# De Moivres Theorem Q

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#1
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...
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4 weeks ago
#2
(Original post by jamiet0185)
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...
Not sure what you've covered? Does (z+/-1/z) mean anything?

If so, just multiply the cos^4 and sin^3 expressions together.
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#3
(Original post by jamiet0185)
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...
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
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#4
(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!
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#5
(Original post by BrandonS15)
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
Ah, thanks for that, I'll give it a try!
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4 weeks ago
#6
(Original post by jamiet0185)
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!
https://www.wolframalpha.com/input/?...sin%5E3%28x%29
Last edited by mqb2766; 4 weeks ago
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#7
(Original post by mqb2766)
Thank you
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4 weeks ago
#8
(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.
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4 weeks ago
#9
(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

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

. If you then expand out the integrand, you're left with a bunch of terms of form , which can be integrated by recognition as .

(This totally fails to work when n is even, unfortunately).
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#10
Thanks everyone for the help.
Ignore the picture below
Last edited by jamiet0185; 4 weeks ago
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#11
I'm just not getting to the right answer. Can someone point out where I've gone wrong please?
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4 weeks ago
#12
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).
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#13
(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!
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4 weeks ago
#14
(Original post by jamiet0185)
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...
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.
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4 weeks ago
#15
(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.
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#16
I'm really sorry but there's another bit I don't know how to approach. Could anyone suggest how to start Q8ii
Thanks again!!
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4 weeks ago
#17
Divide by cos t (which you're told is not zero). RHS is then a quadratic in cos^2 t.
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4 weeks ago
#18
(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
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
0
4 weeks ago
#19
(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.
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