Question

Eliminate x and y from the equations -
$\cos x+\cos y=a,y=a,\cos 2x+\cos 2y=b,\cos 3x+\cos 3y=c$.

Answer 1

$c=4[co{s}^{3}x+co{s}^{3}y]-3(cosx+cosy),$ by third cubing 1st ,
$a^3=co{s}^{3}x+co{s}^{3}y+3cosxcosy.a,$
Equate $co{s}^{3}x+co{s}^{3}y$
$\therefore C+3a=4[a^3-3acosxcosy]$
$4a^3-c=3a[1+4cosxcosy]$
From 2nd $2(co{s}^{2}x+co{s}^{2}y)-1-1=b$
or $2(co{s}^{2}x+co{s}^{2}y)=b+2$
$\therefore 2a^2-b-2=4cosxcosy=\frac{4a^2-c}{3a}-1$
$\therefore 2a^3+c=3a(b+1)$