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Question:

Determine the amplitude and phase shift of $f(x) = \sqrt{3} \cos2x-\sin2x$

Attempted solution:

The amplitude can be calculated by:

$$A = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{4} = 2$$

The sine and cosine of the phase shift then becomes

$$\sin \delta = \frac{\sqrt{3}}{2}, ~~~~~~~~ \cos \delta = -\frac{1}{2}$$

because the factor before cosine goes into the sine equation and vice versa.

The $\delta$ that fulfills these two criteria is $\frac{\pi}{3}$. In other words,

$A = 2$

$\delta = \frac{\pi}{3}$

However, the answer to the phase shift actually turns out to be $\frac{2\pi}{3}$. Where did I go wrong?

How does the factor in front of $x$ inside the sine and cosine functions affect this kind of calculations in the general case? It is clearly tempting to just multiply by 2 (in this case) because it produces the correct answer, but that is obviously not a mathematically sound approach.

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$$\begin{align}\operatorname{arccos}{\frac{-1}{2}}&=\pi-\operatorname{arcos}{\frac{1}{2}}\\&=\frac{2\pi}{3}\end{align}$$

Edit One has $\sin(\pi-x)=\sin{x}$ and $\cos(\pi-x)=-\cos x$ ; solving

$$\begin{align} \cos{\delta}&=-\frac{1}{2}\\\sin{\delta}&=\frac{\sqrt{3}}{2}\end{align}$$

is therefore the same as solvant

$$\begin{align} \cos{(\pi-\delta)}&=\frac{1}{2}\\\sin{(\pi-\delta)}&=\frac{\sqrt{3}}{2}\end{align}$$

And this means $\pi-\delta=\frac{\pi}{3}$ whence $\delta=\frac{2\pi}{3}$

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  • $\begingroup$ I see how this relates to the main question but it is far from clear. Please explain just how this answers the OP's question. $\endgroup$ – Rory Daulton May 13 '15 at 23:54
  • $\begingroup$ If $\sin{x}=\frac{\sqrt{3}}{2}$ and $\cos{x}=-\frac{1}{2}$, knowing that $\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}$ and $\cos{\frac{\pi}{3}}=\frac{1}{2}$, last but least taking into account the two identities $\sin{(\pi-x)}=\sin{x}$ and $\cos{(\pi-x)}=-\cos{x}$ you get to the result outlined in my comment $\endgroup$ – marwalix May 14 '15 at 0:49
  • $\begingroup$ @marwalix The explanation in the above comment should be edited into the answer itself. $\endgroup$ – user642796 May 14 '15 at 3:24

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