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anonymous

  • one year ago

Does... 1. sin(2pi + Ѳ) = -cos Ѳ 2. cos(2pi - Ѳ) = cos Ѳ I don't know what to do with the pi's. My teacher never taught me how to check if equations are identities with pi in them. Please explain.

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  1. johnweldon1993
    • one year ago
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    |dw:1433989500490:dw|

  2. johnweldon1993
    • one year ago
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    The unit circle Let me ask...if you start |dw:1433989606182:dw| And you travel 2π around the circle...where will you end up?

  3. anonymous
    • one year ago
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    it would be in the same place

  4. johnweldon1993
    • one year ago
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    Right...you would end up in the same exact place, meaning we can just ignore them and just worry about 1) \(\large sin(\theta) = -cos(\theta)\) and clearly that is NOT a correct statement Does that make sense?

  5. anonymous
    • one year ago
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    okay but what about if it wasn't 2pi? what would you do if it was something where you couldn't ignore it?

  6. johnweldon1993
    • one year ago
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    It would be the same for 2) as well \(\large cos(2\pi - \theta) = cos(\theta)\) since 2π brings us back to the same place, we ignore it and just have \[\large cos(-\theta) = cos(\theta)\] which *since we know cos is an odd function is correct

  7. anonymous
    • one year ago
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    what do you mean an odd function?

  8. freckles
    • one year ago
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    cos is an even function sin is an odd function

  9. johnweldon1993
    • one year ago
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    Oops, yup my mistake there!

  10. anonymous
    • one year ago
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    okay how does that help?

  11. johnweldon1993
    • one year ago
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    It is an identity \[\large cos(-\theta) = cos(\theta)\]

  12. anonymous
    • one year ago
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    You said since cos in an odd function it is correct? what do you mean by that? How would it being an odd function help

  13. anonymous
    • one year ago
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    i meant it being an even function

  14. johnweldon1993
    • one year ago
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    Yeah I did too, still sorry about that typo lol So here maybe a visual would help with this...the graph of cos(x) is |dw:1433991380473:dw|

  15. johnweldon1993
    • one year ago
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    So for example if you have cos(π) and cos(-π) |dw:1433991571607:dw| So it really helps up simplify things like \[\large cos(-\theta) = cos(\theta)\]

  16. anonymous
    • one year ago
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    What if it was sin(-θ) = sin θ

  17. johnweldon1993
    • one year ago
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    Ahh good question...THAT is an incorrect statement...sin(x) is actually an odd function...and I'll show you why

  18. johnweldon1993
    • one year ago
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    |dw:1433991809831:dw| It has symmetry about the origin...meaning if you were to rotate the right half of the graph about the origin you would end up with the left half

  19. johnweldon1993
    • one year ago
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    SOOOO now lets say you choose x = pi/2 for example If you go to pi/2 and -pi/2 on this graph you will get |dw:1433991965337:dw|

  20. johnweldon1993
    • one year ago
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    Which leads to another good identity that \[\large sin(-\theta) = -sin(\theta)\] Like here we chose theta to me pi/2 \[\large sin(-\pi/2) = -sin(\pi/2) \rightarrow -1 = -1 \checkmark \]

  21. anonymous
    • one year ago
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    Okay. How would you verify something like cos (θ + (π/2)) = -sin θ. I get it with the 2π because it cancels but what about this?

  22. anonymous
    • one year ago
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    @johnweldon1993 sorry that I am asking so many questions. I really want to get this stuff down because I have my final exam tomorrow

  23. johnweldon1993
    • one year ago
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    No that's okay...actually everything we've been doing has been just from the fact that we knew the 2π would go away...so it's good to know the others as well! EVERYTHING we have done so far and can do...is from these next things Remember those identities that say \[\large sin(a \pm b) = sin(a)cos(b) + sin(b)cos(a)\] and \[\large cos(a \pm \ b) = cos(a)cos(b) \mp sin(a)sin(b)\] Well if you have sin(theta + π/2) that would be \[\large sin(\theta + \pi /2) = \cancel{sin(\theta)cos(\pi/2)} + sin(\pi/2)cos(\theta) = (1)(cos(\theta)) = cos(\theta)\]

  24. johnweldon1993
    • one year ago
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    Sorry got cut off there supposed to be \(\large = cos(\theta)\)

  25. anonymous
    • one year ago
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    why do the sin cos(pi/2) cancel

  26. johnweldon1993
    • one year ago
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    What is cos(pi/2)?

  27. anonymous
    • one year ago
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    sinθ cos(π/2)

  28. anonymous
    • one year ago
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    there's a slash so i assume that's supposed to mean it cancels

  29. johnweldon1993
    • one year ago
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    No I know what you were referring to :P I meant evaluate...what is cos(pi/2) ?

  30. anonymous
    • one year ago
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    Oh haha. umm is it 0 since pi/2 is at 90 degrees or (0,1)?

  31. johnweldon1993
    • one year ago
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    Right...and cos(90degrees) = 0 so that makes the \(\large sin(\theta)cos(\pi/2)\) cancel

  32. anonymous
    • one year ago
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    Ahhh that makes sense

  33. johnweldon1993
    • one year ago
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    Good :) So as long as you can remember those sum and difference formulas for these types of problems...they will go by pretty quick!

  34. anonymous
    • one year ago
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    Thanks so much for all your help! :)

  35. johnweldon1993
    • one year ago
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    Not a problem :) \[\bbox [10pt, black, border:5pt solid blue]{\Huge\cal\color{red}{\bigstar}\color{orange}{G}\color{yellow}{O}\color{green}{O}\color{blue}{D} \space \color{indigo}{L}\color{violet}{U}\color{red}{C}\color{orange}{K}\space \color{yellow}{T}\color{green}{O}\color{blue}{M}\color{indigo}{O}\color{violet}{R}\color{red}{R}\color{orange}{O}\color{yellow}{W}\color{blue}{\bigstar}}\]

  36. anonymous
    • one year ago
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    Thanks :)

  37. johnweldon1993
    • one year ago
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    Didn't fully come out...but ehh :P XD

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