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A solution that i found on net but cant figure out why we need to do Sinx + Sin5x

Why does we take the sum of Sin x and Sin 5x ?

Why cant we take the sum of Sin 3x and Sin 5x or Sin x and Sin 3x?

Because when i take the sum of Sin 3x and Sin 5x i get incorrect general solutions, Please help me i cant find a reasonable answer for this question ?

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closed as unclear what you're asking by A---B, Watson, choco_addicted, C. Falcon, colormegone Jun 22 '16 at 20:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ We strongly encourage that you format your posts. Formatting tips here. $\endgroup$ – Em. Feb 9 '16 at 19:22
  • $\begingroup$ What are your steps for taking the sum of $\sin(3x)$ and $\sin(5x)$ first? $\endgroup$ – GoodDeeds Feb 9 '16 at 19:28
  • $\begingroup$ This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials $\endgroup$ – Simply Beautiful Art Feb 9 '16 at 21:41
  • $\begingroup$ @SimplyBeautifulArt I have improved since then :). $\endgroup$ – A---B Feb 15 '17 at 18:24
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The main reason why you take $\sin(x)$ and $\sin(5x)$ together is that you get $\sin (3x) \cos(2x)$, where you can factor $\sin (3x) $ out.

If you take $\sin(x)$ and $\sin(3x)$ together is that you get $2\sin (2x) \cos(x)+\sin(5x)=0$.

If you take $\sin(3x)$ and $\sin(5x)$ together is that you get $2\sin (4x) \cos(x)+\sin(x)=0$.

The first one is horrible. The second one is doable, but not nice. We know that $\sin(4x)=2\cos(2x)\sin(2x)=4\cos(2x)\cos(x)\sin(x)$. Hence we get $$8\cos(2x)\cos^2(x)\sin(x)+\sin(x)=0$$

$$8\cos(2x)\cos^2(x)+1=0 \vee \sin(x)=0$$

$$8(2\cos^2(x)-1)\cos^2(x)+1=0 \vee \sin(x)=0$$

$$16\cos^4(x)-8\cos^2(x)+1=0 \vee \sin(x)=0$$

The former is a quadratic equation in $\cos^2(x)$, which you can solve, and it gives in fact the same roots as you have in your picture.

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  • $\begingroup$ So which is the correct answer because if i do sin 3x + sin5x i get :- 16cos4(x)−8cos2(x)+1=0 and sin(x)=0 as mentioned, therefore, i get x = n * pi where n is any integer or x = 2npi +/- pi/3_which is different from _x = pi *n/3 or x = n * pi +/- pi/3. $\endgroup$ – A---B Feb 10 '16 at 9:20

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