how do i show cos^5=

[page89]

Hi guys,

I am struggling with a question and i was wondering if anyone can help me, I will use Y as theta.

I need to know how you show that,

cos^5 Y = 1/16 (cos5Y + 5cos3Y + 10cosY)

I think it has something to do with euler's formulae but i dunno how to do it. Does anyone know how to do this. Help would be very much appreciated.

Thanks

Nick

 

[Deleted member 4993]

Hi guys,

I am struggling with a question and i was wondering if anyone can help me, I will use Y as theta.

I need to know how you show that,

cos^5 Y = 1/16 (cos5Y + 5cos3Y + 10cosY)

I think it has something to do with euler's formulae but i dunno how to do it. Does anyone know how to do this. Help would be very much appreciated.

Thanks

Nick

You know Euler's formula.

You also know binomial expansion of \(\displaystyle (a+b)^5\)

start with

\(\displaystyle cos x = \frac{e^{ix} + e^{-ix}}{2}\)

Now take it to fifth power and expand.

 

[page89]

I am really sorry but i am getting a little confused when it comes to expanding out when i have to do it 5 times and i have e's involved. Can anyone help?

 

[Deleted member 4993]

\(\displaystyle (x+y)^5 = x^5 + 5x^4y + 10x^3y^2 +10x^2y^3 + 5xy^4 + y^5\)

Now show us what you have tried - even if it is wrong (you think).

remember

\(\displaystyle e^{ix}\cdot\ e^{-ix} = 1\)

 

[page89]

I think i have got it just need to tidy up,

(5/16)(e^ix +e^-ix) + (5/32)(e^3ix +e^-3ix) + (1/32)(e^5ix + e^-5ix)

then using cosx = 1/2(e^ix + e^-ix) and simplifying a few things i got the answer.

Thanks alot subhotosh kahn.

 

[Deleted member 4993]

\(\displaystyle cos^5(x)\)

\(\displaystyle = [\frac{e^{ix}+e^{-ix}}{2}]^5\\)

\(\displaystyle = [\frac{e^{i5x} + 5 e^{i4x}e^{-ix} + 10e^{i3x}e^{-i2x} + 10e^{i2x}e^{-i3x} +5 e^{ix}e^{-i4x} + e^{-i5x}}{32}]\\)

\(\displaystyle = [\frac{e^{i5x} + e^{-i5x} + 5 (e^{i3x} + e^{-i3x}) + 10(e^{ix} + e^{-ix})}{32}]\\)

\(\displaystyle = \frac{1}{16}[\frac{e^{i5x} + e^{-i5x}}{2}\ + 5 \frac {e^{i3x} + e^{-i3x}}{2} + 10\frac{e^{ix} + e^{-ix}}{2}]\\)

\(\displaystyle = \frac{1}{16}\ [cos(5x) + 5 cos(3x) + 10cos(x)]\)