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I have two points emanating two cosine waves,in concrete, this ones: The format is the following: \begin{align} y(x,t) & = A\cos(\omega t-kx + \mathrm{phase}) \\ y_1(x,t) & = 2\cos(9600 \pi t -24 \pi x+\pi/4) \:\mathrm{Pa}\\ y_2(x,t) & = 2\cos(9600 \pi t -24 \pi x +5\pi/4) \:\mathrm{Pa} \\ \end{align}

In a point P, there is a interference between this two waves. The result wave that I calculated is: $$ y_T(x,t) = -2\cos(9600 \pi t -24 \pi x +\pi/4) \:\mathrm{Pa}. $$ In the solution that my teacher gave me, the result wave is the following: $$ y_T(x,t) = 2\cos(9600 \pi t -24 \pi x -3\pi/4) \:\mathrm{Pa}. $$ The minus sign of the cosine in the result wave that I calculated, can I pass it to the phase, adding to the phase $\pi/2$? I know that in complex notation I can do that, but in harmonic notation this equivalence is true? Note that if you add to $\pi/4$ the term $\pi/2$ the result is $3\pi/4$.

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You can pass the minus sign into the phase, but it doesn't give you a phase shift of $\pi/2$. It gives you a phase shift of $\pi$. It's easy to demonstrate to yourself (by using any plotting software or using the angle-addition formulas) that

$$\cos(\theta+\pi)=-\cos(\theta)$$

and also that

$$\cos(\theta-\pi)=-\cos(\theta)$$

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  • $\begingroup$ I edited the question, the result wave phase is $-3*pi/4$. And you know how can I get to the result of my teacher? $\endgroup$ – victor26567 Mar 3 '18 at 19:52
  • $\begingroup$ Turn the minus sign into a $-\pi$ phase shift. $\endgroup$ – probably_someone Mar 3 '18 at 20:55

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