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To find value of $\sin 18^\circ$. Now my textbook gives a proof in which it takes $\theta=18^\circ$ and then multiply it out by 5 and write again as sum of $2\theta+3\theta$ and then taking sin on both sides forms a quadratic equation.

Are there any other elegant ways to find its value?

Thanks

Gathdi
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  • You could always just draw a triangle. – Brevan Ellefsen Feb 12 '17 at 04:26
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    This might be useful http://math.stackexchange.com/questions/438362/evaluate-cos-18-circ-without-using-the-calculator this also http://math.stackexchange.com/questions/1707647/remembering-deriving-the-values-of-sine-and-cosine-of-18-degrees-36-degrees-54 – kingW3 Feb 12 '17 at 04:28

3 Answers3

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Let $x = \sin(18^{\circ})\implies x^2 = \dfrac{1-y}{2}$, whereas $y = \cos(36^{\circ})$. And square again $y^2 = \dfrac{1+\cos(72^{\circ})}{2}= \dfrac{1+x}{2}\implies 2x^2+y = 1 = 2y^2-x\implies 2(x^2-y^2) + (x+y) = 0\implies (x+y)(2x-2y+1) = 0\implies 2x= 2y-1\implies 4x^2=2-2y = 1-2x\implies 4x^2+2x-1=0$. You can now solve for $x$ to get the answer in terms of radical which is the expected form of the answer.

DeepSea
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Consider triangle $ABC$ with $\angle A=36^o$ and $\angle B=\angle C=72^o.$

Let the internal bisector of $\angle B$ intersect $AC$ at $D$ .

We have $\angle DBC=\frac {1}{2}\angle B=36^o$ and $\angle DCB=\angle C=72^o$, and $\angle BCD=\angle C=72^o.$ So triangles $ABC$ and $BDC$ are similar.

And triangle $BDA$ is isosceles because $\angle DBA=\frac {1}{2}\angle B=36^o=\angle A.$

Let $AB=AC=1$ and $BC=x.$ Then $DC =x\cdot BC=x^2$ (by the similarity), and $AD=DB=CB=x$ (because ADB and DBC are isoceles). Therefore $$1=AC=AD+DC=x+x^2.$$ From isoceles triangle $ABC$ we also have $$x=2AB \sin \frac {1}{2}\angle A=2\sin 18^o.$$

DanielWainfleet
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$$S=\cos A+\cos3A=2\cos A\cos2A=\dfrac{(2\cos A\sin A)\cos2A}{\sin A}=\cdots=\dfrac{\sin4A}{2\sin A}$$

Now if $S=\dfrac12,\sin4A=\sin A$ with $\sin A\ne0$

either $4A=m360^\circ+A\iff A=120^\circ m$ with $3\nmid m$ as $\sin A\ne0$ $\implies A \equiv\pm120^\circ\pmod{360^\circ}$

or $4A=(2m+1)180^\circ-A\iff A=(2m+1)36^\circ$ with $m\not\equiv2\pmod5$ as $\sin A\ne0$

$\implies A \equiv\pm36^\circ,\pm108^\circ\pmod{360^\circ}$

All the four cases reduce to $$\cos36^\circ+\cos108^\circ=\dfrac12$$

Now use $\cos36^\circ=1-2\sin^218^\circ$

and $\cos108^\circ=\cos(90+18)^\circ=-\sin18^\circ$ to form a Quadratic Equation in $\sin18^\circ$

Use the fact that $\sin18^\circ>0$

See also : Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$

Community
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lab bhattacharjee
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