simplify using trigonometric functions?

Question

if cos30°=(√3)/2 and cos36°=(1+√5)/4
what is cos3° equal to..
show your solution (showing the simplified radicals)
not using calculator..

Answer 1

You can write cos(3) as cos(36/2 - 30/2).

Using the identity cos(a-b) = cos(a)cos(b) + sin(a)sin(b) you get

cos(3) = cos(36/2)cos(30/2) + sin(36/2)sin(30/2)

Then use the formula cos(a/2) = sqrt(1+cos(a))/2] and sin(a/2) = sqrt[(1-cos(a))/2) get

cos(3) = sqrt[(1+cos(36))/2]*sqrt[(1+cos(30/2))/2) + sqrt[(1-cos(36))/2]*sqrt[(1-cos(30)/2]

cos(3) = .5*sqrt[(1+cos(36))*(1+cos(30))] + .5*sqrt[(1-cos(36))*(1-cos(30))

You now have all the values in the equation: cos(36) and cos(30).

After multiplying all the terms and rationalizing you get

cos(3) =

[sqrt(8)/16]*[sqrt[10+sqrt(15)+5*sqrt(3)+2*sqrt(5)] +

sqrt[6+sqrt(15)-3*sqrt(3)-2*sqrt(5)]]

Answer 2

cos(a-b)=cosa cos b+ sin a sinb
cos 2a=cos ^2 a - sin ^2 a

Find cos 6 from the relation
cos 6=cos(36-30)=cos 36 cos 30 +sin36 sin 30
You need to find sin 36

sin30=1/2
cos 6=2 cos^2 (3)-1
cos 3=sqrt[(1+cos 6)/2]

Answer 3

30`=cos![#(3)/2]
36`=cos![(1+#5)/4]
6`= cos![(1+#5)/4]-cos![#(3)/2]
so
3`= {cos![(1+#5)/4]-cos![#(3)/2]}/2
so
cos3`=cos [{cos![(1+#5)/4]-cos![#(3)/2]}/2]

# Squareroot
! inverse
` degree

Answer 4

First homework assignment of the semester, eh?