Let `a=(pi)/(7)`, then
(a) show that `sin^(2)3a-sin^(2)a=sin2asin3a`
(b) show that `co sec a=co sec 2a+co sec4a`.
(c) Prove that `cos a` is a root of the equation `8x^(3)+4x^(2)-4x+1=0`.

Explanation

(a) `sin^(2)3a-sin^(2)a=sin(3a+a)sin(3a-a)`
`=sin 4a sin 2a`
Now, `sin 4a=sin""(4pi)/(7)`
`=sin(pi-(4pi)/(7))`
`=sin ""(3pi)/(7)` ltbr. `=sin 3a`
`therefore sin^(2)3a-sin^(2)a=sin3asin2a`
(b) `co sec 2a+co sec 4a=(sin 4a+sin2a)/(sin 4a sin 2a)`
`=(2sin 3a cos a)/(sin 4a sin 2a)`
`=(2sin 3a cos a)/(sin 4a(2sin a cos a))`
`=(1)/(sin a)`
`=co sec a`
(c) `a=(pi)/(7)`
`therefore 7a=3a+4a=pi`
`therefore sin 3a =sin 4a`
`therefore sin a(3-4sin^(2)a)=2sin 2a cos 2a`
`=4 sin a cos a cos 2a`,
`therefore 3-4(1-cos^(2)a)=4cos a(2cos^(2)a-1)`
`(because sin a ne0)`
If follows that
`8cos^(3)a-4cos^(2)a-4cos a+1=0`
From equation (i), we can say that `cos a ` is a root of
`8x^(3)-4x^(2)-4x+1=0`.