- #1

Bob19

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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

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- Thread starter Bob19
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- #1

Bob19

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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

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- #3

Bob19

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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

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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

Bob19

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/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

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[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

Bob19

- 71

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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

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