# How to find out the equation whose roots are trigonometric?

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.

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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?

• Sir,can u tell me how to find a polynomial equation whose roots are given & are trigonometric like this question?What will be my approach? – user142971 May 10 '14 at 8:22
• can u plz tell me what is the technique of finding polynomial equation whose roots are trigonometric without using concept of complex number? – user142971 May 10 '14 at 8:24
• @user36790, Do you know how expand $$\cos nx$$ in terms of $\cos x$? Also, may I request you to supply a few concrete Questions? – lab bhattacharjee May 10 '14 at 10:39
• Cos nx can be found by De movire's formula;the question is like this:find the equation whose roots are Cos 2pi/7,Cos 4pi/7,Cos 6pi/7. – user142971 May 10 '14 at 11:17
• How to solve such type of questions,sir?What is the technique to find an equations whose roots are given and are trigonometric? – user142971 May 10 '14 at 11:20

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

• No,it wouldn't be like that...the answer would be an algebric cubic polynomial. – user142971 May 10 '14 at 2:59
• Expand it all out, combine products of trig functions into sums, and use the fact that $\sum_{j=0}^{n-1} \sin(2\pi j/n) = 0$ and $\sum_{j=0}^{n-1} \cos(2\pi j/n) = 0$. – Robert Israel May 10 '14 at 3:02
• Alternatively, note that $\cos(2 \pi j/n)$ is a root of $T_n(x) - 1$ where $T_n$ is a Chebyshev polynomial. – Robert Israel May 10 '14 at 3:09
• @user36790 The given equation answers your question. Why is it a problem that it's a cubic polynomial? – Jack M May 10 '14 at 3:24