How to find the exact value of the cosine of 50 degree angle
I want to know the exact value of $\cos 50^\circ$.
Actually I have already tried lot of times to solve but I cannot find the exact value of $\cos 50^\circ$.
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$\begingroup$one method can be:$$\cos 3x =4\cos^3x-3\cos x$$Putting $x=50$ gives $$\cos 150 =4\cos^3x-3\cos x\tag{*}$$But $\cos 150=-\cos 30=-{\sqrt 3\over 2}$. So you have the LHS of $({}^*)$. Therefore We have to solve the following equation using Cardano's method:$$4t^3-3t+{\sqrt 3\over 2}=0$$
where $t=\cos 50$.
$\endgroup$ 11 $\begingroup$TO use Cardano's method for the cubic, first divide by $4,$ to write $4t^3-3t+\sqrt{3}/2=0$ as $$t^3-\frac{3t}4+\frac{\sqrt{3}}8=0.$$
Now let $t=x+\frac1{4x},$ so that the equation becomes $$\left(x+\frac1{4x}\right)^3-\frac34\left(x+\frac1{4x}\right)+\frac{\sqrt{3}}8=0.$$ Then multiplying by $64x^3$ and expanding, we have $$64x^6+(8\sqrt3)x^3+1=0,$$ where a bunch of terms cancel. Now that we have a quadratic form, using the quadratic formula gives: $$x^3=\frac{-\sqrt3\pm i}{16}.$$ Now this equation has 6 solutions, and all 6 are complex, but when you use $$x=\sqrt[3]{\frac{-\sqrt3+i}{16}},$$ and take $t=x+\dfrac1{4x},$ you get $$\sqrt[3]{\frac{-\sqrt3+i}{16}}+\sqrt[3]{\frac{1}{4(-\sqrt3+i)}},$$ which is a real number approximately equal to $0.6427876\approx\cos(50^\circ).$
Now it would be a real treat if someone could 'Mathematica' the result to write it without depending on $i.$
$\endgroup$ 2 $\begingroup$here is my own calculation, you can use cofunction identities $$\cos 50°=\sin\left(90°-50°\right)=\sin 40°=\sin\frac{2\pi}{9}$$ set $\zeta^{k}$ as ninth roots of unity, we can found using De Moivre Formula that $$\zeta^{k}+\zeta^{9-k}=2\cos\frac{2k\pi}{9}$$ $$\zeta^{k}-\zeta^{9-k}=2i\sin\frac{2k\pi}{9}$$ $$\zeta^{k}-2+\zeta^{9-k}=-4\sin^2\frac{2k\pi}{9}$$ $$-4\sin^2\frac{2\pi}{9}=2\cos\frac{4\pi}{9}-2$$ $$\sin\frac{2\pi}{9}=-\frac{i\sqrt{2\cos\frac{4\pi}{9}-2}}{2}$$ then, we will need to solve $2\cos\frac{4\pi}{9}$ , which is the roots of the polynomial $$\left(x-2\cos\frac{2\pi}{9}\right)\left(x-2\cos\frac{4\pi}{9}\right)\left(x-2\cos\frac{8\pi}{9}\right)=0$$ $$x^3-3x+1=0$$ now we left with the cubic polynomial with the interesting form $$x^3-3px+2q=0$$ where $p=1$ , and $q=\frac{1}{2}$ .
Cubic Formula gives $$x=\frac{1}{3}\left(\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{1}{\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}}\right)$$ consequently $$x=\cos\frac{2\pi}{9}=\frac{1}{3}\left(\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}\right)$$ $$2\cos\frac{4\pi}{9}=\frac{1}{3}\left(\frac{-1-i\sqrt{3}}{2}\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{-1+i\sqrt{3}}{2}\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}\right)$$ $$2\cos\frac{8\pi}{9}=\frac{1}{3}\left(\frac{-1+i\sqrt{3}}{2}\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{-1-i\sqrt{3}}{2}\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}\right)$$ $$\sin\frac{2\pi}{9}=-\frac{i\sqrt{2\cos\frac{4\pi}{9}-2}}{2}$$ $$\sin\frac{2\pi}{9}=-\frac{i\sqrt{\frac{1}{3}\left(-6+\frac{-1-i\sqrt{3}}{2}\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{-1+i\sqrt{3}}{2}\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}\right)}}{2}$$ $$\sin\frac{2\pi}{9}=-\frac{i\sqrt{\frac{3}{9}\left(-6+\frac{-1-i\sqrt{3}}{2}\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{-1+i\sqrt{3}}{2}\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}\right)}}{2}$$ $$\sin\frac{2\pi}{9}=-\frac{i\sqrt{3\left(-6+\frac{-1-i\sqrt{3}}{2}\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{-1+i\sqrt{3}}{2}\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}\right)}}{6}$$ $$\cos 50°=\sin\frac{2\pi}{9}=-\frac{i\sqrt{-18+\frac{-3-3i\sqrt{3}}{2}\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{-3+3i\sqrt{3}}{2}\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}}}{6}$$ however, solving directly using $\cos 3\theta = 4\cos^3 \theta -3\cos \theta$ unexpectedly gives $$\cos 20°=\sqrt[3]{\frac{i-\sqrt3}{16}}+\sqrt[3]{\frac{1}{4(i-\sqrt3)}}$$ consequently $$\cos 20°=\sqrt[3]{\frac{i-\sqrt3}{16}}+\sqrt[3]{\frac{i+\sqrt3}{16}}$$ $$\cos 20°=\sqrt[3]{\frac{4i-4\sqrt3}{64}}+\sqrt[3]{\frac{4i+4\sqrt3}{64}}$$ $$\cos 20°=\frac{1}{4}\sqrt[3]{4i-4\sqrt3}+\frac{1}{4}\sqrt[3]{4i+4\sqrt3}$$ the other roots is what do we wanted $$\cos 50°=\frac{-1-i\sqrt{3}}{8}\sqrt[3]{4i-4\sqrt3}+\frac{-1+i\sqrt{3}}{8}\sqrt[3]{4i+4\sqrt3}$$ $$\cos 80°=\frac{-1+i\sqrt{3}}{8}\sqrt[3]{4i-4\sqrt3}+\frac{-1-i\sqrt{3}}{8}\sqrt[3]{4i+4\sqrt3}(\text{unreliable})$$ all of these expressions are real, but involves complex number to express them in radicals because they are too awesome, so we left with the two expression but with the same value $$\frac{-1-i\sqrt{3}}{8}\sqrt[3]{4i-4\sqrt3}+\frac{-1+i\sqrt{3}}{8}\sqrt[3]{4i+4\sqrt3}=-\frac{i\sqrt{-18+\frac{-3-3i\sqrt{3}}{2}\sqrt[3]{\frac{-1+i\sqrt{3}}{2}}+\frac{-3+3i\sqrt{3}}{2}\sqrt[3]{\frac{-1-i\sqrt{3}}{2}}}}{6}$$ and both were equals $\cos 50°$
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