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October 31st, 2016, 04:54 AM   #1
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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!
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 October 31st, 2016, 05:36 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2266 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: 17,919 Thanks: 1385 Just change the last fully-worked 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
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Quote:
 Originally Posted by skipjack Just change the last fully-worked line to $\dfrac{2\sin(\theta)(-\sin(\theta) + i\cos(\theta))}{2\cos(\theta)(\cos(\theta) + i\sin(\theta))}$ which is obviously $i\tan(\theta)$.
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
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Quote:
 Originally Posted by Appletree 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.
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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