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Please help solving this equation:

$\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$

I used numerical method to solve it and got $x=6^\circ$ but I am not able to solve it by trigonmetry.

Thank! Best Regards, Michael.

Jaideep Khare
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luimichael
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3 Answers3

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Using the same observation as @dantopa we get

\begin{eqnarray} \sin(x) &=& \sin(12^\circ - x )\\ \sin(x) &=& \sin(12^\circ)\cos(x)-\cos(12^\circ)\sin(x)\\ \tan(x) &=& \sin(12^\circ)-\cos(12^\circ)\tan(x)\\ \tan(x)(1+\cos(12^\circ))&=&\sin(12^\circ)\\ \tan(x)&=&\frac{\sin(12^\circ)}{1+\cos(12^\circ)}\\ \tan(x)&=&\tan(6^\circ)\\ x &=& 6^\circ+180^\circ n \end{eqnarray}

John Wayland Bales
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Hint: have you tried using the product-to-sum identities? That is, $$ \sin A\sin B =\frac{1}{2}(\cos(A-B)-\cos(A+B)) $$ is one such formula. Note that there are several formulas of this type that can be used here. The rightmost term $\sin(12^{\circ}-x)$ can also be simplified by using a similar formula and solving for the $\sin(A-B)$ term.

M_B
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  • I attempted to show that $\sin 96^0 \sin 12^0 = \sin 18^0 \sin 42^0$ but I needed to use the values of special angles. Here I will show my work and please let me know if there is a better method. – luimichael Jun 16 '17 at 09:25
  • $\sin 96^0 \sin 12^0 - \sin 18^0 \sin 42^0$ $=\frac 12 ( \cos 84^0 - \cos 108^0)-\frac 12 ( \cos 24^0 -\cos 60^0)$ $=\frac 12 ( \cos 84^0 - \cos 24^0)+ \frac 12 ( \cos 60^0 - \cos 108^0)$ $=-\sin 54^0 \sin 30^0 + \frac 12 (\cos 60^0 - \cos 108^0)$ $=-\frac 12 \sin 54^0+ \frac 12 ( \frac 12 - \cos 108^0)$ $=\frac 14 - \frac 12( \sin 54^0 - \cos 72^0)$ $=\frac 14 - \frac 12 ( \sin 54^0 - \sin 18^0)$ $= \frac 14 - \frac 12 ( \frac {\sqrt{5}+1}2-\frac {\sqrt{5}-1})$ $=\frac 14 - \frac 12 \times \frac 12 = 0$ – luimichael Jun 16 '17 at 09:34
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Observation $$ \sin 96^{\circ} \sin 12^{\circ} = \sin 18^{\circ} \sin 42^{\circ} $$

Implication $$ \sin x = \sin \left(12^{\circ} - x \right) \qquad \Rightarrow \qquad x = 12^{\circ} - x $$

Conclusion

$$x=6^{\circ}$$

dantopa
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  • Implication The sine function is not injective, for example $x=186^\circ$ is a solution as well. – dxiv Jun 16 '17 at 00:19
  • @dxiv: of course we can add and subtract integer multiples of $\pi$. The structure of the problem suggests the answer given. – dantopa Jun 16 '17 at 00:49
  • I don't see what in the problem suggests that $6^\circ$ would be a "better" answer than, say, $186^\circ$. of course we can add and subtract integer multiples of π That belongs into the answer, not as a comment. – dxiv Jun 16 '17 at 00:53
  • @dantopa, I'm really interested in the proof of the $1$st proposition. I could utilize https://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2 to prove it – lab bhattacharjee Jun 16 '17 at 09:20