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

$sin[wt+arctan\frac{R}{\omega L}] = cos[\omega t+arctan\frac{-\omega L}{R})]$

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

  • $\begingroup$ Please learn MathJax. $\endgroup$ – Em. Oct 21 '16 at 5:35


$\cos \theta = \sin \left(\dfrac\pi2-\theta\right)$

$\arctan x + \arctan\left(\dfrac1x\right) = \dfrac\pi2$ (if $x>0$)

$\arctan(-x) = -\arctan x$

$\cos(-\theta) = \cos\theta$

  • $\begingroup$ 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. $\endgroup$ – nbstrong Oct 21 '16 at 16:37
  • $\begingroup$ @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. $\endgroup$ – tilper Oct 21 '16 at 17:17

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