fitzpatrick q's (1 Viewer)

[shkspeare]

arghh im having trouble with 3 of these fitzpatrick questions

here goes :

1. Show that the cubic equation 8x³ -6x + 1 = 0 can be reduced to the form cos3@ = -1/2 by substituting x = cos@

Deduce that cos2pi/9 + cos4pi/9 = cospi/9

2. If tan(@), tan(#), tan(%) are roots the the equation x³ - (a+1)x² + (c-a)x - c = 0

Show that (@) + (#) + (%) = npi + pi/4

3. Expand cos(2A+B) and hence prove that 1/4cos3@ = cos³@ - (3/4)(cos@)

Putting x = kclos@ and giving k a suitable value use the preceeding formula to find the three roots of the equation 27x³ - 9x = 1. Hence write down the value of the product cospi/9 . cos3pi/9 . cos5pi/9 . cos7pi/9

THANKS A LOT!! ive been struggling with these for a while

 

[ND]

Do you want hints or solutions? I'll assume the former:

1.) Expand cos3@ and look at the similarity between that and 8x3 -6x + 1 = 0 (remembering that it is given that cos3@=-1/2). Also use sum of roots.

2.) Expand tan(@+#+%), look at the similarity of that and the sum of roots, sum of roots taken two at a time and product of roots.

3.) Similar to 1., only this time use product of roots. Also remember that 1/2=cos(3pi/9).

 

[kpq_sniper017]

Q1:
8x³-6x+1 = 0
Substituting x=cos@:

8cos³@-6cos@ = -1
.'. cos@(4cos²@-3) = -1/2 (divide through by 2 and factorise)
The LHS is the factorised form of expanded and factorised form of cos3@.

I'll leave it up to CM_Tutor, Affinity or Grey Council etc. to see if that's all that's needed. My only question to them (on your behalf), would be: do you have to prove that cos3@ = LHS....??

 

[ND]

You also have to get the values of @ that make cos3@=-1/2, and do sum of roots to deduce cos2pi/9 + cos4pi/9 = cospi/9.

edit: if it were in an exam you would also have to expand cos3@.

 

[shkspeare]

mmmm i still dont get it >.<"

anyone can provide the full or a brief solution =\

 

[ND]

Ok:

1. Cos3@=4(cos@)^3-3cos@=-1/2
so 8(cos@)^3-6cos@+1=0
letting x=cos@ gets you to 8x^3-6x+1=0.
and cos3@=-1/2
3@=2pi/3, 4pi/3 8pi/3
@=2pi/9, 4pi/9, 8pi/9.
.'. roots of 8x^3-6x+1=0 are cos2pi/9, cos4pi/9, cos8pi/9.
Sum of roots:
cos2pi/9+cos4pi/9+cos8pi/9=0
and cos8pi/9=-cospi/9, so:
cos2pi/9 + cos4pi/9 = cospi/9

2. x^3-(a+1)x^2+(c-a)x-c = 0
for ease, i'll let tan@=p, tan(#)=q, tan(%)=r.
p+q+r=a+1
pq+pr+qr=c-a
pqr=-c
summing these gives:
p+q+r+pq+qr+pr+pqr=1
p+q+r+pqr=1-pq+qr+pr
(p+q+r+pqr)/(1-pq+pr+qr)=1
taking inverse tan of both sides gives:
(@) + (#) + (%) = npi + pi/4 (as the LHS is the expansion of tan(@+#+%)

I think you should be able to do 3 after seeing the solution to 1. (as they are very similar).

 

[:: ck ::]

these q's look like 4u polynomials involving trig... ?