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Given the function f: $$ f(x) = \sqrt{3} \cos(2x) - \sin(2x) $$

Question: What is its amplitude and phase shift?

My attempt:

Let c be the hypothenuse of a triangle with the sides from the expression

$$ c = \sqrt{\sqrt{3}^2 + (-1)^2} = 2 $$

Then f can be written as

$$ f(x) = 2(\frac{\sqrt{3}}{2} \cos(2x) + \frac{-1}{2} \sin(2x)) $$

From trigonometric identities we know that

$$ \sin(α + β) = \sin(α)\cos(β) + \cos(α)\sin(β) $$

Comparing this expression to the one for f(x), one sees that

$$ \{ \sin(α) = \frac{\sqrt{3}}{2}; \quad \cos(α) = \frac{-1}{2} \} \implies α = \frac{2\pi}{3} + 2 \pi n $$

Using this we rewrite f(x) as

$$ f(x) = 2 \sin(\frac{2\pi}{3} + 2x) $$

At this point it becomes unclear what the definition of phase shift is. According to many sources, it would be given by delta in

$$ f(x) = A \sin(k(\delta + x)) $$

So, in this case

$$ f(x) = 2 \sin(2(\frac{\pi}{3} + x)) $$

Clearly then

$$ \delta_{found} = \frac{\pi}{3} $$

However, the correct answer according to the book is

$$ \delta_{book} = \frac{2\pi}{3} $$

Which seems very strange!

Which is the proper definition, is the book really wrong on multiple occasions?

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Depends; the found one is phase shift after time compression; the book one is phase shift before time compression. –  peterwhy Oct 4 '13 at 14:48
    
There's a possible error when you go from the first to the second expression for $f(x)$. It looks like you have exchanged $\sin$ and $\cos$. Could that have to do anything with $\delta_{\text{found}} \neq \delta_{\text{book}}$? –  Daniel R Oct 4 '13 at 15:03
    
you time argument here is $2x$ not just $x$. At the very beginning you can set $t=2x$ and at the end you'd get $t+\frac{2\pi}{3}$, so there'd be no hesitations. :) –  Caran-d'Ache Oct 4 '13 at 16:17
    
That is an interesting point of view, @Caran-d'Ache. I understand phase shift as the shift from $ sin(2x) $, and in a graph it is plain that the actual phase shift $ \frac{\pi}{3} $! –  lericson Oct 4 '13 at 16:38
    
I don't know where you got this definition from but the classical is. –  Caran-d'Ache Oct 4 '13 at 17:16
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