Question

sin36 sin144 sin108 sin72=5/16, prove that

Answer 1

sin(36)=sin(180-36)=sin(144); sin(108)=sin(180-108)=sin(72).

So sin(36)sin(144)sin(108)sin(72)=sin²(36)sin²(72).

According to de Moivre:

e^niθ=cos(nθ)+isin(nθ)=(cosθ+isinθ)ⁿ. Let n=5:

e^5iθ=cos(5θ)+isin(5θ)=(cosθ+isinθ)⁵,

cos(5θ)+isin(5θ)=cos⁵θ+5icos⁴θsinθ-10cos³θsin²θ-10icos²θsin³θ+5cosθsin⁴θ+isin⁵θ (binomial expansion).

Equating imaginary parts:

sin(5θ)=5cos⁴θsinθ-10cos²θsin³θ+sin⁵θ.

cos²θ=1-sin²θ, so cos⁴θ=(1-sin²θ)²=1-2sin²θ+sin⁴θ;

sin(5θ)=5sinθ-10sin³θ+5sin⁵θ-10sin³θ+10sin⁵θ+sin⁵θ,

sin(5θ)=16sin⁵θ-20sin³θ+5sinθ.

[This sine of a multiple angle can also be found by progressively applying the identity sin(A+B)=sinAcosB+cosAsinB until the expression only contains the sine of a single angle. De Moivre offers a quicker way to attain such an expression.]

Since θ=36°, sin(5θ)=sin(180)=0:

0=sinθ(16sin⁴θ-20sin²θ+5), 16sin⁴θ-20sin²θ+5=0, since sinθ≠0.

Applying the quadratic formula:

sin²θ=sin²(36)=(20±√(400-320))/32=(20±4√5)/32=(5±√5)/8.

Only one of these solutions can be correct. Since sin²(30)=¼=0.25 and sin²(45)=½=0.5:

0.25<sin²(36)<0.5. But (5+√5)/8≈0.9, so we can only accept sin²(36)=(5-√5)/8=0.35 approx.

sin²(72)=(2sin(36)cos(36))²=4sin²(36)cos²(36)=

4sin²(36)(1-sin²(36))=½(5-√5)(1-(5-√5)/8),

sin²(72)=(1/16)(5-√5)(3+√5).

sin²(36)sin²(72)=(1/128)(5-√5)²(3+√5)=(1/128)(30-10√5)(3+√5)=

(1/128)(90-50)=40/128=5/16 (QED).