Trigonometry - Solving Trigonometric Equations

[Rectifier]

The problem

I am trying to write $-sin 2x + \sqrt{3} \cos 2x$ as $A \sin(2x+\phi)$ without using this formula below $a sin (x) + b cos (x) = c sin (x+v), \ \ c=\sqrt{a^2+b^2} \ ,\ \ \tan v = \frac{b}{a}$

The attempt

$A \sin(2x+\phi) = A \cos(\phi) \sin(2x) + A \sin(\phi) \cos(2x)$ comparison with the starting expression.

\begin{cases} A \cos(\phi) = -1 \\ A \sin(\phi) = \sqrt{3} \end{cases}

$\phi :$
\frac{A \sin(\phi)}{A \cos(\phi)} = \tan \phi = \frac{\sqrt{3}}{-1} = - \sqrt{3} \\ \tan -\phi = \sqrt{3} \\ \phi = -\frac{\pi}{3} + \pi n, \ \ n \in \mathbb{Z}

For $n=0$ -> $-\frac{\pi}{3}$ gives $$A \sin(-\frac{\pi}{3}) = \sqrt{3} \\ -A \sin(\frac{\pi}{3}) = \sqrt{3} \\ -A \frac{\sqrt{3}}{2}=\sqrt{3} \\ A = -2$$

$n = 1$ from above gives $\phi = \frac{2 \pi}{3}$
$$ A \sin(\phi) = \sqrt{3} \\ A \sin(\frac{2 \pi}{3}) = \sqrt{3}$$

I can rewrite $\sin(\frac{2 \pi}{3})$ as $2 \sin(\frac{\pi}{3}) \cos(\frac{\pi}{3}) = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$

$$ A \sin(\frac{2 \pi}{3}) = \sqrt{3} \\ A \frac{\sqrt{3}}{2} = \sqrt{3} \\ A = 2$$

Which A should I use in my answer? And why?

 

[Irene Kaminkowa]

An easier approach
$$-sin 2x + \sqrt{3}cos 2x = 2\left ( -\frac{1}{2}sin 2x+\frac{\sqrt{3}}{2} cos 2x\right )=2\left ( cos (2\pi/3) sin 2x+sin(2\pi/3)cos 2x\right )=2sin(2x + 2 \pi/3)$$

 

[Rectifier]

Thank you for that elegant solution. I solved the problem myself just now, it does not really matter which A I use as long as I use it with the corresponding angle, since $-2 \sin(2x- \frac{\pi}{3}) = 2 \sin(2x + \frac{2 \pi}{3})$ .