How Do You Simplify cos^4x Using Trigonometric Identities?

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The discussion focuses on simplifying cos^4x using trigonometric identities. The user attempts to express cos^4x in terms of cos2x and cos4x by squaring the identity cos^2x = 1/2(1 + cos2x). They derive the expression 3/8 + 1/2 cos2x + 1/4 cos4x but seek clarification on how to handle (cos2x)^2. The conversation highlights the importance of using established identities to simplify trigonometric expressions efficiently. Overall, the thread emphasizes the application of trigonometric identities in simplifying complex expressions.
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Hi, I get stuck on the 3rd part of this question:

(i) Draw sketches of the functions sin x and sin^2x over the range -360<x<360.
(ii) Explain why, for the range 0 < x < 90=2, sin x is smaller than tan x.
(iii) Using the equality cos^2x=\frac{1}{2}(1+cos2x) or otherwise, express
cos^4xin terms of cos2x and cos4x.

My attempt at part iii:

I squared the left hand side of cos^2x=\frac{1}{2}(1+cos2x) to get cos^4x and therefore squared the right hand side as well, leaving the right hand side as \frac{1}{4}(1+cos2x)^2

I'm presuming I have to square the right hand bracket out but I'm unsure on what (cos2x)^2 becomes.

Any help would be appreciated



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If you know (because it was just given!) that
cos^2(a)= \frac{1}{2}(1+ cos(2a)
for any a, then what is cos^2(2x)?
 
Thanks for the reply, following your method I ended up with \frac{3}{8}+\frac{1}{2}cos2x+\frac{1}{4}cos4x

Just out of interest, if I were to square cos2x would I have to use the knowledge given in the question or does it equal something obvious?
 
Only the "obvious" cos^4(2x)! But then you could use the same identities to reduce that. It's just easier to use the already factored form.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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