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Help with 1 step in proof of d/dx sin x = cos x

  1. Feb 23, 2012 #1
    I understand everything in the proof except the last step I have written here. What comes after, I understand.

    How is it that the cosine and sine are able to be factored out of the fraction? That one step gets me. I was never too good with my trig, and have finally gotten a decent grasp of it, but this one befuddles me.

    Prove: [itex]\frac{d}{dx}[sin (x)] = cos (x) [/itex]

    [itex] \frac{d}{dx}[sin (x)] = \lim_{\Delta x \to 0}\frac{sin (x) cos (\Delta x) + cos (x) sin (\Delta x) - sin (x)}{\Delta x}[/itex]


    [itex] = \lim_{\Delta x \to 0}\frac{cos(x) sin (\Delta x) - (sin (x)) (1- cos(\Delta x))}{\Delta x}[/itex]

    [itex] = \lim_{\Delta x \to 0} \Bigg(cos(x) \frac {sin (\Delta x)} {\Delta x} - sin(x) \frac {(1-cos(\Delta x)} {\Delta x}\Bigg) [/itex]
     
  2. jcsd
  3. Feb 23, 2012 #2

    HallsofIvy

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    I'm not clear what your question is! Obviously we can factor sin(x) and cos(x) as done here because there is a factor of sin(x) and cos(x) in the appropriate terms. The reason we want to do that is to get the limits [itex]\lim_{\Delta x\to 0} sin(\Delta x)/\Delta x[/itex] and [itex]\lim_{\Delta x\to 0} (1- cos(\Delta x)/\Delta x[/itex] which are limits you already know.
     
  4. Feb 23, 2012 #3
    I know it sounds like a stupid question. Mixing trig into calc usually breaks my algebraic intuitions.

    I just don't get how cos (x) and sin(x) get factored out in the last step. I tried watching videos from MIT OCW and youtube where the proof is demonstrated, but they always just write it out. This is just one of those things that is getting me.

    I'd rather understand it, because it seems I am missing one of those building blocks that I should have known.
     
  5. Feb 23, 2012 #4
    Is it true that (ab + cd)/e = ab/e + cd/e = a(b/e) + c(d/e)?
     
  6. Feb 23, 2012 #5
    Ah yes! That makes sense now.

    Thank you.
     
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