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FP2 edexcel exercise 3D question 7(b),(c) help!!!

a)show that sin4theta=4cos^3(theta)sin(theta)-4costhetasin^3theta

b)hence or otherwise show that tan4theta=4tantheta-4tan^3theta divided by 1-6tan^2theta+tan^4theta

c)use your answer to part b to find,to 2 d.p the four solutions of the equation X^4+4X^2-4X+1=0

(edited 13 years ago)

Reply 1

ghostwalker

17

stoneaxe

a)show that sin4theta=4cos^3(theta)sin(theta)-4costhetasin^3theta

b)hence or otherwise show that tan4theta=4tantheta-4tan^3theta divided by 1-6tan^2theta+tan^4theta

c)use your answer to part b to find,to 2 d.p the four solutions of the equation X^4+4X^2-4X+1=0

For part b) work out cos4theta in terms of sin(theta) and cos(theta), then use sin4theta/cos4theta to work out tan4theta.

c) Can't see the methodology of that at present.

Edit: For c; I just ran that through a graphing package, and it has no solutons, so there's either a typo in the question, or it's been copied down wrong.

(edited 13 years ago)

Reply 2

rbnphlp

11

ghostwalker

c) Can't see the methodology of that at present.

he's written out the question incorrectly, it should be x^4+4x^3-6x^2-4x+1=0

Reply 3

DFranklin

18

ghostwalker

c) Can't see the methodology of that at present.

Since the equation posted has no real roots, I am not entirely surprised...

rbnphlp

..

How the heck do you know that from what he's posted? [AKA: Dammit, Jim! I'm a mathematician, not a clairvoyant!]

(edited 13 years ago)

Reply 4

ghostwalker

17

DFranklin

Since the equation posted has no real roots, I am not entirely surprised...

Yep, just updated as you were posting.

Edit: A Trek'er I see.

(edited 13 years ago)

Reply 5

rbnphlp

11

DFranklin

How the heck do you know that from what he's posted? [AKA: Dammit, Jim! I'm a mathematician, not a clairvoyant!]

because I read the thread title :smug:

(edited 13 years ago)

Reply 6

Slowbro93

20

I remember this question, I was confused as to what do do, as I couldn't get in the in form they wanted

Reply 7

ghostwalker

17

@OP So for part c), use the obvious substitution tan(theta) = x, and see if you can arrange your equation to look like the RHS of b)....

(edited 13 years ago)

Reply 8

olipal

1

Great, I read it as show that:

\sin^4\theta=4\cos^3\theta sin\theta -4\cos \theta \sin^3 \theta

Would like the last 20 minutes back plz :frown:

Reply 9

Bumbafool

7b) Equate the real parts of the equation:

cos4θ = cos^θ - 6cos^2θsin^2θ + sin^4θ

So : tan4θ = (sin4θ) / (cos4θ)
= (4cos^3θsinθ - 4cosθsin^3θ)
(cos^4θ - 6cos^2θsin^2θ + sin^4θ)

Divide through by cos^4θ to get the result shown.

7c) First let x = tanθ
So: tan4θ = 4x - 4x^3
1 - 6x^2 + x^4

Then, when tan4θ = 1:
1 - 6x^2 + x^4 = 4x - 4x^3
x^4 + 4x^3 - 6x^2 -4x + 1 = 0

tan4θ = 1
4θ = arctan(1)
θ = ... π/16, 5π/16, 9π/16, 13π/16, ...

But x = tanθ
So x = tan(π/16), tan(5π/16), tan(9π/16), tan(13π/16)
x = 0.20, 1.50, -5.03, -0.67 (2dp)

(The infinite results acquired for theta will cancel, as both tan and arctan have been used on the equation, leaving only 4 results)

Reply 10

L'Evil Wolf

6

@Zacken

@physicsmaths

hey you know for this question, above user said that:

"(The infinite results acquired for theta will cancel, as both tan and arctan have been used on the equation, leaving only 4 results)"

Will this hold for the other trig identities, is this required knowledge that if a trig function and an inverse is used then only 4 solutions will be yielded?

