October 31st, 2016, 04:54 AM  #1 
Newbie Joined: Apr 2016 From: Wonderland Posts: 10 Thanks: 0  Solving for tan
Hi, I got stuck on a question. I have attached my working. I used the trigo formula, but I still couldn't solve it. Any help is appreciated, thank you!

October 31st, 2016, 05:36 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,539 Thanks: 2144 Math Focus: Mainly analysis and algebra 
For your final complete line, you factorised $2\sin \theta$ in the numerator and $2\cos \theta$ in the denominator, but you forgot to write these factors in: $\tan \theta$. The next move is probably to multiply top and bottom by the conjugate of your denominator: $\cos \theta  i\sin\theta$. Last edited by skipjack; October 31st, 2016 at 05:51 AM. 
October 31st, 2016, 06:02 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 16,766 Thanks: 1231 
Just change the last fullyworked line to $\dfrac{2\sin(\theta)(\sin(\theta) + i\cos(\theta))}{2\cos(\theta)(\cos(\theta) + i\sin(\theta))}$ which is obviously $i\tan(\theta)$. 
November 1st, 2016, 03:51 AM  #4 
Newbie Joined: Apr 2016 From: Wonderland Posts: 10 Thanks: 0  Sorry, but I couldn't get the part where the equation equals i. I know how to obtain tanx , but I'm at a loss at converting ( sinx + i cosx) to the required complex form.

November 1st, 2016, 07:08 PM  #5 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Notice $\displaystyle \begin{align*} 1 = \mathrm{i}^2 \end{align*}$, so write $\displaystyle \begin{align*} \sin{(x)} = \mathrm{i}^2\sin{(x)} \end{align*}$, and then you can factor out an $\displaystyle \begin{align*} \mathrm{i} \end{align*}$.


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