\cos \dfrac{π}{7}+ \cos\dfrac{3π}{7}+ \cos\dfrac{π}{57}

Multiply and divide by 2\sin\dfrac{π}{7}

= \dfrac{1}{2\sin\dfrac{π}{7}} [ 2\sin\dfrac{π}{7}\cos \dfrac{π}{7}+ 2\sin\dfrac{π}{7}\cos\dfrac{3π}{7}+ 2\sin\dfrac{π}{7}\cos\dfrac{π}{57}]

In the parentheses, apply double angle identity on the first term and product to sum identity on the second and third terms.

\sin 2\alpha=2\sin \alpha \cos \alpha

\sin \alpha\cos \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) +\sin(\alpha - \beta) ]

Then,

x= \dfrac{1}{2\sin\dfrac{π}{7}} [ \sin\dfrac{2π}{7} + \sin(\dfrac{π}{7}+\dfrac{3π}{7}) + \sin(\dfrac{π}{7}-\dfrac{3π}{7}) + \sin(\dfrac{π}{7}+\dfrac{5π}{7}) + \sin(\dfrac{π}{7}-\dfrac{5π}{7}) ]

= \dfrac{1}{2\sin\dfrac{π}{7}} ( \sin\dfrac{2π}{7} + \sin\dfrac{4π}{7} + \sin\dfrac{-2π}{7} + \sin\dfrac{6π}{7} + \sin(\dfrac{-4π}{7})

= \dfrac{1}{2\sin\dfrac{π}{7}} \sin\dfrac{6π}{7}

= \dfrac{1}{2\sin\dfrac{π}{7}} \sin(π-\dfrac{π}{7})

= \dfrac{1}{2\sin\dfrac{π}{7}} \sin\dfrac{π}{7}

=\dfrac{1}{2}

Therefore, the value of cos(π/7) + cos(3π/7) + cos(5π/7) is 1/2