# De Moivres Theorem Homework Help??

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#1
A Level Further Maths

(i) Express e^ikθ and e^-ikθ in the form a + ib, and show that
e^2iθ - 1 = 2ie^iθ sinθ

My working:

e^ikθ = cos(kθ) + isin(kθ)
e^-ikθ = cos(kθ) - isin(kθ)

e^2iθ - 1 = cos(2θ) + isin(2θ) - 1

From here on I'm a bit stuck... I've done the expansion of (c + is)^2 but I don't know if I'm doing the right thing. Help???
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1 month ago
#2
(Original post by woffle)
A Level Further Maths

(i) Express e^ikθ and e^-ikθ in the form a + ib, and show that
e^2iθ - 1 = 2ie^iθ sinθ

My working:

e^ikθ = cos(kθ) + isin(kθ)
e^-ikθ = cos(kθ) - isin(kθ)

e^2iθ - 1 = cos(2θ) + isin(2θ) - 1

From here on I'm a bit stuck... I've done the expansion of (c + is)^2 but I don't know if I'm doing the right thing. Help???
There are a few ways you could go. If you've come across the identity for sin(theta) in terms of imaginary exponentials, you're being asked to prove a variant on that. Can you think of any way to rearrange/transform the left hand side, or the right hand side to make progress? You can obviously start on either side.

Its not easy to give much more of a hint without telling you what to do, so even if things don't work, describe what you've tried. Your idea of (c+is)^2 is one way to go, what can you do with the 1?
Last edited by mqb2766; 1 month ago
0
1 month ago
#3
(Original post by woffle)
A Level Further Maths

(i) Express e^ikθ and e^-ikθ in the form a + ib, and show that
e^2iθ - 1 = 2ie^iθ sinθ

My working:

e^ikθ = cos(kθ) + isin(kθ)
e^-ikθ = cos(kθ) - isin(kθ)

e^2iθ - 1 = cos(2θ) + isin(2θ) - 1

From here on I'm a bit stuck... I've done the expansion of (c + is)^2 but I don't know if I'm doing the right thing. Help???
Given the way you've started (which is perfectly good), I would suggest subtracting your second line of working from your first line of working, and see where that leads.
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