It's not a problem per se, I'm just trying something, so there's no statement. What I'm trying to do it's to prove the forllowing equation but without using the member of the right.
(cos(36º) + 1)²/(cos(36º)) = 5cos(36º)
There's the trivial answer, using both members, that it's multiply both sides by cos(36º), call it x and solve the 2nd degree equation and get the positive result. I'm not trying to do that. I'm trying to develop (cos(36º) + 1)²/(cos(36º)) and turn it into 5cos(36º).
I'm not sure.
The Attempt at a Solution
For simplicity sake, let 36º = u
(cos(u) + 1)²/(cos(u))
Developing and using sin^2(u) + cos^2(u) = 1 -> [cos^2(u) + 2cos(u) + 1]/cos(u)⇔ 2cos^2(u)+2cos(u)+sin^2(x)/cos(u)
Using the double angle for cosine, cos(2x) = cos^2(u) - sin^2(u):
3cos^2(u)-cos(2u)+2cos(u) / cos(u) ⇔ (3cos^2(u)/cos(u)) + (-cos(2u)+2cos(u))/cos(u)
Here's where I got stuck. I got everything in terms of cosine but the (-cos(2u)+2cos(u))/cos(u)) is giving me some trouble. How do I handle it?