Polynomials (1 Viewer)

[DcM]

I need help...for polynomials from Fitzpatrick
Ex 27 d

Q 25
Show that the cubic equation 8x^3 - 6x +1 =0 can be reduced to the form cos 3theta = -1/2 by substituting x = cos theta.

Deduce that:
a) cos 2pi/9 + cos 4pi/9 = cos pi/9

b) sec 2pi/9 + sec 4pi/9 = 6 + sec pi/9

c) sec pi/9 sec 2pi/9 sec 4pi/9 = 8

d) tan^2 pi/9 + tan^2 2pi/9 + tan^2 4pi/9 = 33

thanks!

 

[J0n]

Q 25
8x³ - 6x + 1 = 0
8cos³@ - 6cos@ + 1 = 0
2cos@(4cos²@ - 3cos@) + 1 = 0
2cos@(2(2cos²@ - 1) - 1) + 1 = 0
2cos@(2cos2@ - 1) + 1 = 0
4cos2@cos@ - 2cos@ + 1 = 0
4*(1/2)(cos3@ + cos@) - 2cos@ + 1 = 0
2cos3@ + 2cos@ - 2cos@ + 1 = 0
cos3@ = -1/2

 

[DcM]

4cos2@cos@ - 2cos@ + 1 = 0
4*(1/2)(cos3@ + cos@) - 2cos@ + 1 = 0
2cos3@ + 2cos@ - 2cos@ + 1 = 0


what happened in these 3 lines?

 

[iambored]

damn i can't believe i have forgotten this already

i think it's got something to do with the forumulas. if you look at the forumulas there must be some that convert the lines as above?

 

[J0n]

Originally posted by DcM
4cos2@cos@ - 2cos@ + 1 = 0
4*(1/2)(cos3@ + cos@) - 2cos@ + 1 = 0
2cos3@ + 2cos@ - 2cos@ + 1 = 0


what happened in these 3 lines?

There is a formula:
cos xcos y = 1/2[cos(x+y) + cos(x-y)]

Proof
LHS = cos xcos y
=(1/2)*2cos xcos y
=1/2[cos xcos y - sin xsin y + cos xcos y + sin xsin y]
=1/2[cos(x+y) + cos(x-y)]
=RHS

 

[DcM]

Thanks JOn

i got another one which is hard..and i cant do as well...
from the same exercise but the next 2 questions

Q26. If tan X, tan Y, tan Z are the roots of the eqn x^3 - (a + 1)x^2 + (c - a)x - c = 0
show that X + Y + Z = npi + pi/4

Q27. Expand cos (2A + B) and hence prove that 1/4 cos 3@ = cos^3 @ - 3/4 cos @
Putting x = k Cos @ and giving k a suitable value use the preceding formula to find the three roots of the equation 27x^3 - 9x = 1. Hence write down the value of the product cos pi/9 cos 3pi/9 cos 5pi/9 cos 7pi/9.

(For this question i have proved the first part..but duno how to do the second part.)

Thanks

 

[ND]

Q26. Try expanding tan(X+Y+Z), then use the sum of roots, sum of roots in 2's, etc. and play with it a little, shouldn't be too difficult...


Q27. Are you sure you're not missing any info? Like cos3@=1/2 ? I'm going to go with that anyway:

cos3@=1/2
3@=pi/3+2npi (where n is an integer)
@=pi/9+2npi/3

Now 1/4 cos 3@ = cos^3 @ - 3/4 cos @ --> 1=8(cos@)^3-4cos@
which means that 1=27x^3-9x where x=2/3*cos@.

So the roots of this are:
when n=0, x=2/3*cos(pi/9),
when n=-1, x=2/3*cos(-5pi/9)=2/3*cos(5pi/9)
n=1, x=2/3*cos(7pi/9)

so product of roots:
(2/3)^3*cos(pi/9)*cos(5pi/9)*cos(7pi/9)=1/27
--> 8*cos(pi/9)*cos(5pi/9)*cos(7pi/9)=1
--> cos(pi/9)*cos(3pi/9)*cos(5pi/9)*cos(7pi/9)=1/16 (because cos(3pi/9)=1/2)

Here's a similar question:

http://www.boredofstudies.org/community/showthread.php?s=&threadid=6197 (looking back on that, the 1st bit i did was pretty round-about)

 

[freaking_out]

Originally posted by J0n
There is a formula:
cos xcos y = 1/2[cos(x+y) + cos(x-y)]

Proof
LHS = cos xcos y
=(1/2)*2cos xcos y
=1/2[cos xcos y - sin xsin y + cos xcos y + sin xsin y]
=1/2[cos(x+y) + cos(x-y)]
=RHS

yeah, i was told, that if in an exam they are gonna have a question which requires u to use this formula- they will make u derive it in the first part of the question. i hope u understood what i mean.