Complex Trigo

Thats totally wrong!
Re{(a+ib)(c+id)} not equal to ac. Its rather equal to (ac-bd)!!

Roots of z⁷ + 1 = 0 are -1, α, α³, α⁵, α',α'³,α'⁵, where α = e^{(iÎ /7)} & α' is the conjugate of α.
Now, use them here......

z^7+1=0
The roots of this eqn are -1.\alpha,\alpha ^3 ,\alpha ^5,\bar{\alpha},\bar{\alpha ^3},\bar{\alpha ^5}
where \alpha =cos\frac{\pi}{7}+isin\frac{\pi}{7} \alpha ^3=cos\frac{3\pi}{7}+isin\frac{3\pi}{7},\alpha ^5=cos\frac{5\pi}{7}+isin\frac{5\pi}{7}

Nowz^7+1=(z+1)(z-\alpha)(z-\bar{\alpha})(z-\alpha ^3)(z-\bar{\alpha ^3})(z-\alpha ^5)(z-\bar{\alpha ^5})

=>z^6-z^5+z^4-z^3+z^2-z+1=(z^2-(\alpha +\bar{\alpha})z+\alpha \bar{\alpha})....(z-(\alpha^5 +\bar{\alpha ^5})z+\alpha^5 \bar{\alpha ^5})Since \ \alpha +\ \bar{\alpha}=2cos\frac{\pi}{7},\alpha \bar{\alpha }=1 \ =>\frac{z^6+1}{z^3}-\frac{z^5+z}{z^3}+\frac{z^4+z^2}{z^3}-\frac{z^3}{z^3}=(\frac{z^2-2zcos\frac{\pi}{7}+1}{z})(\frac{z^2-2zcos\frac{3\pi}{7}+1}{z})(\frac{z^2-2zcos\frac{5\pi}{7}+1}{z})

=>(z^3+\frac{1}{z^3})-(z^2+\frac{1}{z^2})+(z+\frac{1}{z})-1=[(z+\frac{1}{z})-2cos\frac{\pi}{7}][(z+\frac{1}{z})-2cos\frac{3\pi}{7}][(z+\frac{1}{z})-2cos\frac{5\pi}{7}]

=>(z+\frac{1}{z})^3-3(z+\frac{1}{z})-(z+\frac{1}{z})^2+2+(z+\frac{1}{z})-1=(z+\frac{1}{z}-2cos\frac{\pi}{7})(z+\frac{1}{z}-2cos\frac{3\pi}{7})(z+\frac{1}{z}-2cos\frac{5\pi}{7})

Put \ (z+\frac{1}{z})=2x

=>8x^3-6x-4x^2+2+2x-1=8(x-cos\frac{\pi}{7})(x-cos\frac{3\pi}{7})(x-cos\frac{5\pi}{7})

=>8x^3-4x^2-4x+1=8(x-cos\frac{\pi}{7})(x-cos\frac{3\pi}{7})(x-cos\frac{5\pi}{7})

Put \ x=-1\ =>7=64(cos^2\frac{\pi}{14})(cos^2\frac{3\pi}{14})(cos^2\frac{5\pi}{14})

Rejecting \ negative \ root \ ,(cos\frac{\pi}{14})(cos\frac{3\pi}{14})(cos\frac{5\pi}{14})=\frac{\sqrt{7}}{8}

yup nikhil ..i copied the soln...i never denied that..
copied it becoz it was good and surely better than mine..

and btw this copy-paste debate was held loong back...and there is no harm to copy paste if a person has done ques on his own and wants to use the soln from some authentic source [1]