I need help writing the expression $5(\sin 2x- \cos 2x)$ in terms of sine only. I need help writing the expression $5(\sin 2x- \cos 2x)$ in terms of sine only. Using the double angle  identity formula, I was able to get it to $5(\sin 2x - 1 + 2\sin^2x)$, but not sure how to move forward. I know the answer is $$5\sqrt{2}\sin\left(2x+ \frac{7\pi}{4}\right)$$ but I'm not sure how to get there. Thanks.

Write 
$5(\sin2x-\cos2x) = a \sin(bx+c)$
Then
$5\sin2x-5\cos2x = a \sin(bx)\cos(c)+a \sin(c)\cos(bx)$
Choose $b=2$, $a\cos(c)=5$ and $a\sin(c)=-5$. So
$a^2\cos^2(c)+a^2\sin^2(c) = 25+25 = 50 \Rightarrow a^2=50 \Rightarrow a = \pm 5\sqrt{2}$
Choose $a = 5\sqrt{2}$
$\Rightarrow\cos(c) = \dfrac{5}{5\sqrt{2}}=\dfrac{\sqrt{2}}{2}$ and
$\sin(c) = \dfrac{-5}{5\sqrt{2}}=-\dfrac{\sqrt{2}}{2}$
Hence $c = \dfrac{7\pi}{4}$
Then 
$5(\sin2x-\cos2x) = 5\sqrt{2} \sin\left(2x+\dfrac{7\pi}{4}\right)$
