# 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.

-
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
add comment

## 3 Answers

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.

-
add comment

Use the following facts:

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

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

-
And if you append "$= \sin (2 \theta)$" to this, that's the entire proof! –  The Chaz 2.0 Apr 18 '12 at 18:54
add comment

$$\sin2\theta$$ $$2\cdot\sin\theta\cdot \cos\theta$$ multiply and divide by $\cos\theta$ $$2\cdot \dfrac {\sin\theta}{\cos\theta}\cdot \cos^2\theta$$ $$2\cdot\tan\theta\cdot\cos^2\theta$$ $$\dfrac{2\cdot\tan\theta}{\sec^2\theta}$$ $$\dfrac{2\cdot\tan\theta}{1+\tan^2\theta}$$

-
add comment