1. Give a convincing argument that $\sin (2\cdot 3\pi/5) \neq 2\cdot\sin(3\pi/5)$

    This is my proof, can someone verify it.

$$ \sin 2x = \sin x$$ if $$ 2 \sin x \cos x = \sin x$$ if $$ 2 \sin x \cos x - \sin x = 0$$ or $$ (2 \cos x - 1) \sin x = 0 $$ if either $ 2 \cos x - 1 = 0$ or $ \sin x = 0$ if $\cos x = 1/2$ or $\sin x = 0$ if $x = \pi/3 + 2 n \pi$ or $x = m \pi$ for some integers $n$, $m$. Since $3 \pi/5$ does not fall in either of the two sets of values, we can conclude the desired statement.

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    $\begingroup$ Another convincing argument is to note that $\sin(6\pi/5)<0$, while $\sin(3\pi/5)>0$. $\endgroup$ – Peter Košinár Aug 20 '13 at 13:34

You seem to have forgotten that you want to show that $\frac{3\pi}5$ is not a solution to $$\sin(2x)=2\sin x.$$ You switched instead to $$\sin(2x)=\sin x.$$ Your general approach is fine, though, except you forgot solutions of the form $x=-\frac\pi3+2n\pi$.

  • $\begingroup$ I fixed it and I got $2nπ$ for every $n∈ℤ$. Is this correct. I don't think you can get solutions in the form of $x=±π/3+2nπ$ $\endgroup$ – saqqib Aug 20 '13 at 13:25
  • $\begingroup$ That is nearly correct! Using double-angle makes the equation $$2\sin x\cos x=2\sin x,$$ which becomes $$2\sin x(\cos x-1)=0,$$ so we need $\sin x=0$ or $\cos x=1$. The $\cos x=1$ is redundant, here, since that makes $\sin x=0$ by Pythagorean Theorem. Thus, we need the solutions to $\sin x=0,$ so the solutions are $x=n\pi$ for $n\in\Bbb Z$. You are correct that $x=\pm\frac\pi3+2n\pi$ works for $\sin(2x)=\sin x,$ but not for $\sin(2x)=2\sin x$. $\endgroup$ – Cameron Buie Aug 20 '13 at 13:30

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