Finding Coefficients in Trigonometric Functions

[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

 

[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?

 

[Bob19]

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?

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

 

[TD]

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

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.

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

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

 

[Bob19]

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

/Bob

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).

 

[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

 

[Bob19]

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

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

Right ?

/Bob