# Way to prove $\sin 3^{\circ}$ value

I was looking wikipedia ,suddenly I saw this $$\sin {{3}^{{}^\circ }}=\cos {{87}^{{}^\circ }}=\frac{2\left( 1-\sqrt{3} \right)\sqrt{5+\sqrt{5}}+\left( 1+\sqrt{3} \right)\left( \sqrt{10}-\sqrt{2} \right)}{16}$$ can someone help me that : how to find $\sin3^{\circ}$ or how to prove it . Is there a Algebraic expressions for this kind of problem ? Thanks in advanced.

• Repeat the half-angle formula on 24°. Oct 14, 2016 at 8:15
• This ok ,but what's sin or cos of $24^0$? Oct 14, 2016 at 8:17
• $$\sin(18^\circ-15^\circ)$$ math-only-math.com/exact-value-of-cos-18-degree.html and cut-the-knot.org/pythagoras/cos15.shtml Oct 14, 2016 at 8:18
• Try finding trigonometric ratios based on the side/diagonal of a regular pentagon. Oct 15, 2016 at 4:29

First

$$\sin\frac{\pi}{5}=\sin\frac{4\pi}5=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$

Hence

$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$

And

$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac{1}{2}\left(\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}\right)=\frac{1}{4}$$

So, you have the system

$$\left\{\begin{eqnarray}\cos\frac{\pi}{5}\cos\frac{3\pi}{5}&=&-\frac14\\ \cos\frac{\pi}{5}+\cos\frac{3\pi}{5}&=&\frac12\end{eqnarray}\right.$$

Hence $\cos\frac{\pi}5$ and $\cos\frac{3\pi}5$ are the roots of $x^2-\frac12x-\frac14$. For this trinomial, $\Delta=\frac54$ and

$$x=\frac{1\pm\sqrt{5}}{4}$$

Since $\cos\frac{\pi}5>0$ and $\cos\frac{3\pi}5<0$, this means

$$\cos\frac{\pi}5=\frac{1+\sqrt{5}}{4}$$

And

$$\cos\frac{2\pi}{5}=-\cos\frac{3\pi}{5}=\frac{\sqrt{5}-1}{4}$$

Then

$$\sin\frac{\pi}{10}=\sin\left(\frac{\pi}{2}-\frac{2\pi}{5}\right)=\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4}$$

And

$$\cos\frac{\pi}{10}=\sqrt{\frac{1+\cos\frac{\pi}{5}}{2}}=\sqrt{\frac{5+\sqrt5}{8}}$$

Next

$$\cos\frac{\pi}{12}=\cos\left(\frac{\pi}{4}-\frac\pi6\right)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac12\frac{\sqrt2}{2}=\frac{\sqrt2}{4}(\sqrt{3}+1)$$

$$\sin\frac{\pi}{12}=\sin\left(\frac{\pi}{4}-\frac\pi6\right)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac12\frac{\sqrt2}{2}=\frac{\sqrt2}{4}(\sqrt{3}-1)$$

Finally

$$\sin 3^\circ=\sin\frac{\pi}{60}=\sin\left(\frac{\pi}{10}-\frac\pi{12}\right)\\=\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt2}{4}(\sqrt{3}+1)\right)-\left(\sqrt{\frac{5+\sqrt5}{8}}\right)\left(\frac{\sqrt2}{4}(\sqrt{3}-1)\right)\\ =\frac{2(1-\sqrt3)\sqrt{5+\sqrt{5}}+(1+\sqrt{3})(\sqrt{10}-\sqrt{2})}{16}$$

Computability of $\cos\frac{\pi}{n}$ and $\sin\frac{\pi}{n}$ by radicals is related to constructible polygons. A more difficult result arises for $n=17$, see for instance this.

First it's more usual to compute this kind of formula by considering the angle in radians. You can note that : $$3^{\circ}=\frac{\pi}{60}\text{ }\text{rad}=\left[\frac{\pi}{10}-\frac{\pi}{12}\right]\text{rad}$$

Then you can compute the sinus of $\frac{\pi}{12}$ using the formula for the triple angle on the sinus of $\frac{\pi}{4}$. For $\frac{\pi}{10}$ they present a method with a pentagon but you can also try to use the Chebyshev polynomials ... Take a look at this article.

• Can you explain in more detail ?please Oct 14, 2016 at 8:21
• See the link, they explain the addition formulae used to reach your result using $sin(a-b)=sin(a)cos(b)-cos(a)sin(b)$ with $a=\frac{\pi}{10}$ and $b=\frac{\pi}{12}$ Oct 14, 2016 at 8:26
• No, do not use the triple angle formula to attack $\sin(\pi/12)$. You will get a casus irreducibilis cubic equation. You are better off to render $\pi/12=\pi/4-\pi/6$ and use the aine of a difference. Oct 14, 2016 at 10:41