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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
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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.
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October 31st, 2016, 06:02 AM   #3
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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 $i\tan(\theta)$.
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November 1st, 2016, 03:51 AM   #4
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Quote:
Originally Posted by skipjack View Post
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.
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November 1st, 2016, 07:08 PM   #5
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Quote:
Originally Posted by Appletree View Post
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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