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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) $

same graph

So how can you convert between them?

1 Answers 1

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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} $

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    @pharmine you're right, I've fixed it.2011-11-05