De Moivre's Theorem: cos4x in the form: cos4x = Acos(4x) + Bcos(2x) + C

[Morgz10]

Hey there, the question I am struggling with is this:
a) By considering De Moivre's Theorem with n=4 and n=2, express cos⁴x in the form: cos⁴x = Acos(4x) + Bcos(2x) + C,

where A, B and C are constants which you should determine.

b) Derive a similar expression, also in terms of cos(4x) and cos(2x), for sin⁴x.

My current solution:
Using the Theorem I have derived that:
cos(2x) = cos²x - sin²x
cos(4x) = cos⁴x - 6cos²xsin²x + sin⁴x

When using trig. identities I can't seem to be able to turn the 'cos(4x)' equation into the form the question requires because I keep heading down ends?

Any help would be amazing, thank you in advance!

 

[sinx]

Hey there, the question I am struggling with is this:
a) By considering De Moivre's Theorem with n=4 and n=2, express cos⁴x in the form: cos⁴x = Acos(4x) + Bcos(2x) + C,

where A, B and C are constants which you should determine.

b) Derive a similar expression, also in terms of cos(4x) and cos(2x), for sin⁴x.

My current solution:
Using the Theorem I have derived that:
cos(2x) = cos²x - sin²x
cos(4x) = cos⁴x - 6cos²xsin²x + sin⁴x

When using trig. identities I can't seem to be able to turn the 'cos(4x)' equation into the form the question requires because I keep heading down ends?

Any help would be amazing, thank you in advance!

this is not using de moivre's, but it gets you there
cos⁴x=(1-sin²)²=1-2sin²x+sin⁴x

now use double angle formulas,
cos2x=1-2sin²x.
cos4x=1-2sin⁴x
solve for sin² and sin⁴ and substitute

 

[pka]

Hey there, the question I am struggling with is this:
a) By considering De Moivre's Theorem with n=4 and n=2, express cos⁴x in the form: cos⁴x = Acos(4x) + Bcos(2x) + C,
where A, B and C are constants which you should determine.
b) Derive a similar expression, also in terms of cos(4x) and cos(2x), for sin⁴x.
My current solution:
Using the Theorem I have derived that:
cos(2x) = cos²x - sin²x
cos(4x) = cos⁴x - 6cos²xsin²x + sin⁴x
When using trig. identities I can't seem to be able to turn the 'cos(4x)' equation into the form the question requires because I keep heading down ends?
Any help would be amazing, thank you in advance!

\(\displaystyle \begin{align*}\cos(4x)&=\cos[2(2x)]\\&=\cos^2(2x)-\sin^2(2x)\\&=[\cos^2(x)-\sin^2(x)]^2-[2\sin(x)\cos(x)]^2\\&=\cos^4-2\sin^2(x) \cos^2(x)+\sin^4(x) -4\sin^2\cos^2(x)\\&=\sin^4(x)+\cos^4(x)-6\sin(x)\cos(x)\end{align*}\)

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