(cos2x)^2


by justine411
Tags: cos2x2
justine411
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#1
Apr11-07, 10:11 AM
P: 16
1. The problem statement, all variables and given/known data

(cos2x)^2

2. Relevant equations



3. 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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neutrino
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#2
Apr11-07, 12:13 PM
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What is there to solve???

(cos2x)^2 is just an expression.
HallsofIvy
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#3
Apr11-07, 02:02 PM
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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.

Rhythmer
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#4
Apr12-07, 06:46 AM
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(cos2x)^2


Quote Quote by HallsofIvy View Post
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) ?
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Apr12-07, 06:54 AM
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Quote Quote by Rhythmer View Post
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).
HallsofIvy
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#6
Apr12-07, 07:49 AM
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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).


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