Verifying Solution for cos^4 x in Terms of cos 4x & cos 2x

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The discussion focuses on expressing cos^4 x in terms of cos 4x and cos 2x, starting from the identity cos^2 x = 0.5(1 + cos 2x). An initial attempt yielded the expression cos^4 x = 0.25 + (cos 2x)/2 + (cos 4x + 1)/8, which was checked against Wolfram Alpha. Participants confirmed that the expression can be simplified further, leading to the correct form of cos^4 x = 3/8 + (cos 2x)/2 + (cos 4x)/8. Graphing the function was suggested as a method to verify the correctness of the derived expression. The conversation concludes with a consensus on the correct formulation of cos^4 x.
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Homework Statement


express cos^4 x in terms of cos 4x and cos 2x given that

cos^ x = 0.5(1 + cos 2x)


The Attempt at a Solution



i did some playing around for a minute and came to this;

cos^4 x = 0.25 + (cos2x)/2 + (cos 4x +1)/8

and thought, great! now i'll just check it on wolfram however i got this;

http://www.wolframalpha.com/input/?i=y+=0.25+++(cos2x)/2+++(cos+4x++1)/8

as opposed to

http://www.wolframalpha.com/input/?i=cos+^4+x

now, just looking at the graphs it seems okay however none of the alternate forms or expansions are the same, i would love it if someone could just verify that I'm right it's quite an important question!

thanks again
 
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your answer is correct

if you group everything over a common denominator it might become more apparent
 
thanks a lot for that, it was really bugging me :D
 
Chewy0087 said:
thanks a lot for that, it was really bugging me :D

You can always try to graph as a way to check your work:

y = \cos ^ 4 x - \left( 0.25 + \frac{\cos (2x)}{2} + \frac{\cos(4x) + 1}{8} \right)

to see if it turns out to be the x axis. If it does, then, everything should be fine. :)

Btw, your expression can be further simplified to:

\cos ^ 4 x = {\color{red}\frac{5}{8}} + \frac{\cos (2x)}{2} + \frac{\cos(4x)}{8}
 
\cos ^ 4 x = {\color{red}\frac{5}{8}} + \frac{\cos (2x)}{2} + \frac{\cos(4x)}{8}

I got
\cos ^ 4 x = {\color{red}\frac{3}{8}} + \frac{\cos (2x)}{2} + \frac{\cos(4x)}{8}
 
that's a good idea actually, thanks

i'm sure he meant 3/8
 
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