# Thread: Deduce the value in surd form

1. ## Deduce the value in surd form

If $\displaystyle A=36^o$, show that $\displaystyle \sin 3A= \sin 2A$, and deduce that $\displaystyle \cos 36^o=\frac{\sqrt{5}+1}{4}$

I have already shown $\displaystyle \sin 3A= \sin 2A$ when $\displaystyle A=36^o$, since $\displaystyle 108^o=180^o-72^o$.
But I don't know how to deduce that $\displaystyle \cos 36^o=\frac{\sqrt{5}+1}{4}$.
Thanks!

2. Hello arze
Originally Posted by arze
If $\displaystyle A=36^o$, show that $\displaystyle \sin 3A= \sin 2A$, and deduce that $\displaystyle \cos 36^o=\frac{\sqrt{5}+1}{4}$

I have already shown $\displaystyle \sin 3A= \sin 2A$ when $\displaystyle A=36^o$, since $\displaystyle 108^o=180^o-72^o$.
But I don't know how to deduce that $\displaystyle \cos 36^o=\frac{\sqrt{5}+1}{4}$.
Thanks!
Four steps:

• Use the identities $\displaystyle \sin2A=2\sin A \cos A$ and $\displaystyle \sin3A=3\sin A -4\sin^3A$ in the equation $\displaystyle \sin 3A= \sin 2A$

• Divide through by $\displaystyle \sin A$ (it's non-zero, so that's OK)

• Replace $\displaystyle \sin^2A$ by $\displaystyle 1 - \cos^2A$

• Solve the quadratic for $\displaystyle \cos A$, taking the positive root.

Can you complete it now?