# If tanx= -1/3, cos>0, then how do you find sin2x?

Aug 1, 2016

$\sin \left(2 x\right) = - 0.6$

#### Explanation:

Any trigonometric function of some angle can be easily expressed in terms of a tangent of half of this angle.

We can express $\sin \left(2 x\right)$ in terms of $\tan \left(x\right)$ as follows:
$\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right) = 2 \sin \frac{x}{\cos} \left(x\right) \cdot {\cos}^{2} \left(x\right) = 2 \tan \left(x\right) \cdot {\cos}^{2} \left(x\right)$

In its turn,
$\frac{1}{\cos} ^ 2 \left(x\right) = \frac{{\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right) = 1 + {\sin}^{2} \frac{x}{\cos} ^ 2 \left(x\right) = 1 + {\tan}^{2} \left(x\right)$

Therefore,
$\sin \left(2 x\right) = \frac{2 \tan \left(x\right)}{1 + {\tan}^{2} \left(x\right)}$

Using this and given value $\tan \left(x\right) = - \frac{1}{3}$, we conclude
$\sin \left(2 x\right) = 2 \frac{- \frac{1}{3}}{1 + \frac{1}{9}} = - \frac{6}{10} = - 0.6$