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 $\cos 50^\circ$.
 A: 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 $t=\cos 50$.
A: 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 $i.$
