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apple_pi

  • 2 years ago

Differentiate sin(x) from first principles

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  1. hartnn
    • 2 years ago
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    use sin A-sin B formula, do u know it ?

  2. apple_pi
    • 2 years ago
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    the sin(A+B)?

  3. hartnn
    • 2 years ago
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    nopes, sin A - sin B on sin(x+h) - sin x.

  4. apple_pi
    • 2 years ago
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    I was thinking on starting out like this: \[\lim_{h \rightarrow 0}\frac{ \sin (x+h) - \sin(x) }{ h }\]

  5. hartnn
    • 2 years ago
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    thats the correct start, i was giving a hint on how would u start simplifying numerator. using sin A-sin B with A=x+h,B=x

  6. apple_pi
    • 2 years ago
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    us, would you expand sin(x+h)?

  7. apple_pi
    • 2 years ago
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    *um

  8. hartnn
    • 2 years ago
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    i think that would also work, there u have to separate out denominator's h to all numerator terms.....i personally find it easier to do with sin A-sin B...and less lengthy

  9. apple_pi
    • 2 years ago
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    how does sinA-sinB work?

  10. hartnn
    • 2 years ago
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    sin A - sin B = 2 cos (A+B)/2 sin (A-B)/ 2 right ? so sin (x+h)-sin x = 2 cos (x+h/2) sin (h/2) ok ? now put h=0 directly in cos term and use sin t/t = 1 in sin term...

  11. apple_pi
    • 2 years ago
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    sorry, don't know sinA-sinB = 2 cos((A+B)/2)sin((A+B)/2)

  12. Callisto
    • 2 years ago
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    It's... actually \[sinA - sinB = 2cos\frac{A+B}{2}sin\frac{A-B}{2}\]

  13. Skaematik
    • 2 years ago
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    Q: Is that the product to sums formula?

  14. hartnn
    • 2 years ago
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    then u better go with sin(x+h) = sin x cos h + cos x sin h ....

  15. Callisto
    • 2 years ago
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    sum-to-product indeed.

  16. apple_pi
    • 2 years ago
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    lim... [ sin(x)cos(h)+cos(x)sin(h)-sin(x) ] / h lim... [ sin(x)(cos(h)-1)+cos(x)sin(h) ] / h then what?

  17. hartnn
    • 2 years ago
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    can u solve lim x->0 (1-cos x)/x = 1/2 ?? pretty standard. the other limit is pretty easy, just use sin h/h =1

  18. Callisto
    • 2 years ago
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    Hmmm... Isn't it lim x->0 (1-cos x)/x = 0?

  19. apple_pi
    • 2 years ago
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    isn't it lim h->0?

  20. hartnn
    • 2 years ago
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    made a mistake....wait.yup its 0,sorry.

  21. hartnn
    • 2 years ago
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    but basically u need to separate the numerator in 2 limits.

  22. hartnn
    • 2 years ago
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    so your 1st limit will be 0 as pointed out correctly and 2nd limit will be just cos x *1 = cos x

  23. Callisto
    • 2 years ago
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    \[\lim_{h \rightarrow 0}\frac{sin(x)(cos(h)-1)+cos(x)sin(h) }{h}\]\[=\lim_{h \rightarrow 0}(\frac{sin(x)(cos(h)-1)}{h}\ \ +\frac{cos(x)sin(h) }{h})\]\[=\lim_{h \rightarrow 0}sin(x)\frac{(cos(h)-1)}{h} \ +\lim_{h \rightarrow 0}cos(x)\frac{sin(h) }{h}\]\[=...\]

  24. Callisto
    • 2 years ago
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    Perhaps one more step for you... \[=sin(x)\lim_{h \rightarrow 0}\frac{(cos(h)-1)}{h} \ +cos(x) \lim_{h \rightarrow 0}\frac{sin(h) }{h}\] I think you can evaluate the limit now..

