Double / Half & Area Trig Help

I have two questions:

1. Given that sin 3pi/10 = sqrt(5)+1/4 find an exact expression for cos (3pi/5)

I used cos(2theta) = 1 - 2sin^2(theta)

cos(3pi/5) = 1- 2sin^2(3pi/10)

= 1 - 2(sqrt(5)+1/4)^2

= 1- 2(3+sqrt(5)/8) = 4-3+sqrt(5)/4 = 1+sqrt(5)/4 <- i got that as the final answer but the answer key says that it's 1-sqrt(5)/4

2. What is the area of the triangle whose sides all have length r

i used A = 1/2absin(theta)

= 1/2 r^2sin(theta)

i got to that part and now i'm confused as to what to do next. the answer is sqrt(3)r^2/4

would it be right if i said that sin(theta) = 1/2 = pi/6 and then cos(pi/6) = sqrt(3)/2 and then use that to get sqrt(3)r^2/4?