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Derive cos'x from cos(A)-cos(B)

  1. Jul 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Use ##~\cos\alpha-\cos\beta~## to develop the formula for the derivative of cos(x) from the definition:
    $$\frac{d(\cos x)}{dx}=\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}$$

    2. Relevant equations
    $$\cos\alpha-\cos\beta=(-2)\sin\left( \frac{\alpha+\beta}{2} \right)\cdot \sin\left( \frac{\alpha-\beta}{2} \right)$$
    $$\lim\frac{\sin x}{x}=1$$

    3. The attempt at a solution
    $$\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}=...=(-2)\lim\frac{\sin\left( \frac{2x+\Delta x}{2} \right)}{\Delta x}\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
    Even if i could use ##~\lim\frac{\sin x}{X}=1~##, which would eliminate the second member, i can't deal with the first member.
  2. jcsd
  3. Jul 14, 2017 #2


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    You divide by Δx too many times.
  4. Jul 14, 2017 #3
    $$\lim\frac{\cos(x+\Delta x)-\cos x}{\Delta x}=(-2)\lim\frac{\sin\left( \frac{x+\Delta x+x}{2}\right)\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
    $$=(-2)\lim \sin\left( \frac{2x+\Delta x}{2} \right)\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}$$
    $$u=\frac{\Delta x}{2}, ~~\lim\frac{\sin\left( \frac{\Delta x}{2} \right)}{\Delta x}=\lim\frac{\sin u}{2u}=\frac{1}{2}$$
    $$\cos'x=-\sin x$$
    Last edited: Jul 14, 2017
  5. Jul 14, 2017 #4


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