# Cos 3pi/5

• Dec 11th 2012, 01:29 PM
Eraser147
Cos 3pi/5
I don't understand the step where 2Sin^2(3pi/10) = 2((sqrt5+1)/4) Attachment 26188
• Dec 12th 2012, 01:07 AM
chiro
Re: Cos 3pi/5
Hey Eraser147.

Try setting up a right angled triangle or using sums or differences formula to get the value.
• Dec 13th 2012, 08:42 AM
BobP
Re: Cos 3pi/5
Consider the right-angled triangle with acute angles $\displaystyle \frac{3\pi}{10}$ and $\displaystyle \frac{2\pi}{10}$ and note that $\displaystyle \sin{\left(\frac{3\pi}{10}\right)}=\cos{ \left( \frac{2\pi}{10}\right) }.$

To ease typing, let $\displaystyle \frac{\pi}{10}=A,$ then $\displaystyle \sin 3A=\cos 2A.$

Use the standard identities, $\displaystyle \sin 3A =3\sin A-4\sin^{3} A .............(1) \text{ and } \cos 2A= 1-2\sin^{2} A ............(2)$

and we have the cubic $\displaystyle 4\sin^{3} A -2\sin^{2}A-3\sin A +1 = 0.$

$\displaystyle (\sin A -1)$ is a factor, remove this and we are left with the quadratic

$\displaystyle 4\sin^{2} A +2\sin A -1=0,$ and solving this, (for the angle in the $\displaystyle 0,\pi/2$ range),

$\displaystyle \sin A = \frac{-1+\sqrt{5}}{4}.$

Substituting this into (1) and simplifying leads to $\displaystyle \sin 3A = \sin{ \ \left( \frac{3\pi}{10} \right)} = \frac{1+\sqrt{5}}{4}.$