How to find out the equation whose roots are trigonometric?

Question

Find out the equation whose roots are $\sin^2 (2\pi/7), \sin^2 (4\pi/7), \sin^2 (8\pi/7)$. I wanted to solve this by firstly finding the equation ,the roots of which are $\sin 2\pi/7,\sin 4\pi/7, \sin 8\pi/7$ .Then i would find the equation wanted in the equation by using $Y = x^2$ or, $X = Y^1/2$; $X$ is the root of the 2nd equation i.e. $\sin 2\pi/7\ldots$ & $Y$ is the root of 1st equation. Now i would put the value of $X$ in the 2nd equation i.e. $Y^1/2$ & wld get the equation required. But i don't know how to find an equation whose roots are trigonometric. Plz help.

Answer 1

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factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$

or

Prove that $\frac{1}{4-\sec^{2}(2\pi/7)} + \frac{1}{4-\sec^{2}(4\pi/7)} + \frac{1}{4-\sec^{2}(6\pi/7)} = 1$

$\displaystyle z^3+z^2-3z-1=0\ \ \ \ (1),$ has the roots $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$

Now using $\displaystyle\cos2A=2-\sin^2A\iff\sin^2A=\frac{1-\cos2A}2,$

$\displaystyle\sin^2\frac{8\pi}7=\dfrac{1-\cos\frac{16\pi}7}2$ and again$\displaystyle\cos\frac{16\pi}7=\cos\left(2\pi+\frac{2\pi}7\right)=\cos\frac{2\pi}7$ etc.

Can you take it home form here?

See also: Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$

Answer 2

You mean $(x - \sin^2(2 \pi/7))(x - \sin^2(4 \pi/7))(x - \sin^2(8\pi/7)) = 0$?