# Prove $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$

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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You have $1+\tan^2 u=\frac1{\cos^2 u}$ and $\sin\,2u=2\sin\,u\cos\,u$. Can you take it from there? –  Ｊ. Ｍ. Apr 18 '12 at 18:53
@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. –  dotslash May 5 at 2:38

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! –  The Chaz 2.0 Apr 18 '12 at 18:54
$$\sin2\theta$$ $$2\cdot\sin\theta\cdot \cos\theta$$ multiply and divide by $\cos\theta$ $$2\cdot \dfrac {\sin\theta}{\cos\theta}\cdot \cos^2\theta$$ $$2\cdot\tan\theta\cdot\cos^2\theta$$ $$\dfrac{2\cdot\tan\theta}{\sec^2\theta}$$ $$\dfrac{2\cdot\tan\theta}{1+\tan^2\theta}$$