0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Please learn MathJax. $\endgroup$ – Em. Oct 21 '16 at 5:35
1
$\begingroup$

Hints:

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

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

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.