Question

sin square 18+ cos sqare 36=?​

Answer 1

Answer :

Answer is 3/4

Step by step Explanation

$sin^218 + cos ^236$

⇒   $sin^2 18 + ( 1 - 2sin^2 18)^2 - - - - - - (I)$

Now

$sin 36 = cos 54 ---- (compl. rule )$

⇒ $sin 2A = cos 3A ---- (Where A = 18 )$

⇒ $2 sin A cos A = 4 cos^3 A - 3 cos A$

⇒ $2 sin A = 4 cos^2 A - 3 - - - (Divide by cos A )$

⇒ $2 sin A = 4 - 4 sin^2 A - 3$

⇒ $4 sin^2 A + 2 sin A - 1 = 0$

⇒ $4 sin^2 (18) + 2 sin (18) - 1 = 0$

Also

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

⇒ $sin 18 = (\sqrt{5}  - 1)/4$

⇒ $sin^2 18 = [5 + 1 - 2\sqrt{5} ] /16$

⇒ $sin^2 18 = (3 - \sqrt{5} )/8$

Finally

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

$sin^2 18 + ( 1 - 2sin^2 18)^2 = (3 - \sqrt{5} )/8 + [1 - 2 (3 - \sqrt{5} )/8]^2$

⇒ $(3 - \sqrt{5} )/8 + [1 - (3 - \sqrt{5}  )/4]^2$

⇒ $(3 - \sqrt{5} )/8 + [(4 - 3 + \sqrt{5} )/4]^2$

⇒  $(3 - \sqrt{5} )/8 + [(1 + \sqrt{5} )/4]^2$

⇒ $(3 - \sqrt{5} )/8 + (1+ 5+ 2\sqrt{5} )/16$

⇒ $(3 - \sqrt{5}) /8 + (3 + \sqrt{5} )/8$

⇒ $(3 - \sqrt{5}  + 3 +\sqrt{5} )/8$

⇒  $3/4$

Hence 3/4 is the answer