(a) Prove that sin 3A = 3 sin A - 4 sin^{3 }A.

(b) If A = 36^{o}, show that sin 3A = sin 2A.

(c) Deduce that cos 36^{o}= (1 + 5^{1/2})/ 4.

I've managed to solve (a) & (b), but I'm stuck at (c). I've figured out (c) has something to do with the fact that cos 36^{o}= cos (63 - 27)^{o}and that the angles 63^{o}, 27^{o}and 90^{o}form a right angle, which matches the hypotenuse (5^{1/2}), adjacent (1) and opposite (2) somehow. I tried substitution of (a) and (b) into the working for (c) too, and I get the left hand side as 0.5(3 - 4 sin^{2 }A) and the right hand side as 0.5(sin A - 1)/ cos A. Thank you!