How to go from this equation in terms of cosine and sine to this one in terms of only one.

Question

$$x=A\cos(\omega t)+B\sin(\omega t )\equiv\mu \cos(\omega t+\phi)$$

I'm thinking it must have something to do with the double angle forumla. Any help?

Answer 1

$$A\cos(\omega t)+B\sin(\omega t)=\sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\cos(\omega t)+\frac{B}{\sqrt{A^2+B^2}}\sin(\omega t)\right)$$ Then let's denote $\frac{A}{\sqrt{A^2+B^2}}$ as $\cos\phi$ and $-\frac{B}{\sqrt{A^2+B^2}}$ as $\sin\phi$. Such $\phi$ always exists because $\left(\frac{A}{\sqrt{A^2+B^2}}\right)^2+\left(\frac{B}{\sqrt{A^2+B^2}}\right)^2=1$.
Then $\cos\phi\cos(\omega t)-\sin\phi\sin(\omega t)$ forms $\cos(\omega t+\phi)$.

Answer 2

$\mu \cos(\omega t+\phi)=\mu(\cos(\omega t) \cos(\phi)-\sin(\omega t) \sin( \phi))= [\mu \cos(\phi)] \cdot \cos(\omega t)+[-\mu sin(\phi)] \cdot \sin(\omega t)=A \cos(\omega t)+B \sin(\omega t) \\ \text{ when} A=\mu \cos( \phi) \text{ and } B=-\mu \sin(\phi) \\ A^2+B^2=\mu^2(\cos^2(\phi)+\sin^2(\phi))=\mu^2(1)=\mu^2 $

Answer 3

you can also get there by substituting handy values for $t$

$t=0$ gives $A = \mu \cos(\phi)$

$t=\frac {\pi}{2\omega}$ gives $B = - \mu \sin(\phi)$

( using $cos(\frac{\pi}{2} + x) = -\sin(x)$ )