Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

FP2 De-Moivre

Could somebody help me out with the following please. Its part (c). Not even sure where to start, so any pointers would be appreciated. Thanks in advance.

Original post by kingm

Could somebody help me out with the following please. Its part (c). Not even sure where to start, so any pointers would be appreciated. Thanks in advance.

You need to use the previous part, bearing A, B and C in mind.

If you're still unsure, state the values of A, B and C

(edited 11 years ago)

how do u do the whole question? how do u do part a?

Original post by study beats

how do u do the whole question? how do u do part a?

Seriously? :eek:

:eek:

For part (a), use the fact that

\dfrac{1}{z^n} = z^{-n}

and note that

z = r(\cos x + i\sin x) \Rightarrow z^n = r^n (\cos nx + i\sin nx)

(edited 11 years ago)

Original post by Indeterminate

Seriously? :eek:

:eek:

For part (a), use the fact that

\dfrac{1}{z^n} = z^{-n}

and note that

z = r(\cos x + i\sin x) \Rightarrow z^n = r^n (\cos nx + i\sin nx)

whats 1/z

1/ cosx+sinx ?

Original post by study beats

whats 1/z

1/ cosx+sinx ?

Well, yes, but if you raise z to any power you multiply the argument by the same.

OP

Original post by Indeterminate

You need to use the previous part, bearing A, B and C in mind.

If you're still unsure, state the values of A, B and C

Yeah, i'm still unsure. A, B and C are 1/8, 1/2 and 3/4 respectively. I get that you need to use the previous parts, but i'm not sure how and why.

Original post by kingm

Yeah, i'm still unsure. A, B and C are 1/8, 1/2 and 3/4 respectively. I get that you need to use the previous parts, but i'm not sure how and why.

Well,

8\cos^4 2\theta = \cos 8\theta + 4\cos 4\theta + 6

(edited 11 years ago)

OP

Original post by Indeterminate

Well,

8\cos^4 \theta = \cos 8\theta + 4\cos 4\theta + 6

lool. That was rather easy in the end :-)

Cheers dude.

OP

Original post by Indeterminate

Well,

8\cos^4 2\theta = \cos 8\theta + 4\cos 4\theta + 6

Actually wait as sec. I'm sure that this is even more simple, but how is it that they then get k is equal to 1/12, 5/12, 7/12 and 11/12??

Original post by kingm

Actually wait as sec. I'm sure that this is even more simple, but how is it that they then get k is equal to 1/12, 5/12, 7/12 and 11/12??

Well, you may have made an error in working out A, B and C as they must have had

4 \theta = \frac{\pi}{3}

OP

Original post by Indeterminate

Well,

8\cos^4 2\theta = \cos 8\theta + 4\cos 4\theta + 6

It should be +3 here instead of +6. So thats where you may be getting mixed up.

Original post by Indeterminate

Well, you may have made an error in working out A, B and C as they must have had

4 \theta = \frac{\pi}{3}

The answer is definitely cos4(theta)=1/2. I'm just confused about how to get to their answers now!

(edited 11 years ago)

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