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Trig derivatives.

  1. Oct 10, 2005 #1
    Okay. i know that
    d/dx of SIn(x)=Cos(X)

    d/dx of COs(x) = -Sin(x)

    and the rest of them you get by product rule, quotent rule etc using the rules of the derivatives and setting the fractions up correctly in regard to sin and cos.

    but i have three questions:

    1) If a sin is a trig function of the smae power as the cos becasue they have the same number of direction changes...how can a cos be the derivative of sin.

    2) if d/dx of Sin(x)=cos(x) why is D/dx of Cos(x) not = to Sin(x) since sin and cos are same behavior equations.

    3) How do you get the derivative of a sin or cos...i mean how did they come up with them? did they just made them up and make the match around it match?

    Like i saw how they came up with 0!=1 and i think they kinda "made it fit"...with the whole 5!=5*4! explanation...
     
  2. jcsd
  3. Oct 10, 2005 #2
    When you say you know what the derivatives of sin(x) and cos(x) are, what do you mean? Do you mean you memorized them? Do you know how to show what their derivatives are? If you do, the answers to your questions should be apparent. For the sine, use this (similar can be done for cosine):

    [tex]\lim_{h\to 0}\frac{\sin{\left(x+h\right)}-\sin{\left(x\right)}}{h}[/tex]

    Alex
     
    Last edited: Oct 10, 2005
  4. Oct 10, 2005 #3

    arildno

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    I'll take 3), which includes answers to 1) and 2) as well:

    Let us find the derivative of sin(x).
    x is measured in radians. That is IMPORTANT!!
    By the definition of the derivative, it must be given by (if existing)
    (\sin(x))'=\lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}[/tex]
    We now use the trigonometric identity:
    [tex]\sin(x+h)=\sin(x)\cos(h)+\sin(h)\cos(x)[/tex]
    that is, we have:
    [tex]\frac{\sin(x+h)-\sin(x)}{h}=\sin(x)\frac{\cos(h)-1}{h}+\cos(x)\frac{\sin(h)}{h}=\frac{\sin(h)}{h}(\cos(x)-\sin(x)\frac{\sin(h)}{1+\cos(h}})[/tex]
    Now, assuming sin(x), cos(x) is continuous at h=0, we see that the ugly term within the big parenthesis must go to zero as h goes to zero.

    Hence, the derivative of sin(x) should be given by:
    [tex](sin(x))'=\lim_{h\to{0}}\frac{\sin(h)}{h}\cos(x)[/tex]

    It is relatively simple to show that we have the limit [tex]\lim_{h\to{0}}\frac{\sin(h)}{h}=1[/tex], for example by geometry.

    Do you follow that?
     
  5. Oct 10, 2005 #4
    Good point. Usually (at least how I've seen it), the limit is shown geometrically using the fact:

    [tex]\sin{x}\cos{x}\leq x\leq \tan{x}[/tex]

    Alex
     
  6. Oct 10, 2005 #5
    i'm looking at it and understand the work. you used the limit and used the product sisusoid rule to break sin(ab) into Sin(a)Cos(B)+Sin(b)Cos(a) and you simplified it.

    it makes sense in a way...but still...although logically it works, out, it's hard to swallow that sin(x) derives into Cos(x)...i mean...you know what i'm saying, right?
     
  7. Oct 10, 2005 #6

    arildno

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    Not really.

    Look at a few values: At x=0, cos(x)=1, that is, it predicts that the slope of sin(x) at the origin is 1
    At [itex]x=\frac{\pi}{2}[/itex] the sine function has its maximum value, and thus its derivative should be 0, which agrees with [itex]\cos\frac{\pi}{2}=0[/itex]
    and so on..
     
  8. Oct 10, 2005 #7

    Tom Mattson

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    That could be immediately seen as wrong just by looking at the graphs of the sine and cosine functions. Start with cosine at x=0+ (that means slightly to the right of zero). Is it increasing or decreasing? What does this imply for the sign of the derivative? And what is the sign of sin(x) at x=0+?

    If you answer those questions correctly you'll see why sin(x) cannot be the derivative of cos(x).
     
  9. Oct 11, 2005 #8
    yea...i know some basic limits properties. i follow you. i get it now. Thank you guys.

    also, i succesfully worked out the other 4 of them and memorised them becasue it saves me time...It looks like an interesting chapter in math, especially when you know the Chain Rule.
     
  10. Oct 12, 2005 #9

    arildno

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    Memorizing saves time, understanding saves brain capacity.
     
  11. Oct 15, 2005 #10
    the power of the mind is infinite. not properly focused...but infinite.
     
  12. Oct 16, 2005 #11

    Tide

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    Robokapp,
    This may help:
    If
    [tex]\frac {d \sin x}{dx} = \cos x[/tex]
    and
    [tex]\frac {d \cos x}{dx} = -\sin x[/tex]
    then
    [tex]\sin x \frac {d \sin x}{dx} + \cos x \frac {d \cos x}{dx} = 0[/tex]
    and it follows that
    [tex]\frac {d}{dx} \left(\sin^2 x + \cos^2 x\right) = 0[/tex]
    with the obvious conclusion that the sine, cosine and their derivatives are rather intimately related.

    P.S. The power of the mind is not infinite since only a finite number of neural connections are possible with a finite number of neurons! :)
     
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