Question

cos^{​​​​6} π/16 + cos^{​​​​6} 3π/16 + cos^{​​​​6} 5π/16 + cos^{​​​​6} 7π/16 = ?

Answer 1

cos⁶(π/16) + cos⁶(3π/16) + cos⁶(5π/16) + cos⁶(7π/16)
= cos⁶(π/16) + cos⁶(3π/16) + cos⁶(π/2 - 3π/16) + cos⁶(π/2 - π/16)
= cos⁶(π/16) + cos⁶(3π/16) + sin⁶(3π/16) + sin⁶(π/16), via cofunction identities

= [cos⁶(π/16) + sin⁶(π/16)] + [cos⁶(3π/16) + sin⁶(3π/16)]
= [cos²(π/16) + sin²(π/16)] * [cos⁴(π/16) - cos²(π/16) sin²(π/16) + sin⁴(π/16)] + [cos²(3π/16) + sin²(3π/16)] * [cos⁴(3π/16) - cos²(3π/16) sin²(π/16) + sin⁴(3π/16)], via sum of cubes

= [cos⁴(π/16) - cos²(π/16) sin²(π/16) + sin⁴(π/16)] +
[cos⁴(3π/16) - cos²(3π/16) sin²(π/16) + sin⁴(3π/16)]

= [cos²(π/16) + sin²(π/16)]² - 3 cos²(π/16) sin²(π/16) +
[cos²(3π/16) + sin²(3π/16)]² - 3 cos²(3π/16) sin²(3π/16)

= 1² - 3 cos²(π/16) sin²(π/16) + 1² - 3 cos²(3π/16) sin²(3π/16)
= 2 - (3/4)(2 sin(π/16) cos(π/16))² - (3/4)(2 sin(3π/16) cos(3π/16))²
= 2 - (3/4) sin²(2π/16) - (3/4) sin²(2 * 3π/16), double angle identity
= 2 - (3/4) [sin²(π/8) + sin²(3π/8)]
= 2 - (3/4) [sin²(π/8) + sin²(π/2 - π/8)]
= 2 - (3/4) [sin²(π/8) + cos²(π/8)], via cofunction identity
= 2 - (3/4) * 1
= 5/4