# Analysis Question 2

## Homework Statement

Suppose f:R→R and g:R→R are both differentiable and that f'(x)=g(x) and g'(x)=-f(x) for all x ∈ R; f(0)=0 and g(0)=1.
Prove : (f(x))²+(g(x))²=1 for all x ∈ R.

## The Attempt at a Solution

I know I need the find d/dx[f(x)²+g(x)²]=d/dx, but I am not sure what that is going to help me find and how to use the result.

Proceed with d/dx[f(x)²+g(x)²]=d/dx and use the fact that f'(x)=g(x) and g'(x)=-f(x) for all x ∈ R to get an equality that is clearly true. Then see if you can make all of your steps reversible.

Prove that both f and g are twice differentiable.

Obtain the differential equations: f'' = -f and g'' = -g

Then you know that the function sinx, with initial conditions sin(0) = 0 and d(sin0)/dx = 1 and the funtion cosx with initial condition cos(0) = 1 and d(cos0)/dx = 0 are the solutions to the differential equations.

If you aren't familiar with defining cosine and sine with a differential equation, then just do what snipez advised. Assume that
d(f^2(x) + g^2(x))/dx = d(1)/dx until you get an equality that is true and try to reverse it. Also remember that if y' = 0 then y = c, where c is some real number.

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