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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

$$ 1 \angle 0 + 2 \angle 45 ^\circ + 3 \angle 90^\circ $$

$$ = 1 + \sqrt{2} + j \sqrt{2} + 3 i $$

$$ = 1 + \sqrt{2} + j ( 3 + \sqrt{2} ) $$

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

How can you convert between them?

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1 Answer 1

up vote 3 down vote accepted

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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$\sqrt{14+8\sqrt{2}}$, not $\sqrt{14+\sqrt2}$ –  pharmine Nov 5 '11 at 9:03
    
@pharmine you're right, I've fixed it. –  Heike Nov 5 '11 at 9:56
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