So, here's a question:
$ \cos( \omega t ) + 2 \cos( \omega t + \frac{\pi}{4} ) + 3 \cos( \omega t + \frac{\pi}{2} ) $
To add these together, I figure there should be at least 2 ways:
1) Cosine addition laws:
$$
\cos( \omega t ) +
2 \left(
\cos( \omega t ) \cos( \frac{\pi}{4} ) –
\sin( \omega t ) \sin( \frac{\pi}{4} )
\right)
+ 3 \left(
\cos( \omega t ) \cos( \frac{\pi}{2} ) –
\sin( \omega t ) \sin( \frac{\pi}{2} )
\right) \\
=\cos( \omega t )
\left(
1 + \sqrt{2}
\right)
–
\sin( \omega t )
\left(
3 + \sqrt{2}
\right)
$$
2) Phasors / complex addition
$$
1 \angle 0 + 2 \angle 45 ^\circ + 3 \angle 90^\circ
$$
$$
= 1 + \sqrt{2} + j \sqrt{2} + j 3
$$
$$
= 1 + \sqrt{2} + j ( 3 + \sqrt{2} )
$$
Which has
$ A = \sqrt{ 14 + 8 \sqrt{2} } \approx 5.03 $
$ \phi = \arctan{ \left( \frac{ 3 + \sqrt{2} }{ 1 + \sqrt{2} } \right) } \approx 1.07 rad \approx 61 ^\circ $
Thus answer is $ 5 \angle 61^\circ $, or $5 \cos( \omega t + 1.07 )$
If you graph them, $5 \cos( \omega t + 1.07 )$ produces the same graph as $ \cos( \omega t )
\left( 1 + \sqrt{2} \right) – \sin( \omega t ) \left( 3 + \sqrt{2} \right) $
So how can you convert between them?
Best Answer
You have $$ \tan\phi = \frac{\sin\phi}{\cos\phi}= \frac{3+\sqrt{2}}{1+\sqrt{2}} $$ Since $0<\phi<\pi/2$ we know that $\sin\phi,\cos\phi>0$. Therefore, $\sin\phi$ and $\cos\phi$ are equal to $$ \sin\phi = \frac{3+\sqrt{2}}{\sqrt{(3+\sqrt{2})^2+(1+\sqrt{2})^2}} =\frac{3+\sqrt{2}}{\sqrt{14+8\sqrt{2}}}$$ and $$ \cos\phi = \frac{1+\sqrt{2}}{\sqrt{(3+\sqrt{2})^2+(1+\sqrt{2})^2}} =\frac{1+\sqrt{2}}{\sqrt{14+8\sqrt{2}}}$$ Therefore $$ \begin{split} \cos(\omega t)(1+\sqrt{2}) - \sin(\omega t)(3+\sqrt{2}) &= \sqrt{14+8\sqrt{2}}\left( \cos(\omega t)\cos\phi - \sin(\omega t) \sin\phi \right)\\& = \sqrt{14+8\sqrt{2}} \cos(\omega t+\phi) \end{split} $$