Question

Help me year 11 trig question
(a) prove(i have done this part already)
cos3a = 4cos^3 a - 3cosa
(b) Show that cos40 is a root of the equation 8x^3 - 6x +1 = 0

I need help with part (b) only

Answer 1

@jim_thompson5910 @ParthKohli help please :)

Answer 2

well you have part a) already done, so why not replace 'a' with 40 and see what happens

cos3a = 4cos^3 a - 3cosa

cos(3*40) = 4cos^3(40) - 3cos(40)

cos(120) = 4cos^3(40) - 3cos(40)

-1/2 = 4cos^3(40) - 3cos(40)

Answer 3

You could just plug in x=cos40

Answer 4

all you need to do is find the cosine of 40 degrees and replace x in the equation.
cos40=0.7660

Answer 5

Now multiply both sides of the last equation by 2 to get

-1/2 = 4cos^3(40) - 3cos(40)

2*(-1/2) = 2*(4cos^3(40) - 3cos(40) )

-1 = 8cos^3(40) - 6cos(40)

0 = 8cos^3(40) - 6cos(40) + 1

8cos^3(40) - 6cos(40) + 1 = 0

so that confirms x = cos(40) is a root of 8x^3 - 6x +1 = 0

Answer 6

oh :) thank you so much. didnt have any clue what to do :)

Answer 7

@jim_thompson5910 nice ... @AonZ we can also find by replacing x with cos40(0.7660) but this is very long method.
when you replace x in your equation with the stored value of .7660, the result is 0 indicating that cosine of 40 degrees is a root of the equation.
and by this, jim had proven this!!!

good job @jim_thompson5910

Answer 8

ty

Answer 9

@jim_thompson5910 way is the best :) cause it uses trig which is the topic i am currently doing :D