- #1

- 71

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Bob19
- Start date

- #1

- 71

- 0

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

LeonhardEuler

Gold Member

- 859

- 1

- #3

- 71

- 0

LeonhardEuler said:

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

TD

Homework Helper

- 1,022

- 0

No, these are just coefficients so they can be higher than 1.Bob19 said:Since its trigonometry then A,B,C,D can't higher than one?

It's cos(a) (and sin(a))) which are bounded between -1 and 1.

Bob19 said: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).

- #5

- 71

- 0

/Bob

TD said: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).

- #6

TD

Homework Helper

- 1,022

- 0

[tex]\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}[/tex]

Rearranging gives:

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

We want this to equal [itex]F'\left( x \right) = x\cos x[/itex]

So:

[tex]\left( {A + D} \right)\cos x + \left( {B - C} \right)\sin x - Ax\sin x + Bx\cos x = x\cos x[/tex]

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

- #7

- 71

- 0

TD said:

[tex]\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}[/tex]

Rearranging gives:

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

We want this to equal [itex]F'\left( x \right) = x\cos x[/itex]

So:

[tex]\left( {A + D} \right)\cos x + \left( {B - C} \right)\sin x - Ax\sin x + Bx\cos x = x\cos x[/tex]

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

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

Right ?

/Bob

Share: