# Thread: rewriting sin^4(x)*cos^4(x) in terms of cos^1(x)

1. ## rewriting sin^4(x)*cos^4(x) in terms of cos^1(x)

This problem has driven me mad for 2 days now. The directions are to rewrite the function in terms of the first power of the cosine (that is, write it such that there are no sines, tangents, etc., and such that there are no powers on any of the cosines greater than 1).

The original problem: sin^4 (x) * cos^4(x)

testing the math tags: $sin^4 (x) * cos^4 (x)$

I have two pages of running in circles- any and all help is very much appreciated.

2. Use a few identities.

Rewrite: $sin^{4}(x)cos^{4}(x)=(sin^{2}(x))^{2}(cos^{2}(x))^ {2}$

$sin^{2}(x)=1-cos^{2}(x)$

and

$cos^{2}(x)=\frac{1+cos(2x)}{2}$

3. you could also try using euler's formula:

(sinx * cosx) = [ (e^(ix) - e^(-ix) )/2i ] * [ (e^(ix) + e^ (-ix) )/2 ]

= (e^(2ix) - e^(-2ix)) / 4i

Raise this to the 4th power and multiply it out and you'll get:

3/128 - (1/64)*[e^(4ix) + e^(-4ix)] + (1/256)*[e^(-8ix) + e^(8ix)]

then remember from above:

cos(x) = (e^(ix) + e^ (-ix) )/2

then:

e^(4ix) + e^(-4ix) = 2*cos (4x)

and

e^(-8ix) + e^(8ix) = 2*cos(8x)

then you have:

3/128 - (1/64)*(2*cos (4x)) + (1/256)*(2*cos(8x))

then

(sinxcosx)^4 = 3/128 - (1/32)*cos(4x) + (1/128)*cos(8x)

4. Oh that's how I started. You'll notice that it gets ugly fast when the $cos^4 (x)$ becomes $[1+2cos2x+(1+cos4x)/2]/4$ and that when working with $sin^ (4)$ gives you that and two other terms of slightly less magnitude.

Those two then get multiplied together, giving us the first term I wrote out, added to one twice its size, added to another 4 times its length. The problem, however, is that the multiplication leading to these terms gives us, among other, simpler terms, $-cos^4 (x) + 6cos^2 (x) - sin^4 (x) +2cos^6 (x) -12cos^4 (x) sin (x) + 2sin^4 (x) cos^2 (x)$ and that crops up at least twice, in both ways I've gone about the problem. (that is, working with the sines and cosines seperately, then combining, or combining right away into cosines)

The algebra is burring me. Certainly I'm not taking the simplest rout, but it's the only one I see. Our teacher took 2 pages to realize she needed to do it differently, and her final copy was only just over half a page. I'm at one and a half right now.

Belated edit- thanks much to mentia, who's post avoids the mountains of symbols and characters and is very easy to follow.

5. I like to try a different route,

$sin^{4}(x)*cos^{4}(x)$
$=(\frac{sin(2x)}{2})^4$
$=\frac{1}{16}*(sin^{2}(2x))^{2}$
$=\frac{1}{16}*(\frac{cos(4x)-1}{2})^{2}$
$=\frac{1}{64}*(cos^{2}(4x)-2cox(4x)+1)$
$=\frac{1}{64}*(\frac{cos(8x)+1}{2}-2cox(4x)+1)$

etc.

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### sin^4.cos^4 = 1 / 128

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