We have, sinα = k/(2k) = BC/AB (For some value of k)

By Pythagoras theorem, (hypotenuse)² = (perpendicular)² + (base)²

∴AB² = BC² + AC²

= (2k)² = (k)² + AC²

= 4k² = k² + AC²

= AC² = 3k²

→ AC = k√3

∴ cosα = AC/AB = (k√3)/(2k) = √3/2

Then, 3cosα - 4cos³α =

=

= 0

LHS = RHS

HENCE PROVED