
Differentiating (1) gives \(s''(x)=c'(x)\). Substituting into (2) gives
\[s''(x)=-s(x)~~\iff~~s''(x)+s(x)=0\]
which is a second order ODE with characteristic equation \(r^2+1=0\), roots \(r=\pm i\), which generate the fundamental solution set \(\{\cos x,\sin x\}\), like you said.

Alternatively, you can solve this as you would a system of ODEs.
\[\begin{pmatrix}s(x)\\c(x)\end{pmatrix}'=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}s(x)\\c(x)\end{pmatrix}\]
The coefficient matrix has eigenvalues \(\lambda=\pm i\), which give eigenvectors \(\vec{\eta}=\begin{pmatrix}\mp i\\1\end{pmatrix}\), again giving you the same solution set.

Answer 5

Slight mistake, the bottom left 1 in the coefficient matrix should be negative.

Answer 6

i got what you did but how would we prove
the identity (this is not a DE question)
it is cal2 problem

Answer 7

you know i kind of went to that DE too but idk how it could help that way

Answer 8

my prof said do implicit differentiation to the identity we want to prove
and some other stuff, but i disagree with him
we need to prove that holds so we can use the identity itself we need to arrive at it

Answer 9

can't*

Answer 10

Suppose we try a reverse approach to what your professor suggested.
\[\begin{align*}\begin{cases}s'(x)=c(x)\\c'(x)=-s(x)\end{cases}~~\implies~~-s(x)s'(x)&=c(x)c'(x)\\
-2s(x)s'(x)&=2c(x)c'(x)\\
0&=2c(x)c'(x)+2s(x)s'(x)\\
0&=\frac{d}{dx}\left[(c(x))^2+(s(x))^2\right]\\
C&=(c(x))^2+(s(x))^2
\end{align*}\]
Now when \(x=0\), you have
\[\begin{align*}C&=(c(x))^2+(s(x))^2\\
C&=1^2+0^2\\
C&=1\end{align*}\]

Answer 11

yes that's good!
that is what i was looking for
i thought of summing both equations but i didn't go further

Answer 12

thanks

Answer 13

yw

Answer 14

i was trying with a similar problem hehe
\[f(x+y)=f(x)f(y)-g(x)g(y)~~and ~~g(x+y)=f(x)g(y)+g(x)f(y)\]
and again i have to prove
\[[f(x)]^2+[g(x)]^2=1\]

Answer 15

there are similar we have f(0)=1 g(0)=0

Answer 16

i will see how i would solve this one!

Answer 17

I'm not too experienced with functional equations, but if I were to guess you might be able to find a form of a partial derivative if you play around with the terms.

Answer 18

hmm partial derivative! unfortunately i forgot how to play with calc3 stuff
i will see what i can do :)
thanks

Answer 19

my thought for now is to let x=y
and then see what i can pull from that

Answer 20

trying this
\[\begin{cases}f(x+y)=f(x)f(y)-g(x)g(y)\\g(x+y)=g(x)f(y)+g(y)f(x)\end{cases} \Longrightarrow \begin{cases}[f(x+y)]^2=[f(x)f(y)-g(x)g(y)]^2\\ [g(x+y)]^2=[g(x)f(y)+g(y)f(x)]^2\end{cases}\]

Answer 21

but this is not using differentiation lol

Answer 22

if i add them up i should get
\[[f(x+y)]^2+[g(x+y)]^2=[f(x)+g(x)][f(y)+g(y)]\]
since g(0)=0
and f(0)=1
then we 1

Answer 23

eh but this a flow didn't pay attention to left hand side lol

Answer 24

how about using differentiation from that point?

Answer 25

eh that must be incorrect
if i let y=0
\[[g(x)]^2+[f(x)]^2=f(x)+g(x)\]
this wouldn't make any sense darn

Answer 26

hello maki san :)

Answer 27

by the way the question we are doing is on the comment 14th

Answer 28

@Kainui have a look at this :)