Question

cos²36°+cos²72°

don't spam​

Answer 1

Answer is in attachment

thank you

Answer 2

Answer:

Cos²36+cos²72

=>cos²36+cos²(90-18)

=>cos²36+sin²18 ...[ cos²(90-theta)=sin²theta]

=>sin² 18 + cos² 36

=> sin² 18 + ( 1 - 2sin² 18)² - - - - - - (I)

Now, sin 36 = cos 54 - - {compl. rule }

=> sin 2A = cos 3A - - {Where A = 18 }

=> 2 sin A cos A = 4 cos^3 A - 3 cos A

=> 2 sin A = 4 cos² A - 3 - - - {Divide by cos A }

=> 2 sin A = 4 - 4 sin² A - 3

=> 4 sin² A + 2 sin A - 1 = 0

=> 4 sin² (18) + 2 sin (18) - 1 = 0

=> sin 18 = [-2 ± √(4 + 16) ] / 8 - - - - { By Quadratic Formula }

=> sin 18 = (√5 - 1)/4

=> sin² 18 = [5 + 1 - 2√5] /16 

=> sin² 18 = (3 - √5)/8

Substituting this in - - (i), we get,

sin² 18 + ( 1 - 2sin² 18)² = (3 - √5)/8 + [1 - 2 (3 - √5)/8]²

=> (3 - √5)/8 + [1 - (3 - √5)/4]²

=> (3 - √5)/8 + [(4 - 3 + √5)/4]²

=> (3 - √5)/8 + [(1 + √5)/4]²

=> (3 - √5)/8 + (1+ 5+ 2√5)/16

=> (3 - √5)/8 + (3 + √5)/8 

=> (3 - √5 + 3 + √5)/8

=> 3/4