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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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Like

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 $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$

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  • $\begingroup$ 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? $\endgroup$ – user142971 May 10 '14 at 8:22
  • $\begingroup$ can u plz tell me what is the technique of finding polynomial equation whose roots are trigonometric without using concept of complex number? $\endgroup$ – user142971 May 10 '14 at 8:24
  • $\begingroup$ @user36790, Do you know how expand $$\cos nx$$ in terms of $\cos x$? Also, may I request you to supply a few concrete Questions? $\endgroup$ – lab bhattacharjee May 10 '14 at 10:39
  • $\begingroup$ 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. $\endgroup$ – user142971 May 10 '14 at 11:17
  • $\begingroup$ How to solve such type of questions,sir?What is the technique to find an equations whose roots are given and are trigonometric? $\endgroup$ – user142971 May 10 '14 at 11:20
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You mean $(x - \sin^2(2 \pi/7))(x - \sin^2(4 \pi/7))(x - \sin^2(8\pi/7)) = 0$?

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

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