1. ## Trigo surds question

(a) Prove that sin 3A = 3 sin A - 4 sin3 A.
(b) If A = 36o, show that sin 3A = sin 2A.
(c) Deduce that cos 36o = (1 + 51/2)/ 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 36o = cos (63 - 27)o and that the angles 63o, 27o and 90o form a right angle, which matches the hypotenuse (51/2), 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 sin2 A) and the right hand side as 0.5(sin A - 1)/ cos A. Thank you!

2. ## Re: Trigo surds question

Using parts a) and b), we may write:

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

Since $\sin A\ne0$ we may divide through by this to get:

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

Now, using a Pythagorean identity, we have:

$3-4(1-\cos^2A)=2\cos A$

Now, arrange this as a quadratic in $\cos A$ and take the positive root.