My circuit analysis teacher is asking us to prove for extra credit:

sin[wt+arctanRωL]=cos[ωt+arctan−ωLR)]sin[wt+arctan\frac{R}{\omega L}] = cos[\omega t+arctan\frac{-\omega L}{R})]

w = omega

t = time

R = resistant

L = inductance

Ive been working at it for a couple of hours and I cannot make any headway. Would anyone be able to point me in the correct direction?

Thank you for your time

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Please learn MathJax.

– Max

2 days ago

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1 Answer

1

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Hints:

cosθ=sin(π2−θ)\cos \theta = \sin \left(\dfrac\pi2-\theta\right)

arctanx+arctan(1x)=π2\arctan x + \arctan\left(\dfrac1x\right) = \dfrac\pi2 (if x>0x>0)

arctan(−x)=−arctanx\arctan(-x) = -\arctan x

cos(−θ)=cosθ\cos(-\theta) = \cos\theta

Thank you, I think the negative in the cos’s arctan is incorrect though. Without it I am able to make them equal but with it the two arctan’s just cancel themselves.

– MushinZero

2 days ago

@MushinZero the negative sign in the problem is correct. I’ll add another hint related to it. Let me know if you still need help with it.

– tilper

2 days ago

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Thank you, that did it

– MushinZero

2 days ago