Question #113664
(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.
1
Expert's answer
2020-05-08T15:23:39-0400

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

since cos(Π /5) lies between 0 and 1

so,cos( Π/5) = ((5\sqrt{5} +1)/4)


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

16 x4 - 20 x2 + 5 = 0

x = (55)/8\mp \sqrt{\smash[b]{(5 \mp \sqrt{\smash[b]{5}})/8}}

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

sin( Π /5) = (55)/8\sqrt{\smash[b]{(5-\sqrt{\smash[b]{5}})/8}}


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

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



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