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Solve the equation $$\frac{\sin(18°+x)}{\sin(x)}=\frac{\sin48°}{\sin18°}$$

If we use a computer we quickly note that $x=12°$, which can be easily proven: $$\sin18°=2\sin48°\sin12°=\cos36°-\cos60°$$ $$\iff 2\sin18°=2\cos36°-1$$ $$\iff2\sin18°=1-4\sin^218°$$

The last equation is true because if $x=18°$ we have $$5x=90°\implies 3x=90-2x$$ $$\implies 3\sin x-4\sin^3 x=\cos2x=1-2\sin^2x$$ $$\implies (1-\sin x)(4\sin^2x+2\sin x -1)=0$$ Since $\sin 18°\neq1$, we have the desired equality. But how can we possibly solve it constructively?

chubakueno
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  • sin(A+B) = sinAcosB + sinBcosA...I think haha – BCLC Oct 29 '14 at 05:21
  • @chubakueno, See https://math.stackexchange.com/questions/951464/trigonometric-equation-help-is-sought – lab bhattacharjee Oct 29 '14 at 18:01
  • @labbhattacharjee This is one of the limited cases, but how would you suppose that $\sin(18+x)=\frac{1}{2}$?(with a chain of reasoning) – chubakueno Oct 29 '14 at 22:31
  • @chubakueno, Have you noticed https://math.stackexchange.com/questions/951464/trigonometric-equation-help-is-sought/952871#952871? Here $\sin x,\sin(18^\circ+x)$ can be $\pm\dfrac12$. Also here the accepted answer does not satisfy your requirement either – lab bhattacharjee Oct 30 '14 at 04:52

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I found that $$\cot x =\frac{\sin 48^{\circ}-\sin 18^{\circ}\cos 18^\circ}{\sin^2 18^{\circ}}$$ Let $y=12^\circ$. Note that $$\begin{aligned} \sin y(2\sin 4y-\sin 3y)-\cos y(1-\cos 3y) & =\cos 3y-\cos 5y +\cos 4y -\cos y \\ & = 1-2\sin^2\frac{3y}{2}-\frac12-2\sin\frac{5y}{2}\sin\frac{3y}2 \\ & = \frac12(1-2\sin\frac{3y}{2}-4\sin^2\frac{3y}{2})\\ & =0\end{aligned}$$ The last step is true because $\frac{3y}2=18^{\circ}$ and due to the relation derived by the original poster. Hence $$\frac{2\sin 4y-\sin 3y}{1-\cos 3y} =\cot y$$

Can you finish it?

Sayan
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  • But how would you suppose that $\cos(5y)=1/2$(or that $y=12°$, from the start) – chubakueno Nov 01 '14 at 21:16
  • If you cant solve a trig equation $f(x)=0$ just by simplification, a good method is to guess first a solution, say $c$ satisfies $f(c)=0$, then simplify $f(x)-f(c)$. I have done the same thing here. – Sayan Nov 02 '14 at 03:04