Prove cos^2(x)+sin^2(x)=1 using IVP's

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Shadowl08
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Ok so for this problem I have to use IVPs to prove that cos^2(x)+sin^2(x)=1. I know the end result is suppose to be:
du/dt= - v, u(0)=1
dr/dt= u, v(0)=0
but I have no idea how to go about getting to this point.
 
Are you aware of Euler's formula, or identity, or have any idea on how to derive Euler's?

Oh, I think you might have to use a Taylor expansion. I remember doing something similar last quarter, but it isn't quite coming back to me yet.
 
I have used Euler's method before but it was based on a program. As for deriving Euler's I am unsure of that. I don't think it would be tayler expansion solely because we learned that after I received an assignment with this question.
 
They ask you to do it by integration by parts? What you can do is take the derivative of sin^2x +cos^2x and finding that it is a constant function, then plugging in those values to verify, but as for doing it by IVP I am unsure, I'm sorry.
 
yea sadly instead of integration by parts it specifies IVP-Initial value problem. Thanks for trying though :)
 
Shadowl08 said:
Ok so for this problem I have to use IVPs to prove that cos^2(x)+sin^2(x)=1. I know the end result is suppose to be:
du/dt= - v, u(0)=1
dr/dt= u, v(0)=0
but I have no idea how to go about getting to this point.
Do you know how to solve a differential equation in matrix form?

Your system can be written as
$$\begin{bmatrix} u \\ r \end{bmatrix}'= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} u \\ v\end{bmatrix}$$
with the initial condition
$$\begin{bmatrix} u(0) \\ r(0)\end{bmatrix} = \begin{bmatrix} 1 \\ 0\end{bmatrix}$$
Solving this matrix equation involves diagonalizing the matrix by finding its eigenvalues and eigenvectors.
 
Shadowl08 said:
Ok so for this problem I have to use IVPs to prove that cos^2(x)+sin^2(x)=1. I know the end result is suppose to be:
du/dt= - v, u(0)=1
dr/dt= u, v(0)=0
but I have no idea how to go about getting to this point.

That's the starting point. The quantity you want to show is constant is [itex]u^2 + v^2[/itex]. So what does the chain rule give you for [tex] \frac{d}{dt}(u^2 + v^2)[/tex]