Question 847512
Hi, I have been stuck on this part of a question for a while.

I have got to the stage;
 4cos^2(36) - 2cos(36) -1 = 0 of a trig. identity but need to simplify this down into the form a+b√5 where a and b are found. 
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It's not clear what you're trying to do.
That's not an identity.  It has no variable.
If it is true, you can solve for cos(36).
Sub x for cos(36)
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{{{4x^2 - 2x - 1 = 0}}}
*[invoke solve_quadratic_equation 4,-2,-1]
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{{{cos(36) = 1/4 + sqrt(5)/4}}}