Answer to Question #113664 in Trigonometry for Caylin

Question

(De Moivres theorem) - Use Cos5A and sin4A as polynomials in sinA and CosA to evaluate:

1) sin pi/5
2) sin2*pi/5
3) cos pi/5.

Answer 1

De Moivres theorem states that if z=r cosA+ i r sinA

then, zⁿ = cos (nA) + i sin (nA)

so, using de moivres theorem, (cosA + i sinA)⁵ = cos(5A) + i sin(5A)

using binomial theorem ,

cos(5A) + i sin(5A) = (cosA)⁵ - 10 cos³A sin²A + 5 cosA sin⁴A + i( 5cos⁴A sinA + 10 cos²A sin³A + sin⁵A)

comparing both sides,

cos(5A) = (cosA)⁵ - 10 cos³A sin²A + 5 cosA sin⁴A

using sin²x + cos²x = 1

cos(5A) = 16 cos⁵A - 20 cos³A + 5 cosA equation1

sin(5A) = 5cos⁴A sinA + 10 cos²A sin³A + sin⁵A

using sin²x + cos²x = 1

sin(5A) = 16 sin⁵A - 20 sin³A + 5 sinA equation2

now, let y=(Π /5 ) ,x= cosy in equation 1

-1 = 16 x⁵ - 20x³ + 5x +1

(x+1)[x- (1-$\sqrt{\smash[b]{5}} )/4$ )[(x-(1+$\sqrt{\smash[b]{5}})/4$ ]=0

since cos(Π /5) lies between 0 and 1

so,cos( Π/5) = (($\sqrt{5}$ +1)/4)

let y= (Π /5) and x= siny in equation2

16 x⁴ - 20 x² + 5 = 0

x = $\mp \sqrt{\smash[b]{(5 \mp \sqrt{\smash[b]{5}})/8}}$

since sin(Π/5) lies between 0 and $\sqrt{\smash[b]{3}}$ /2

sin( Π /5) = $\sqrt{\smash[b]{(5-\sqrt{\smash[b]{5}})/8}}$

similarly, if y = 2$\Pi$ /5 and x=siny in equation2 and sin(2$\Pi$ /5) lies between $\sqrt{\smash[b]{3}}$ /2 and 1,

so sin(2$\Pi$ /5) = $\sqrt{\smash[b]{(5+\sqrt{\smash[b]{5}})/8}}$