De Moivres Theorem Q

Hi
Could anyone help me get started with this question please?
I understand how to do it for example with cos^5x or sin^3x but I'm not sure what to when they're multiplied together as in question 6 below...
Thanks in advance! image-76994783-9bc7-46f0-a301-4baa1f76aa468252122157487141-compressed.jpg.jpeg

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Reply 1

mqb2766

Original post by jamiet0185

Not sure what you've covered? Does (z+/-1/z) mean anything?

http://www.cosmicriver.net/uploads/5/3/0/7/5307578/algebra8b.pdf

If so, just multiply the cos^4 and sin^3 expressions together.

(edited 3 years ago)

Reply 2

BrandonS15

Original post by jamiet0185

Note that cos^4(theta) can be written as (1-sin^2(theta))^2
Maybe manipulation using this may help you express the integrand in terms of powers of sines and then you may be able to integrate term by term

Reply 3

jamiet0185

Original post by mqb2766

Not sure what you've covered? Does (z+/-1/z) mean anything?. If so, just multiply the cos^4 and sin^3 expressions together.

Thanks for that
Yeah, it does mean something to me.
I thought about multiplying them but didn't because I thought it would give something horrible, very possibly with things being squared again (which obviously is not the objective of the question) but I will give it a go!

Reply 4

jamiet0185

Original post by BrandonS15

Ah, thanks for that, I'll give it a try!

Reply 5

mqb2766

Original post by jamiet0185

Added link with some examples.
https://www.wolframalpha.com/input/?i=+cos%5E4%28x%29sin%5E3%28x%29

(edited 3 years ago)

Reply 6

jamiet0185

Original post by mqb2766

Added link with some examples.

Thank you

Reply 7

mqb2766

Original post by jamiet0185

Thank you

The wolfram answer (previous post) tends to suggest using brandons s idea and mapping everything to sin then expand. Obviously there are several routes to an answer.

Reply 8

DFranklin

Original post by jamiet0185

Ah, thanks for that, I'll give it a try!

It should work, but I don't see it as using de Moivre (I'm not clear whether you are supposed to be using it for all these questions).

If you're aiming to get to a sum of single trig functions, I'd use the (z+1/z)^4(z-1/z)^3 form, note that (z+1/z)(z-1/z) = (z^2 - 1/z^2) to rewrite as (z^2-1/z^2)^3 (z+1/z), and multiply out. It's not too bad. (It does feel there "should" be a slightly shorter way, but it's short enough to be fairly acceptable).

Incidentally, if we're talking about "not using de Moivre", it's worth noting that if you want to find

\int \sin^n x\,dx

where n is odd, then the fastest solution is almost certainly to rewrite the integrand as

\int \sin x (1 - \cos^2 x)^{(n-1)/2}\, dx

. If you then expand out the integrand, you're left with a bunch of terms of form

\sin x \cos^k x

, which can be integrated by recognition as

- \frac{1}{k+1} \cos^{k+1} x

.

(This totally fails to work when n is even, unfortunately).

Reply 9

jamiet0185

Thanks everyone for the help.
Ignore the picture below

(edited 3 years ago)

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Reply 10

jamiet0185

I'm just not getting to the right answer. Can someone point out where I've gone wrong please? image-869ade6b-e1d2-45d7-bec1-539666e24b893700568741020597600-compressed.jpg.jpeg

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Reply 11

DFranklin

When you "convert back" from your powers of z to sin/cos, you've gone from z^k -1/z^k to sin kz when you should have gone to 2i sin kz. (Not sure that's the only error - on mobile).

Reply 12

jamiet0185

Original post by DFranklin

When you "convert back" from your powers of z to sin/cos, you've gone from z^k -1/z^k to sin kz when you should have gone to 2i sin kz. (Not sure that's the only error - on mobile).

Thanks very much!

Reply 13

boulderingislife

Original post by jamiet0185

Have you tried the identity: cos^2 x = 1-sin^2 x?

My hunch is that once you do that, you can proceed as per normal.

Reply 14

DFranklin

Original post by boulderingislife

Have you tried the identity: cos^2 x = 1-sin^2 x?

My hunch is that once you do that, you can proceed as per normal.

Although it's a viable method, it's not the desired approach if you are supposed to use De Moivre's theorem.

Reply 15

jamiet0185

I'm really sorry but there's another bit I don't know how to approach. Could anyone suggest how to start Q8ii image-bc127458-f36b-441e-8fe2-7891d9f36cbb7022474636687117602-compressed.jpg.jpeg

image-bc127458-f36b-441e-8fe2-7891d9f36cbb7022474636687117602-compressed.jpg.jpeg

Thanks again!!

Reply 16

DFranklin

Divide by cos t (which you're told is not zero). RHS is then a quadratic in cos^2 t.

Reply 17

boulderingislife

Original post by jamiet0185

I'm really sorry but there's another bit I don't know how to approach. Could anyone suggest how to start Q8ii image-bc127458-f36b-441e-8fe2-7891d9f36cbb7022474636687117602-compressed.jpg.jpeg

Thanks again!!

Let cos theta = t, factorise out one t, then you are left with t^4, a t^2 and a constant. Now if you let t^2=y, you should be able to factorise the resulting expression and sub back t.

Should end up with something like:
t(t^2+k)(t^2+h) =0. Then you can solve for theta

Reply 18

boulderingislife

Original post by DFranklin

Although it's a viable method, it's not the desired approach if you are supposed to use De Moivre's theorem.

Didn’t read the question properly, you’re right. Using the identity would give you a single theta expression, whereas a multiple theta expression is required. So demoivre is the way forward.