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How to prove: $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$ Help please. Don't know where to start.

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    @Pheter I think your question has been answered now. Could you please select one of the answers that you think is the best? (click on the arrow to the left of the answer start, which will turn green). This will close the question on the community and stop it from popping to the front of the list.2014-05-05

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Set $a=b=\theta $ in the identity $\begin{equation*} \sin (a+b)=\sin a\cdot \cos b+\cos a\cdot \sin b \end{equation*}$ to get this one $\begin{equation*} \sin 2\theta =2\sin \theta \cdot \cos \theta . \end{equation*}$ Then divide the RHS by $\sin ^{2}\theta +\cos ^{2}\theta =1$ and afterwards both numerator and denominator by $\cos ^{2}\theta \neq 0$ $\begin{equation*} \sin 2\theta =\dfrac{2\sin \theta \cdot \cos \theta }{\sin ^{2}\theta +\cos ^{2}\theta }=\dfrac{2\dfrac{\sin \theta \cdot \cos \theta }{\cos ^{2}\theta }}{ \dfrac{\sin ^{2}\theta +\cos ^{2}\theta }{\cos ^{2}\theta }}, \end{equation*}$ and simplify.

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Use the following facts:

  • $\sin(A+B)= \sin{A} \cdot \cos{B} + \cos{A} \cdot \sin{B}$.

  • $\displaystyle\frac{2 \tan\theta}{1+\tan^{2}\theta} = 2 \cdot \frac{\sin\theta}{\cos\theta} \cdot \cos^{2}\theta = 2 \cdot \sin\theta \cdot \cos\theta$

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    And if you append "$ = \sin (2 \theta)$" to this, that's the entire proof!2012-04-18
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$\sin2\theta$ $2\cdot\sin\theta\cdot \cos\theta$ multiply and divide by $\cos\theta$ $ 2\cdot \dfrac {\sin\theta}{\cos\theta}\cdot \cos^2\theta$ $2\cdot\tan\theta\cdot\cos^2\theta$ $\dfrac{2\cdot\tan\theta}{\sec^2\theta}$ $\dfrac{2\cdot\tan\theta}{1+\tan^2\theta}$