- #1
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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.
[tex]\begin{cases} A \cos(\phi) = -1 \\ A \sin(\phi) = \sqrt{3} \end{cases}[/tex]
## \phi : ##
[tex]\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}[/tex]
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?
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.
[tex]\begin{cases} A \cos(\phi) = -1 \\ A \sin(\phi) = \sqrt{3} \end{cases}[/tex]
## \phi : ##
[tex]\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}[/tex]
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?