# Trigonometri Question (HELP)

1. Aug 31, 2005

### Bob19

Hi

I have this here Trigonometri function:

F(x) = Ax cos(x) + Bx sin(x) + C cos(x) + D sin(x)

where F'(x) = x cos(x)

I'm suppose to find A, B, C , D any hints or idears on how I do that?

Sincerely
Bob

2. Aug 31, 2005

### LeonhardEuler

Take the derivative of F(x). You know that F'(x)=x cos(x). This means the coefficient of (x cos(x)) should be 1. What can you say about the coefficients of all other terms?

3. Aug 31, 2005

### Bob19

Since its trigonometry then A,B,C,D can't higher than one?

F'(x) = A (cos(x) - x *sin(x)) + B(x *cos(x) + sin(x)) + C ( -sin(x)) + D(cos(x) = x cos(x)

But how I proced from here ?

/Bob

4. Aug 31, 2005

### TD

No, these are just coefficients so they can be higher than 1.
It's cos(a) (and sin(a))) which are bounded between -1 and 1.

Your derivative seems correct, now just check what the coefficients have to be to get xcos(x).

5. Aug 31, 2005

### Bob19

The best way of doing this is that to make the left side of the expression shorter ?

/Bob

6. Aug 31, 2005

### TD

Well, we have:

$$\begin{array}{l} F\left( x \right) = Ax\cos x + Bx\sin x + C\cos x + D\sin x \\ F'\left( x \right) = A\left( {\cos x - x\sin x} \right) + B\left( {\sin x + x\cos x} \right) - C\sin x + D\cos x \\ \end{array}$$

Rearranging gives:

$$F'\left( x \right) = \left( {A + D} \right)\cos x + \left( {B - C} \right)\sin x - Ax\sin x + Bx\cos x$$

We want this to equal $F'\left( x \right) = x\cos x$

So:

$$\left( {A + D} \right)\cos x + \left( {B - C} \right)\sin x - Ax\sin x + Bx\cos x = x\cos x$$

That should give a very simple system for the unknown coefficients A -> D

7. Aug 31, 2005

### Bob19

Thanks then A = 0 , D = 0, B = C = 1

Right ?

/Bob