  25. apple_pi
    • 2 years ago
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    how are theses limits calculated?

  26. Callisto
    • 2 years ago
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    *limits

  27. Callisto
    • 2 years ago
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    As @hartnn mentioned, use these: \[\lim_{x \rightarrow 0}\frac{1-cos(x)}{x} =0\]\[\lim_{x \rightarrow 0}\frac{sin(x)}{x} =1\]

  28. hartnn
    • 2 years ago
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    actually the 1st one isn't standard, u need to solve it by multiplying and dividing by (1+cos x) which gives sin^2 x in numerator and using 2nd limit to get answer as 0

  29. apple_pi
    • 2 years ago
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    yeah um, how are these proven? (Sorry if I'm being a bit fussy)

  30. Callisto
    • 2 years ago
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    @apple_pi Do you have time to watch a video for explanations?

  31. apple_pi
    • 2 years ago
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    maybe later... just send me the links

  32. hartnn
    • 2 years ago
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    there are many ways to prove lim sin x/x =1 the simplest one is the use of L'Hospitals rule, but since u are starting limits, i guess u don't know about this rule..

  33. apple_pi
    • 2 years ago
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    Oh yeah, MIT OCW is good. Does this show the whole d/dx sin(x)?

  34. Callisto
    • 2 years ago
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    Yup!

  35. apple_pi
    • 2 years ago
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    thanks...turns out I downloaded this ages ago, but stopped after lecture 2 and jumped about places

  36. Callisto
    • 2 years ago
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    Here it goes. The notes I've copied from that lecture. IMG.pdf should contain all three pages.

  37. apple_pi
    • 2 years ago
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    Thanks a heap. Can you show me how sinA-sinB = ...

  38. Callisto
    • 2 years ago
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    Do you know how to work out the product-to-sum formula?

  39. apple_pi
    • 2 years ago
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    whats that?

  40. Callisto
    • 2 years ago
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    Example: sinAcosB = (1/2) [ sin(A+B) +sin(A-B)]

  41. hartnn
    • 2 years ago
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    that u can get using sin (x-y) and sin (x+y) formula, u know that right ? just solve sin(x+y) - sin (x-y) = ....... and then put x+y = A x-y = B so x= (A+B)/2 y=(A-B)/2.

  42. apple_pi
    • 2 years ago
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    sorry, no

  43. apple_pi
    • 2 years ago
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    no @Callisto

  44. apple_pi
    • 2 years ago
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    yes @hartnn

  45. Callisto
    • 2 years ago
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    hartnn explained, I think :)

  46. hartnn
    • 2 years ago
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    so whats sin(x+y) - sin (x-y)= ?

  47. apple_pi
    • 2 years ago
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    sin(x)cos(y)+cos(x)sin(y)-sin(x)cos(y)+cos(x)sin(y) = 2cos(x)sin(y)

  48. hartnn
    • 2 years ago
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    thats correct, now just put values of x and y as i mentioned, u understood, how i found x and y in terms of A and b .. ?

  49. apple_pi
    • 2 years ago
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    um, no

  50. hartnn
    • 2 years ago
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    x+y=A x-y=B Adding x+x=A+B --->x=(A+B)/2 i think y can can find now...

  51. apple_pi
    • 2 years ago
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    (A+B)/2+y=A y=(A-B)/2

  52. hartnn
    • 2 years ago
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    yup, any more doubts?

  53. apple_pi
    • 2 years ago
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    uh, so does that mean sin(A)-sin(B)=2cos((A+B)/2)sin((A-B)/2)

  54. apple_pi
    • 2 years ago
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    @hartnn

  55. hartnn
    • 2 years ago
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    exactly, yes.

  56. apple_pi
    • 2 years ago
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    OK, thank you

  57. hartnn
    • 2 years ago
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    similarly u can find sin A+sin B cos A + cos B cos A - cos B and remember them as they are standard formulas.

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