Reply 11

Zacken

22

Original post by L'Evil Wolf

@Zacken

@physicsmaths

hey you know for this question, above user said that:

"(The infinite results acquired for theta will cancel, as both tan and arctan have been used on the equation, leaving only 4 results)"

Will this hold for the other trig identities, is this required knowledge that if a trig function and an inverse is used then only 4 solutions will be yielded?

Huh? You should know that tangent is pi-periodic and sine/cosine is 2pi-periodic.

Reply 12

L'Evil Wolf

6

Original post by Zacken

Huh? You should know that tangent is pi-periodic and sine/cosine is 2pi-periodic.

yeah that holds, but for tangent then as all possible soltuions tend to positive and negative infinity will there only ever be 4 solutions?

Reply 13

Zacken

22

Original post by L'Evil Wolf

yeah that holds, but for tangent then as all possible soltuions tend to positive and negative infinity will there only ever be 4 solutions?

What? The infinite solutions for theta only give four possible values for tan theta because tangent is a pi-periodic function.

Reply 14

username1162972

18

Original post by L'Evil Wolf

@Zacken

@physicsmaths

hey you know for this question, above user said that:

"(The infinite results acquired for theta will cancel, as both tan and arctan have been used on the equation, leaving only 4 results)"

Will this hold for the other trig identities, is this required knowledge that if a trig function and an inverse is used then only 4 solutions will be yielded?

Well I am pretty good at trig but have not heard what he is talking about. Unless it is trivial and I have misunderstood.
We take the principal values and add pi tbh. Nothing more.

Reply 15

username1162972

18

I understand the question now,
Zack hes saying why dont we have add more pis sonce there are infinte solutions.
Reason is we take the 4 dstinct solutions, after this we just add more pi and get the same solutions so we can jjst stop as there are inifinite distinct theta but only atmost 4 real x satisfying the equation(fundemental theorem of somethin)

Reply 16

Zacken

22

Original post by physicsmaths

I understand the question now,
Zack hes saying why dont we have add more pis sonce there are infinte solutions.
Reason is we take the 4 dstinct solutions, after this we just add more pi and get the same solutions so we can jjst stop as there are inifinite distinct theta but only atmost 4 real x satisfying the equation(fundemental theorem of somethin)

Yeah, like I said, any other theta found will be pi + (one of the thetas already found) so taking the tangent of that will generate the same solution.

(Fundamental theorem of algebra ensures that there are 4 solutions to a fourth degree polynomial with real coefficients in the complex field)

Reply 17

username1162972

18

Original post by Zacken

Yeah, like I said, any other theta found will be pi + (one of the thetas already found) so taking the tangent of that will generate the same solution.

(Fundamental theorem of algebra ensures that there are 4 solutions to a fourth degree polynomial with real coefficients in the complex field)

Indeed mate, indeed.

Posted from TSR Mobile

Reply 18

L'Evil Wolf

6

Original post by Zacken

What? The infinite solutions for theta only give four possible values for tan theta because tangent is a pi-periodic function.

Original post by physicsmaths

I understand the question now,
Zack hes saying why dont we have add more pis sonce there are infinte solutions.
Reason is we take the 4 dstinct solutions, after this we just add more pi and get the same solutions so we can jjst stop as there are inifinite distinct theta but only atmost 4 real x satisfying the equation(fundemental theorem of somethin)

Original post by Zacken

Yeah, like I said, any other theta found will be pi + (one of the thetas already found) so taking the tangent of that will generate the same solution.

(Fundamental theorem of algebra ensures that there are 4 solutions to a fourth degree polynomial with real coefficients in the complex field)

Thank you both for your help again.

The question did say there were only four solutions, I just wanted to understand why.

FP2 edexcel exercise 3D question 7(b),(c) help!!!

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