Proving Integration: Solving the Cosine and Sine Proofs for High School Students

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Homework Help Overview

The discussion revolves around proving the integral of the difference between cosine and sine raised to an even power over a specified interval, specifically \(\int_0^{2\pi} (cos^{2k}x - sin^{2k} x) \ dx = 0\). The subject area is calculus, focusing on integration and properties of trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of the variable \(k\) and its potential values, questioning whether it includes all integers, all reals, or specific ranges. There are attempts to apply symmetry in the functions and to relate the cosine and sine functions through transformations. Some participants express difficulty in formalizing their reasoning into a proof.

Discussion Status

Participants are actively engaging with the problem, sharing hints and exploring different approaches. Some have suggested using symmetry and transformations to aid in the proof, while others are still grappling with how to express their findings formally. There is no explicit consensus on a method yet, but the discussion is productive with various lines of reasoning being explored.

Contextual Notes

There is uncertainty regarding the definition and limitations of \(k\), which may affect the approach to the problem. Additionally, some participants mention previous related exercises that may influence their understanding of the current proof.

synkk
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I'm almost out of high school and have been trying to do some "harder" proofs, anyone, I'm not quite sure on how to proceed on this:

prove that for all k:

[tex]\displaystyle \int_0^{2\pi} (cos^{2k}x - sin^{2k} x) \ dx = 0[/tex]

If anyone could start me off as I'll I have in my head is "try to express it into something you can integrate", but having no luck.
 
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It's not clear what k is limited to. Is k all integers, all reals, etc? If k is an integer, will its range include negative values? What about k = 0?
 
SteamKing said:
It's not clear what k is limited to. Is k all integers, all reals, etc? If k is an integer, will its range include negative values? What about k = 0?

The parts before this question asked me to integrate cos^6(x) and sin^6x using identities from cos^6(x) - sin^6(x) and cos^6x + sin^6x which I done successfully. At the end of the question it says: "You might like to consider how you would prove that [tex]\displaystyle \int_0^{2\pi} (cos^{2k}x - sin^{2k} x) \ dx = 0[/tex] for all k, without having to derive a new identity for each value of k.
 
hint: use symmetry :wink:
 
tiny-tim said:
hint: use symmetry :wink:

Could you expand on that? I tried rewriting cos in terms of sin, but still pretty stuck

thanks
 
cos2k(π/2 - x) - sin2k(π/2 - x) = … ? :wink:
 
tiny-tim said:
cos2k(π/2 - x) - sin2k(π/2 - x) = … ? :wink:


= [itex]sin^{2k}x - cos^{2k}x = -(cos^{2k}x - sin^{2k} x)[/itex]

I feel so dumb for not getting this :(

I tried for around 15 minutes playing around with what you gave me to show it's 0 but to no avail... I'll try again tomorrow and post if I've made progress. Thanks for your patience.
 
EDIT: Ignore my previous post, I'd made an error.
 
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synkk said:
Could you expand on that? I tried rewriting cos in terms of sin, but still pretty stuck

thanks

Look at the graphs of sin(x) and cos(x). Can you get one of the graphs by shifting the other one to the right or the left?
 
  • #10
π
Ray Vickson said:
Look at the graphs of sin(x) and cos(x). Can you get one of the graphs by shifting the other one to the right or the left?

yes cos(x) = sin(x + π/2)
also sin(x) = cos(x - π/2)

I can see why the integral is 0, by drawing both of the graphs and shading in the required area, but I'm having real trouble formalising a proof to actually put onto paper :\
 
  • #11
synkk said:
I can see why the integral is 0, by drawing both of the graphs and shading in the required area, but I'm having real trouble formalising a proof to actually put onto paper :\

try substituting y = π - x, or y = π/2 - x (and dy = -dx), and seeing what it does to the integral of one of your shaded parts :smile:
 

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