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In physics problems involving 2 wave sources on axis, the extrema of an equation of the form

$$y = \sin x + \sin(x + a)$$

have significance as related to the amplitude of the superposition of the two wave sources. In order to solve for these, I have personally used numerical methods. I have also seen that:

$$\sin x + \sin(x + a) = 2 \cos(a/2) \sin(x + a/2)$$

can be used to make these calculations easier. The truth of this can be verified easily with wolfram alpha or a graphing calculator, but what is the easiest way to prove such an identity?

rjm27trekkie
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2 Answers2

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Identities like these can always be proven using the definitions of the trigonometric functions. These definitions are:

$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2} \qquad \text{and} \qquad \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

So, in order to prove the relation that was posted by the OP, we can proceed as follows:

\begin{align*} \sin(x) + \sin(x+a) &= \frac{e^{ix} - e^{-ix}}{2i} + \frac{e^{i(x+a)} - e^{-i(x+a)}}{2i} \\ &= \frac{e^{i(x+\frac{a}{2})}e^{-i\frac{a}{2}} - e^{-i(x+\frac{a}{2})}e^{i\frac{a}{2}} + e^{i(x+\frac{a}{2})}e^{-i\frac{a}{2}} - e^{-i(x+\frac{a}{2})}e^{i\frac{a}{2}}}{2i} \\ &= \frac{(e^{i\frac{a}{2}} + e^{-i\frac{a}{2}})(e^{i(x+\frac{a}{2})} - e^{-i(x+\frac{a}{2})})}{2i} \\ &= 2\cos\left(\frac{a}{2}\right)\sin\left(x+\frac{a}{2}\right) \end{align*} Agreed, depending on the amount of heavy machinery one is prepared to use, there might be different, faster, or more elegant ways to provide proofs of these kinds of identities. But this method always works, never takes more than six lines, and uses no heavily involved theorems, and therefore usually is the method I prefer.

arriopolis
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Notice that $y''(x) = y(x)$ so you must have $$\sin(x) + \sin(x + a) = f(a)\sin(x) + g(a)\cos(x).$$ Can you do anything with this?

ncmathsadist
  • 49,383