Why is the phase constant negative in part B of this graph?

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neoking77
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all the solution is there, but i just don't understand the answer to part B.
CIMG1818.jpg

for part B, why is the sign negative?
 
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The graph is a cosine wave shifted to the right by an angle less than 90...

So of your two solutions (+pi/3 -pi/3) for arccos, you need the one with the minus, for a right shift, as opposed to a left shift...
 
Remember that there are an infinite number of solutions to trigonometric equations, so in your case;

[tex]\cos\phi = \frac{1}{2} \Rightarrow \phi = -\frac{\pi}{3}, \frac{\pi}{3}, \frac{5\pi}{3}...[/tex]

[tex]\phi = \frac{\pi}{3} \pm 2\pi\cdot k \hspace{1cm}; k\in\mathbb{Z}[/tex]

But the thing to remember here is that the phase angle is simply a translation along the x axis. A positive phase angle translates the curve in the negative x direction (to the left) and a negative phase angle translates the curve in the positive x direction (to the right).

Now, in which direction do we need to translate the cosine curve so that is matches yours?

Edit: Several mintues to slow ... :frown:
 
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