Combining Sine Functions: Simplifying with Trigonometry

AI Thread Summary
The discussion centers on simplifying the expression sin(2x) + sin(2[x + π/3]) into a single sine function using trigonometric identities. The user expresses confusion about merging two sine functions, questioning whether the result would be more complex. It is clarified that since both sine functions share the same frequency, they can indeed be combined into a single sinusoidal function. Participants recommend reviewing trigonometric identities and suggest using resources like Wolfram Alpha for assistance. The conversation emphasizes the importance of understanding basic trigonometric identities for simplification.
Benhur
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I have the expression sin(2x) + sin(2[x + π/3]) and I have to write this in terms of a single function (a single harmonic, rather saying). But I don't know how to do this, and... it seems a little bit weird for me, because I'm merging two sine-wave functions into one. Doesn't the sum of sines result in a more complex body than a simple sine alone?

The exercise that I'm trying to solve says that I must use a trigonometry formula to solve.
 
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Thank you, DrClaude. Now I got it.
 
Wolfram Alpha sometimes helpful to remind yourself about trig identities: sin(a)+sin(b). (you might have to 'wade' through a lot of extraneous information before you find the required identity)
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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