F4 Trigo Question

a. Since 5θ=90 degree when θ=18 degree, thus
sin3θ= cos(90 - 3θ) = cos2θ
i.e. θ=18 degree is a solution of the equation sin3θ= cos2θ

b. Now from sin3θ= cos2θ, we have
3 sinθ - 4 sin^3 θ = 1 - 2 sin^2 θ
4 sin^3 θ - 2 sin^2 θ - 3 sinθ + 1 = 0
[sinθ - 1][4 sin^2 θ + 2 sinθ - 1] = 0
4 sin^2 θ + 2 sinθ - 1 = 0 since sin18≠1
sinθ = [-2 +/- √20]/8 = [-1 +/- √5]/4

But since sin18>0, the negative one is rejected,
and therefore sin18 = [-1 + √5]/4

(A)

sin 54=sin(90-54)=cos36

So X=18 is a solution of the equation sin3x = cos2x

(b)

sin3x=3sinx-4sin^3 x

cos36
=1-2(sin18)^2
cos36 =sin54
=3sin18- 4(sin18)^3

1-2(sin18)^2=3sin18- 4(sin18)^3

4(sin18)^3-2(sin18)^2+3sin18+1=0

( sin18-1)[4(sin18)^2+2sin18+1]=0

sin18=( 5^(1/2) - 1) / 4