Thread: finding the exact trigonometric ratios of an equation

Find the exact trigonometric roots of
given that

heres how I did it

let x= cos@
cos5@ = 16x^5 - 20x^3 + 5x + 1
cos5@ = 0, 16x^5 -20x^3 + 5x + 1 = 0

5@ = 2(pi)k +/- pi/2 where k is any integer
= (2(pi)k +/- pi)/10

x = cos@
= cos {2(pi)k +/- pi]/10}
where k is any integer

but how do I proceed to find the five roots of the equation. and is the above correct

Hello differentiate

Originally Posted by differentiate

Find the exact trigonometric roots of
given that

heres how I did it

let x= cos@
cos5@ = 16x^5 - 20x^3 + 5x + 1
cos5@ = 0, 16x^5 -20x^3 + 5x + 1 = 0

5@ = 2(pi)k +/- pi/2 where k is any integer
= (2(pi)k +/- pi)/10

x = cos@
= cos {2(pi)k +/- pi]/10}
where k is any integer

but how do I proceed to find the five roots of the equation. and is the above correct

You have more-or-less the right idea, but are wrong in one or two details.

Let

Then (not, as you said, )

So









So the roots are

(Note that these are not all distinct. The equation has two pairs of equal roots, because and .)

Grandad