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anonymous

  • 5 years ago

Derive sinx using first principles

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  1. amistre64
    • 5 years ago
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    sin(x) = y/r .....

  2. anonymous
    • 5 years ago
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    Wait May I show you how far I got?

  3. anonymous
    • 5 years ago
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    go get em!

  4. amistre64
    • 5 years ago
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    you can

  5. anonymous
    • 5 years ago
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    you need the addition angle forumla\[sin(x+h)-sin(x)=sin(x)cos(h)+cos(x)sin(h)\]

  6. anonymous
    • 5 years ago
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    Ok, imagine this is the limit h -> 0 f(x + h) - f(x) / h sin(x + h) - sin(x) / h Applied compound angle formula:

  7. amistre64
    • 5 years ago
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    ohhh... i dint know we could use the sin function itself lol

  8. anonymous
    • 5 years ago
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    lim h-> 0 [(sinxcosh + cosxsinh - sinx) / h]

  9. anonymous
    • 5 years ago
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    Now what do I do?

  10. anonymous
    • 5 years ago
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    I'm not sure what we are deriving

  11. amistre64
    • 5 years ago
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    factor out a sin

  12. anonymous
    • 5 years ago
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    that is your numerator. your denominator is just h. you get \[\frac{sin(x)cos(h)+cos(x)sin(h)}{h}=\frac{sin(x)cos(h)}{h} + \frac{cos(x)sin(h)}{h}\]

  13. anonymous
    • 5 years ago
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    ohhh derivative of sin(x) haha

  14. anonymous
    • 5 years ago
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    so you need \[lim_{h->0}\frac{sin(h)}{h} = 1\]

  15. anonymous
    • 5 years ago
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    and \[lim_{h->0}\frac{cos(h)}{h}=0\]

  16. anonymous
    • 5 years ago
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    leaving just \[cos(X)\]

  17. anonymous
    • 5 years ago
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    Oh really? cosh/h = 0?

  18. anonymous
    • 5 years ago
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    I didn't know that

  19. anonymous
    • 5 years ago
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    Ok wait so

  20. anonymous
    • 5 years ago
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    it is \[lim_{x->0}\frac{cos(h)}{h}=0\]

  21. anonymous
    • 5 years ago
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    you need to know this to prove your result. probably comes in the text right before this.

  22. anonymous
    • 5 years ago
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    lim cos(h)/h as h goes to 0 doesnt exist

  23. anonymous
    • 5 years ago
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    actually it does and it is 0

  24. anonymous
    • 5 years ago
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    So so this is what I have now: lim h-> 0 [(sinx(cosh - 1) + sinhcosx) / h]

  25. anonymous
    • 5 years ago
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    lets start again.

  26. anonymous
    • 5 years ago
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    how, from the left it approaches -infinity from the right it approaches +infinity

  27. anonymous
    • 5 years ago
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    the numerator is \[sin(x+h)-sin(x)=sin(x)cos(h) +sin(h)cos(x) - sin(x)=sin(x)(cos(h)-1)+sin(h)cos(x)\]

  28. anonymous
    • 5 years ago
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    oops sorry rsvitale is right and i am wrong

  29. anonymous
    • 5 years ago
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    my mistake i apologize

  30. anonymous
    • 5 years ago
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    (cosh-1)/h goes to 0 as h goes to 0 though

  31. anonymous
    • 5 years ago
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    you need two limits: \[lim_{h->0}\frac{sin(h)}{h}=1\] and \[lim_{h->0}\frac{cos(h)-1}{h}=0\]

  32. anonymous
    • 5 years ago
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    yes you are right. my fault

  33. anonymous
    • 5 years ago
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    Sorry but I am getting a little confused as to what is right and what is wrong here.. :\

  34. anonymous
    • 5 years ago
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    i am wrong. but if you like i will write out the correct version since i messed up

  35. anonymous
    • 5 years ago
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    hey I found the derivation online and it's pretty good if you just want to look at that. http://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/sine.html

  36. anonymous
    • 5 years ago
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    \[lim_{h->0}\frac{sin(x+h)-sin(x)}{h}\] \[lim_{h->0}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\] \[=lim_{h->0}\frac{sin(x)(cos(h)-1) + cos(x)sin(x)}{h}\]

  37. anonymous
    • 5 years ago
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    break into two parts. \[sin(x)lim_{h->0}\frac{cos(h)-1}{h} + cos(x)lim_{h->0}\frac{sin(h)}{h}\]

  38. anonymous
    • 5 years ago
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    the first limit is 0, the second limit is 1 leaving\[cos(x)\]

  39. anonymous
    • 5 years ago
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    satellite73, you wrote sinx(cosh - 1) + cosxsinx <-- isn't it suppose to be sinhcosx?

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