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December 31st, 2016, 06:08 AM  #1 
Newbie Joined: Mar 2016 From: UK Posts: 23 Thanks: 1  Rearranging Equation Where Unknown Appears Twice
Hello, I'm struggling to rearrange an equation where I have an unknown appear twice. The example says: $\displaystyle \sin \left ( \phi _m \right )=\frac{\alpha1}{\alpha+1}$ $\displaystyle \therefore \alpha =\frac{1+\sin \left ( \phi_m \right )}{1\sin \left ( \phi_m \right )}$ When I rearrange I get: $\displaystyle \sin \left ( \phi _m \right )=\frac{\alpha1}{\alpha+1}$ $\displaystyle \sin \left ( \phi _m \right )\left ( \alpha+1 \right )=\alpha1$ Multiply out LHS: $\displaystyle \alpha\sin \left ( \phi _m\right )+\sin\left(\phi _m\right)=\alpha1$ Move $\displaystyle \alpha$ over to LHS: $\displaystyle \alpha\sin \left ( \phi _m\right )+\sin\left(\phi _m\right)\alpha=1$ Move $\displaystyle \sin\left(\phi_m\right)$ over to RHS: $\displaystyle \alpha\sin \left ( \phi _m\right )\alpha=\sin\left(\phi _m\right)1$ Now the $\displaystyle \alpha$ terms can be factorised: $\displaystyle \alpha\left(\sin\left(\phi_m\right)1\right)=\sin\left(\phi_m\right)1$ And finally, divide LHS by $\displaystyle \sin\left(\phi_m\right)$: $\displaystyle \alpha=\frac{\sin\left(\phi_m\right)1}{\sin\left(\phi_m\right)1}$ So I suppose I'm asking, how is: $\displaystyle \frac{1+\sin \left ( \phi_m \right )}{1\sin \left ( \phi_m \right )}\equiv\frac{\sin\left(\phi_m\right)1}{\sin\left(\phi_m\right)1}$ If I plug $\displaystyle \phi_m=45$, I get the same for each one. I look forward to hearing from you. Kind Regards Edward 
December 31st, 2016, 06:35 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 998 Thanks: 410 
$GIVEN:\ \alpha = \dfrac{\ \sin( \theta_m)  1}{\sin( \theta_m )  1} \implies$ $\alpha = \dfrac{( 1) * \{ \sin( \theta_m) + 1\}}{\sin( \theta_m )  1} \implies$ $\alpha = \dfrac{( 1) * \{ \sin( \theta_m) + 1\}}{(\ 1) * \{\ \sin( \theta_m ) + 1\}} \implies$ $\alpha = \dfrac{\cancel {(\ 1)} * \{ \sin( \theta_m) + 1\}}{\cancel {(\ 1)} * \{\ \sin( \theta_m ) + 1\}} \implies$ $\alpha = \dfrac{\sin( \theta_m) + 1}{\ \sin( \theta_m ) + 1} \implies$ $\alpha = \dfrac{1 + \sin( \theta_m)}{1 \ \sin( \theta_m )}.$ Last edited by skipjack; December 31st, 2016 at 05:00 PM. 
December 31st, 2016, 07:09 AM  #3 
Newbie Joined: Mar 2016 From: UK Posts: 23 Thanks: 1 
@JeffM1 Thank you Jeff, I forgot that: $\displaystyle \frac{1}{1}=1$ And therefore, by multiplying the expression I had by it, you could arrive at the same equation that the example had. Kind Regards Edward 
December 31st, 2016, 03:46 PM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,579 Thanks: 541 Math Focus: Yet to find out. 
If this is from Ogata (5E), the examples on page 507 and 558 might help.

January 2nd, 2017, 01:54 PM  #5  
Newbie Joined: Mar 2016 From: UK Posts: 23 Thanks: 1  Quote:
EDIT: I've been able to find a copy of the eBook and the solutions on Academia.edu. Kind Regards Edward Last edited by edwardholmes91; January 2nd, 2017 at 02:10 PM.  
January 2nd, 2017, 04:33 PM  #6  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,579 Thanks: 541 Math Focus: Yet to find out.  Quote:
Dorf also uses some different, albeit simpler methods for designing lead/lag compensators. At the time it really helped me to switch between these two books for understanding the design process better.  
January 2nd, 2017, 10:49 PM  #7 
Newbie Joined: Mar 2016 From: UK Posts: 23 Thanks: 1 
@Joppy, thank you for the other reference, I really appreciate it 

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