How Do You Simplify (cos(2x))^2 Using Trigonometric Identities?

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The discussion revolves around simplifying the expression (cos(2x))^2 using trigonometric identities. Participants clarify that (cos(2x))^2 means to first compute 2x, then find the cosine of that value, and finally square the result, emphasizing that it remains (cos(2x))^2, not cos^2(4x). There is confusion about whether it can be expressed as cos^2(2x) or if it relates to cos^2(4x), but it is established that the exponent applies only to the cosine function, not the argument. The conversation highlights the importance of understanding trigonometric identities and the correct interpretation of notation. Ultimately, the expression can be simplified to cos^2(2x) for further analysis.
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Homework Statement



(cos2x)^2

Homework Equations





The Attempt at a Solution



I'm not sure if it is cos^2(2x) or cos^2(4x) or what. Should I use an identity to simplify it to make it easier to solve? Please help! :)
 
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What is there to solve? :confused:

(cos2x)^2 is just an expression.
 
In what sense is (cos(2x))2 a "problem"? What do you want to do with it?

I will say that (cos(2x))2 means: First calculate 2x, then find cosine of that and finally square that result. Notice that it is still 2x, not 4x. The fact that 2 is outside the parentheses means that it only applies to the final result.
 
HallsofIvy said:
In what sense is (cos(2x))2 a "problem"? What do you want to do with it?

I will say that (cos(2x))2 means: First calculate 2x, then find cosine of that and finally square that result. Notice that it is still 2x, not 4x. The fact that 2 is outside the parentheses means that it only applies to the final result.

Doesn't (cos(2x))2 = cos2(2x)2 = cos2(4x2) ?
 
Rhythmer said:
Doesn't (cos(2x))2 = cos2(2x)2 = cos2(4x2) ?

No. 'Cos' is a particular operation and 2x is the argument. The exponent of 2 operates on cos, not on the argument.

cos2y = cos y * cos y.

There are also particular trigonometric identites with which one should be familiar, i.e. cos (x+y) and sin (x+y).
 
You still haven't told us what the problem was! Was it to write (cos(2x))^2 in terms of sin(x) and cos(x)? I would simply be inclined to write (cos(2x))^2 as cos^2(2x).
 
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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