De Moivres (1 Viewer)

[ezzy85]

I saw this q in 2 past papers and cant get it. I tried the usual approach with de moivres but this one seems different.

Solve the equation sin5@ = 1 for 0<@<2pi, and hence show that the equation 16x⁴ + 16x³ - 4x² - 4x + 1 = 0 are x=[sin(4r+1)pi]/10, where r = 0,2,3,4

Thanks

 

[ND]

Too late for maths.

First part:

If sin5@ = 1
isin5@ = i
z^5 = i
= cis(pi/2)
.'. 5@ = pi/2 + 2kpi
@ = pi/10 + 2kpi/5
= pi(4k + 1)/10
where k = 1, 2, 3, 4

 

[ND]

Ok, back for more:

sin5@ = 16(sin@)^5 - 20(sin@)^3 + 5sin@ = 1
and sin(pi(4k + 1)/10) = 1
so 16x^5 - 20x^3 + 5x - 1 = 0 where x = sin(pi(4k + 1)/10)

Now one of the factors is (x - 1), upon dividng by long division:

16x^4 + 16x^3 - 4x^2 - 4x + 1 = 0

shown (i hope i gave enough explanation).

edit: after reading spice girl's soln, i realize i could explain it much better, if anyone doesn't understand anything, i'll explain it tomorrow.

 

[ND]

What happened to spice girl's post?

 

[spice girl]

i liked your solution better than my own...

 

[ezzy85]

cool, thanks for that