Question

Show that $cos^{4}(\frac{\pi}{8}) + cos^{4}(\frac{3\pi}{8}) + cos^{4}(\frac{5\pi}{8}) + cos^{4}(\frac{7\pi}{8}) = \frac{3}{2}$.

Answer 1

we know, cos(π/2 - A) = cosA

so, cos( π/8 ) = cos(π/2 - 3π/8) = sin(3π/8)

also we know, cos(3π/2 - A) = -sinA

so, cos( 5π/8 ) = cos(3π/2 - 7π/8) = -sin(7π/8)

now, LHS = cos⁴(π/8) + cos⁴(3π/8) + cos⁴(5π/8) + cos⁴(7π/8)

= {sin(3π/8)}⁴ + cos⁴(3π/8) + {-sin(7π/8)}⁴ + cos⁴(7π/8)

= sin⁴(3π/8) + cos⁴(3π/8) + sin⁴(7π/8) + cos⁴(7π/8)

= {sin²(3π/8) + cos²(3π/8)}² - 2sin²(3π/8)cos²(3π/8) + {sin²(7π/8) + cos²(7π/8)}² - 2sin²(7π/8)cos²(7π/8)

= 1 - 2/4{2sin(3π/8) cos(3π/8)}² + 1 - 2/4{2sin(7π/8) cos(7π/8)}²

we know, sin2x = 2sinx cosx

so, 2sin(3π/8) cos(3π/8) = sin(3π/4)

and 2sin(7π/8) cos(7π/8) = sin(7π/4)


= 1 - 1/2 sin²(3π/4) + 1 - 1/2 sin²(7π/4)

= 1 - 1/2 × (1/√2)² + 1 - 1/2 × (-1/√2)²

= 1 - 1/4 + 1 - 1/4

= 2 - 1/2 = 3/2 = RHS

Answer 2

HELLO DEAR,



we know, cos(π/2 - A) = cosA

so, cos( π/8 ) = cos(π/2 - 3π/8) = sin(3π/8)

also , cos(3π/2 - A) = -sinA

so, cos( 5π/8 ) = cos(3π/2 - 7π/8) = -sin(7π/8)

now,
cos⁴(π/8) + cos⁴(3π/8) + cos⁴(5π/8) + cos⁴(7π/8)

=> {sin(3π/8)}⁴ + cos⁴(3π/8) + {-sin(7π/8)}⁴ + cos⁴(7π/8)

=> sin⁴(3π/8) + cos⁴(3π/8) + sin⁴(7π/8) + cos⁴(7π/8)

=> {sin²(3π/8) + cos²(3π/8)}² - 2sin²(3π/8)cos²(3π/8) + {sin²(7π/8) + cos²(7π/8)}² - 2sin²(7π/8)cos²(7π/8)

=> 1 - 2/4{2sin(3π/8) cos(3π/8)}² + 1 - 2/4{2sin(7π/8) cos(7π/8)}²

[as, sin2x = 2sinx cosx ]

so, 2sin(3π/8) cos(3π/8) = sin(3π/4)

and 2sin(7π/8) cos(7π/8) = sin(7π/4)


=> 1 - 1/2 sin²(3π/4) + 1 - 1/2 sin²(7π/4)

=> 1 - 1/2 × (1/√2)² + 1 - 1/2 × (-1/√2)²

=> 1 - 1/4 + 1 - 1/4

=> 2 - 1/2 = 3/2


I HOPE IT'S HELP YOU DEAR,
THANKS