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 ?


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

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  • $\begingroup$ 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

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.

  • $\begingroup$ 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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