Question

Formation of equations whose roots are given :
Find the equation whose roots are
$cos\frac{2\pi }{7},cos\frac{4\pi }{7}andcos\frac{8\pi }{7}$
or $cos\frac{2\pi }{7},cos\frac{4\pi }{7}andcos\frac{6\pi }{7}$
Deduction :
$cos\frac{2\pi }{4}cos\frac{4\pi }{7}cos\frac{8\pi }{7}=S_3=\frac{1}{8}$
$cos\frac{2\pi }{4}+cos\frac{4\pi }{7}+cos\frac{8\pi }{7}=S_1=-\frac{1}{2}$
The equation whose roots are
$sec\frac{2\pi }{7},sec\frac{4\pi }{7},sec\frac{8\pi }{7}(=sec\frac{6\pi }{7})$

Answer 1

Let $\theta =\frac{2\pi }{7},cos\frac{4\pi }{7}andcos\frac{8\pi }{7}$
$\therefore 7\theta =Even\pi or4\theta =Even\pi -3\theta$
$\therefore cos4\theta =cos3\theta$
$(2co{s}^{2}2\theta -1)=4co{s}^3\theta -3cos\theta$
Put $cos\theta =x$
$2(2x^2-1{)}^2-1=4x^3-3x$
$2(4x^4-4x^2+1)-1=4x^3-3x$
or $8x^4-4x^3-8x^2+3x+1=0$
Clearly x = 1 is a root of above
$(x-1)(8x^3+4x^2-4x-1)=0$
x = 1 or $cos\theta =1\Rightarrow \theta =0or2\pi$ rejected
$\therefore 8x^3+4x^2-4x-1=0$ ......(3)
is the equation whose roots are given by (1) and (2) of the equation.
Deduction : from (3) $S_1=-\frac{4}{8}=-\frac{1}{2},S_3=\frac{1}{8}$
$\because sec\theta =\frac{1}{cos\theta }.$ Hence putting $x=\frac{1}{2}$ in (3) the required equation is
$y^3+4y^2-4y-8=0$ ........(4)