Does the Pythagorean Identity Hold for sin^2(3x) + cos^2(3x)?

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The discussion centers on whether the Pythagorean identity sin^2(3x) + cos^2(3x) equals 1. Participants express uncertainty about applying the identity to sin^2(3x) and cos^2(3x), questioning if it holds for all values of x. It is clarified that there are no restrictions on x, suggesting the identity does indeed hold true. Examples are provided to illustrate that sin^2(x) + cos^2(x) equals 1 for various values of x. Ultimately, the consensus is that the Pythagorean identity applies universally, including for sin^2(3x) + cos^2(3x).
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Homework Statement


sin2x + cos2x = 1

but would sin23x + cos23x = 1?

Homework Equations


none.

The Attempt at a Solution


[/B]
I'm pretty sure sin23x + cos23x can't equal 1 otherwise the identity would probably be written as sin2cx + cos2cx = 1 and I've never seen it written like this.

I was doing a homework problem and i ended up in a situation where, if i could use sin2cx + cos2cx = 1, i'd get my answer. But I am doing it differently than the book, so my way might be wrong
 
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When you write \sin^2(x) + \cos^2(x) = 1 are there restrictions on the values x can take? For example, could x = 3 * y?
 
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no there are no restrictions, so that must mean it does work, or actually, you said x = 3y. I am not sure... there's no restrictions I am aware of
 
AMan24 said:
no there are no restrictions, so that must mean it does work, or actually, you said x = 3y. I am not sure... there's no restrictions I am aware of

Right. It's true for any value of x.
 
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AMan24 said:

Homework Statement


sin2x + cos2x = 1

but would sin23x + cos23x = 1?

Homework Equations


none.

The Attempt at a Solution


[/B]
I'm pretty sure sin23x + cos23x can't equal 1 otherwise the identity would probably be written as sin2cx + cos2cx = 1 and I've never seen it written like this.

I was doing a homework problem and i ended up in a situation where, if i could use sin2cx + cos2cx = 1, i'd get my answer. But I am doing it differently than the book, so my way might be wrong

Take ##x=10##. Would you agree that ##\sin^2 10 + \cos^2 10 = 1?## Do you really think that makes ##\sin^2 30 + \cos^2 30 ## come out different from 1? What about ##\sin^2 37 + \cos^2 37?## That would be ##\sin^2 cx + \cos^2 cx## with ##c = 3.7##.
 
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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