Polynomials trig q. (1 Viewer)

[namburger]

Need help:

Given cos 4t = 8x^4 - 8x^2 + 1 and cos 3t = 4x^3 - 3x where cos t = x
a) Show that the equation cos 4t = cos 3t can be expressed as
8x^4 - 4x^3 - 8x^2 + 3x + 1 = 0.
b) By considering the solutions to cos 4t = cos 3t, deduce that:
1/2 + cos 2pi/7 = cos pi/7 + cos 3pi/7

a is straightforward
for b, you can re-arrange the equation to 1/2 = cos pi/7 + cos 3pi/7 + cos 5pi/7, if that helps you..

 

[Affinity]

hint:
cos(3t)-cos(4t)=2sin(7t/2)sin(t/2)

 

[Trefoil]

Interestingly, those identities are derived from the trigonometric expression a complex number (cis@) and applications of de Moivre's theorem.

E.g.
cis(@)^2=cis(2@)=cos^2(@)-sin^2(@)+2isin(@)cos(@)
Thus cos(2@)=cos^2(@)-sin^2(@) and sin(2@)=2sin(@)cos(@)

 

[namburger]

hint:
cos(3t)-cos(4t)=2sin(7t/2)sin(t/2)

I need further guidance.
sin(7t/2)sin(t/2) = 0
sin(7t/2) =0
(7t/2) = kpi
t = 2kpi/7

OR
sin(t/2) = 0
t = 2kpi (which is not required as the solution to the top equation cover this one)

So the solutions of t are 0, 2pi/7, 4pi/7, 6pi/7, -2pi/7, -4pi/7, -6pi/7

I figured out how to get to the answer but don't understand some parts:
SO x = cos of the above. so x = 1, cos 2pi/7, cos 4pi/7, cos 6pi/7
I know that cos 2pi/7 = cos -2pi/7 but how come you can't take it as another root like a double root or something?

1 + cos 2pi/7 + cos 4pi/7 + cos 6pi/7 = 1/2
1/2 + cos 2pi/7 + cos 4pi/7 + cos 6pi/7 = 0
1/2 + cos 2pi/7 = cos 3pi/7 + cos pi/7

Is there any other method in doing this.. this was in a polynomials test and i dont think we are suppose to use additions and subtractions of trig atm..

 

[Trebla]

I need further guidance.

Note that:
cos (4t) - cos (3t) = cos (7t/2 + t/2) - cos (7t/2 - t/2)
Expand the RHS and you should get 2sin(7t/2)sin(t/2)

Alternatively you could use the general solutions of cosine:
cos 4t = cos 3t
4t = 2kπ ± 3t where k is an integer
If we take the positive case, we just get multiples of 2π which is pretty pointless. We are more interested in the negative case:
4t = 2kπ - 3t
=> t = 2kπ/7
Let k = 0,1,2,3,4,5,6 to get values of t and use sum of roots with a few trigonometric identities [cos x = cos (2π - x)] to get the result

Also, I think you typed the original polynomial equation wrong...