If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?


I know how to solve a problem like

"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$" 

by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:

Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$  to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$

Where I can not make progress with the question 

"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".

Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?

What I have tried:

If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by  by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
We have 
$$s^2=(\cos\alpha+\sin\alpha)^2=1+2\cos\alpha\sin\alpha=(0.2)^2$$
so we find
$$p=\sin\alpha\cos\alpha=-0.48$$
hence $\sin\alpha$ and $\cos\alpha$ are roots of the quadratic equation:
$$x^2-sx+p=x^2-0.2x-0.48=0$$
\begin{align}
  \sin \alpha+\cos \alpha &amp;= 0.2 \\
  (\sin \alpha+\cos \alpha)^2 &amp;= 0.04 \\
  \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos\alpha &amp;= 0.04 \\
  1+\sin 2\alpha &amp;= 0.04 \\
  \sin 2\alpha &amp;= -0.96
\end{align}

and that would be straight forward to proceed