
Trigo surds question
(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! :)

Re: Trigo surds question
Using parts a) and b), we may write:
$\displaystyle 3\sin A4\sin^3A=2\sin A\cos A$
Since $\displaystyle \sin A\ne0$ we may divide through by this to get:
$\displaystyle 34\sin^2A=2\cos A$
Now, using a Pythagorean identity, we have:
$\displaystyle 34(1\cos^2A)=2\cos A$
Now, arrange this as a quadratic in $\displaystyle \cos A$ and take the positive root.