# Thread: Rewrite Acos(wt)+Bsin(wt) as Ccos(wt+D)

1. ## Rewrite Acos(wt)+Bsin(wt) as Ccos(wt+D)

Ok, so I keep seeing textbooks handwave over this:

Constants $A$, $B$, $C$, and $D$ exist such that
$A \cos{(\omega t)}+B \sin{(\omega t)}$
can be rewritten as
$C \cos{(\omega t -D)}$

But I haven't bee able to find/figure out the derivation of this. Can someone help me fill in the details of this?

P.S. I found one book that added $C=\sqrt{A^2+b^2}$ and $D=\tan^{-1}{(B/A)}$.

2. Originally Posted by teuthid
Ok, so I keep seeing textbooks handwave over this:

Constants $A$, $B$, $C$, and $D$ exist such that
$A \cos{(\omega t)}+B \sin{(\omega t)}$
can be rewritten as
$C \cos{(\omega t -D)}$

But I haven't bee able to find/figure out the derivation of this. Can someone help me fill in the details of this?

P.S. I found one book that added $C=\sqrt{A^2+b^2}$ and $D=\tan^{-1}{(B/A)}$.
Let A = C*cos(D) and B = C*sin(D)

Then $A^2 + B^2 = C^2$

$C = \sqrt{A^2 + B^2}$

Acos(ωt) + Bsin(ωt) = $(\sqrt{A^2 + B^2})(\frac{A}{\sqrt{A^2 + B^2}}\cos(\omega{t}) + \frac{B}{\sqrt{A^2 + B^2}}\sin(\omega{t}))$

Now proceed.

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# differentiate y= Acoswt Bsinwt

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