# De Moivres Theorem - Help please

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#1
Cos^4 + sin^4 = 1/4(cos4(theta)+3)

Prove the following Trig Identity using De Moivre's Theorem

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1 month ago
#2
(Original post by Herbie0209)
Cos^4 + sin^4 = 1/4(cos4(theta)+3)

Prove the following Trig Identity using De Moivre's Theorem

Let z = e^ia. You should already know that cos(a) = (1/2)(z + 1/z) and sin(a) = (1/2i)(z - 1/z).

You could usefully start by expanding the fourth powers of those two expressions.
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1 month ago
#3
De Moivre: Note that By taking real part in both sides and using De Moivre, Im too lazy to type the rest, but use double angle formula twice on and rearrange the terms to get the answer.
Last edited by cflau_; 1 month ago
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#4
(Original post by old_engineer)
Let z = e^ia. You should already know that cos(a) = (1/2)(z + 1/z) and sin(a) = (1/2i)(z - 1/z).

You could usefully start by expanding the fourth powers of those two expressions.
thanks for the help, any chance you could show me the first line so i understand what you mean. I have little experience with this method.

I know the formula will be:

1+j(theta)+(j(theta)^2/2!+(j(theta)^3/3!+(j(theta)^4/4!

Im unsure how I will implement the z+1/z within the equation

Thanks
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1 month ago
#5
(Original post by Herbie0209)
thanks for the help, any chance you could show me the first line so i understand what you mean. I have little experience with this method.

I know the formula will be:

1+j(theta)+(j(theta)^2/2!+(j(theta)^3/3!+(j(theta)^4/4!
Why would you think that's the formula?

Im unsure how I will implement the z+1/z within the equation
If cos theta = (z+1/z)/2, then cos^k theta = (z+1/z)^k / 2^k. Expand (z+1/z) using the binomial theorem.
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1 month ago
#6
Last line should have been: Expand (z+1/z)^k using the binomial theorem. (Stupid TSR not letting me edit my post...)
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1 month ago
#7
(Original post by Herbie0209)
thanks for the help, any chance you could show me the first line so i understand what you mean. I have little experience with this method.

I know the formula will be:

1+j(theta)+(j(theta)^2/2!+(j(theta)^3/3!+(j(theta)^4/4!

Im unsure how I will implement the z+1/z within the equation

Thanks
OK if you're happy that cos(a) = (1/2)(z + 1/z), that means that cos^4(a) = (1/16)(z^4 + 4z^2 + 6 + 4/(z^2) + 1/(z^4))
You need to do the same for sin^4(a) and add to the expansion of cos^4(a)
Can you see the way this is going now?
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#8
(Original post by old_engineer)
OK if you're happy that cos(a) = (1/2)(z + 1/z), that means that cos^4(a) = (1/16)(z^4 + 4z^2 + 6 + 4/(z^2) + 1/(z^4))
You need to do the same for sin^4(a) and add to the expansion of cos^4(a)
Can you see the way this is going now?
Ive got:

(1/8)(cos4(theta)-4cos2(theta)+3)+(1/8)(cos4(theta)+4cos2(theta)+3)

Im sure they are both correct I just dont know how to add them together as stupid as it sounds
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1 month ago
#9
(Original post by Herbie0209)
Ive got:

(1/8)(cos4(theta)-4cos2(theta)+3)+(1/8)(cos4(theta)+4cos2(theta)+3)

Im sure they are both correct I just dont know how to add them together as stupid as it sounds
Well, (1/8)cos([email protected]) + (1/8)cos([email protected]) = (1/4)cos([email protected])
Continue with the other two terms in each bracket.
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#10
(Original post by old_engineer)
Well, (1/8)cos([email protected]) + (1/8)cos([email protected]) = (1/4)cos([email protected])
Continue with the other two terms in each bracket.
Thanks for your help, appreciate it. Completed the question
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