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I found this interesting result. Show that

$$\dfrac{1+\sin{6^\circ}+\cos{12^\circ}}{\cos{6^\circ}+\sin{12^\circ}}=\sqrt{3}.$$ See this Wolfram Alpha output.

My attempt is very ugly. We know that $$\sin{6^\circ}=\dfrac{1}{8}(-1-\sqrt{5})+\dfrac{1}{4}\sqrt{\dfrac{3}{2}(5-\sqrt{5})}$$

$$\cos{6^\circ}=\dfrac{1}{4}\sqrt{\dfrac{1}{2}(5-\sqrt{5})}+\dfrac{1}{8}\sqrt{3}(1+\sqrt{5})$$ $$\sin{12^\circ}=2\sin{6^\circ}\cos{6^\circ}$$$$\cos{12^\circ}=2\cos^2{6^\circ}-1$$ and it is very ugly, maybe there exist simple methods.

Thank you.

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$$1+\sin 6^\circ +\cos 12^\circ = \sqrt{3}(\cos 6^\circ +\sin 12^\circ)$$ $$ \Uparrow $$ $$ \dfrac{1}{2}+\dfrac{1}{2}\sin 6^\circ+\dfrac{1}{2}\cos12^\circ = \dfrac{\sqrt{3}}{2}\cos 6^\circ+\dfrac{\sqrt{3}}{2}\sin 12^\circ $$ $$ \Uparrow $$ $$ \dfrac{1}{2}+ \left( \dfrac{1}{2}\sin 6^\circ - \dfrac{\sqrt{3}}{2} \cos 6^\circ\right) + \left(\dfrac{1}{2}\cos12^\circ -\dfrac{\sqrt{3}}{2}\sin 12^\circ\right)=0 $$ $$ \Uparrow $$ $$ \dfrac{1}{2}+ \sin(6^\circ-60^\circ) + \cos(12^\circ+60^\circ)=0 $$ $$ \Uparrow $$ $$ \dfrac{1}{2}- \sin54^\circ + \cos72^\circ=0 $$ $$ \dfrac{1}{2}- \cos36^\circ + \sin18^\circ=0 $$ $$ \dfrac{1}{2}-\dfrac{\sqrt{5}+1}{4} + \dfrac{\sqrt{5}-1}{4}=0 $$ $$ \dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{4}=0 $$ Less ugly :)

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Here is a proof based on the cosine-sum formula $\cos(A+B)=\cos A\cos B-\sin A\sin B$. After multiplying through by the denominator, the RHS minus the LHS is $$\sqrt3\cos6^\circ-\sin6^\circ+\sqrt3\sin12^\circ-\cos12^\circ-1$$ $$=2\cos30^\circ\cos6^\circ-2\sin30^\circ\sin6^\circ-(2\cos60^\circ\cos12^\circ-2\sin60^\circ\sin12^\circ)-1$$ $$=2\cos(30^\circ+6^\circ)-2\cos(60^\circ+12^\circ)-1$$ $$=2\frac{\sqrt5+1}{4}-2\frac{\sqrt5-1}{4}-1$$ $$=0.$$ To see that $\cos36^\circ=(\sqrt5+1)/4$: Using the cosine-sum formula and the fact that $\cos^2\theta+\sin^2\theta=1$, we obtain $\cos2\theta=2\cos^2\theta-1$ and then, using also the formula $\sin2\theta=2\sin\theta\cos\theta$, we derive $\cos3\theta=4\cos^3\theta-3\cos\theta$. For our value $\theta=36^\circ$, note that $\cos3\theta=-\cos2\theta$. This reduces to a cubic equation in $\cos\theta$:$$4\cos^3\theta+2\cos^2\theta-3\cos\theta-1=0.$$We can easily spot one (irrelevant) solution $\theta=\pi$, or $\cos\theta=-1$; so, factoring out $\cos\theta+1$ leaves the quadratic $4\cos^2\theta-2\cos\theta-1=0$, with the positive root yielding $\cos36^\circ$. The value of $\cos72^\circ$ is easily obtained now from $\cos2\theta=2\cos^2\theta-1$.

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\begin{align} x&=\frac{1+\sin6+\cos12}{\cos6+\sin12} \\ &=\frac{2\sin30+\sin78+\sin6}{\cos78+\cos6} \\ &=\frac{2(\sin54-\sin18)+2\sin42\cos36}{2\cos42\cos36} \\ &=\frac{4\cos36\sin18+2\sin42\cos36}{2\cos42\cos36} \\ &=\frac{2\cos36(2\sin18+\sin42)}{2\cos42\cos36} \\ &=\frac{2\sin18+\sin42}{\cos42} \\ &=\frac{\sin42+\sin18+\sin18}{\cos42} \\ &=\frac{2\sin30\cos12+\cos72}{\cos42} \\ &=\frac{\cos72+\cos12}{\cos42} \\ &=\frac{2\cos42\cos30}{\cos42} \\ &=2\cos30 \\ &=\sqrt 3 \end{align}

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  • $\begingroup$ Welcome to Math SE! Please take a look at MathJax for information about formatting your answer to make it more readable. $\endgroup$ – DMcMor May 30 '17 at 20:59

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