Proving Acosx+Bsinx=sqrt(A^2+B^2)sin(x+alpha) | Trig Proof Homework Help

[blue mango]

Homework Statement

prove that Acosx+Bsinx=sqrt(A^2+B^2)sin(x+alpha) where tan(alpha)=A/B

The Attempt at a Solution

none so far except that sin(x+alpha)=sinx cos(alpha)+cosx sin(alpha)

Any help is appreciated. This is due tomorrow (It was just assigned yesterday).
I've taken a year off school so I'm kind of out of the swing of stuff like this.

Thanks

 

[Mark44]

I would start on the right side and see if you can show that it's equal to the left side. It might be useful to note that tan(alpha) = sin(alpha)/cos(alpha) = A/B = (A/sqrt(A^2 + B^2))/(B/sqrt(A^2 + B^2)).

I haven't worked this out, but that's where I would start.

 

[ehild]

Homework Statement

prove that Acosx+Bsinx=sqrt(A^2+B^2)sin(x+alpha) where tan(alpha)=A/B

The Attempt at a Solution

none so far except that sin(x+alpha)=sinx cos(alpha)+cosx sin(alpha)
Thanks

Denoting sqrt(A^2+B^2)=C, you have the equation

Acosx + Bsinx =(C*sin(alpha))cos x +(C*cos(alpha))sinx This has to be an identity, valid for any x. If x=0, sinx=0, cosx=1,
A=C*sin(alpha). If x=pi/2, sinx=1, cosx=0, and B=C*cos(alpha).

So you have two equations for C and alpha:

A=C*sin(alpha)
B=C*cos(alpha)

Square both of equations and add the together. Use that cos²x+sin²x =1 what do you get for C?

Divide the first equation by the second one, what do you get for alpha?

ehild

 

[Mark44]

Denoting sqrt(A^2+B^2)=C, you have the equation

Acosx + Bsinx =(C*sin(alpha))cos x +(C*cos(alpha))sinx


This has to be an identity, valid for any x.

But that's what the OP needs to show. He/she can't just assume that it is true for all x.

If x=0, sinx=0, cosx=1,
A=C*sin(alpha). If x=pi/2, sinx=1, cosx=0, and B=C*cos(alpha).

So you have two equations for C and alpha:

A=C*sin(alpha)
B=C*cos(alpha)

Square both of equations and add the together. Use that cos²x+sin²x =1 what do you get for C?

Divide the first equation by the second one, what do you get for alpha?

ehild

 

[ehild]

Mark, you start with the right side, I would start with the left one, and show that it can be written in the form of C*sin(x+alpha) (that means it is identical to C*sin(x+alpha)) for a certain C and alpha. If it comes out that C^2=A^2+B^2 and tan(alpha)=A/B it means that the left side is identical to the right side of the original equation that had to be proved.

ehild

 

[Mark44]

OK, I misunderstood what you were saying.

 

[blue mango]

thank you for your suggestions but I'm still a bit confused. It has been a long time since I've done stuff like this so its a bit harder for me to understand than usual. I'd appreciate if someone spelled out a few steps at least to get me started (if you don't mind) and hopefully I can figure out the rest. Thanks again.

 

[Mark44]

Basically what you should do is to start from the left side, as ehild suggests, and work with Acosx + Bsinx to make it look like C(A/C *cosx + B/C*sin x). Then you want to write A/C as sin(alpha) and B/C as cos(alpha) so that you end up with Csin(x + alpha).

 

[blue mango]

I GOT IT! Thank you so much...you were extraordinarily helpful :)