Proving that (d/d)x sin(x) = 
        cos(x).

To prove that (d/d)x sin(x) = 
        cos(x), we first prove that \(TraditionalForm\`lim\_\(θ -> 0\)\(sin(θ)\)\/θ = 1\) and \(TraditionalForm\`lim\_\(θ -> 0\)\(cos(θ) - 1\)\/θ = 0\).

Claim A: \(TraditionalForm\`lim\_\(θ -> 0\)\(sin(θ)\)\/θ = 1\).

To prove Claim A, we will use i.) trigonemetric geometry, ii.) the squeeze theorem, and iii.) we will call upon the symmetry of ((sin(θ))/θ).

i.) Trigonometric Geometry

Finding an upper bound to ((sin(θ))/θ).

Consider the unit circle - specifically the sector of the circle with center O, central angle 0 < θ < (π/2), and radius 1.

What are the coordinates of the point B: 

How long is Arclength(AB):

Zooming in on the sector of the circle:


What is length(BC): 

And since length(BC) < arclength(AB), we have that sin(θ) < θ and hence ((sin(θ))/θ) < 1.

Finding a lower bound to ((sin(θ))/θ).

Let the tangents at A and B intersect at point E. 

 Now, consider the following lengths in relation to eachother and to θ.

length(EB)          length(ED)

length(ED)          length(AE)

length(AD)          length(AE) + length(ED)

θ = 
        arclength(AB) <  length(AE) + length(EB).

θ < tan(θ) -> θ < ((sin(θ))/(cos(θ))) -> cos(θ) < ((sin(θ))/θ).

ii.) The Squeeze Theorem

In part i.) we showed that cos(θ) < ((sin(θ))/θ) < 1,  
      for  0 < θ < (π/2).  Since (lim_(θ -> 0))cos(θ) = 1and (lim_(θ -> 0)) 1 = 1, by the squeeze theorem we have that \(TraditionalForm\`lim\_\(θ -> \(0\^+\)\)\(sin(θ)\)\/θ = 1\).

iii.) Symmetry Argument.

If f(θ) = ((sin(θ))/θ), we have that \(TraditionalForm\`f(\(-θ\)) = \(\(sin(\(-θ\))\)\/\(-θ\) = \(\(-\(sin(θ)\)\)\/\(-θ\) = \(\(sin(θ)\)\/θ = f(θ)\)\)\)\).  Hence, f(θ) = ((sin(θ))/θ)is an even function.  This means the left and right limits must be equal and so \(TraditionalForm\`lim\_\(θ -> 0\)\(sin(θ)\)\/θ = 1\).

Claim B: \(TraditionalForm\`lim\_\(θ -> 0\)\(cos(θ) - 1\)\/θ = 0\).

To find, \(TraditionalForm\`lim\_\(θ -> 0\)\(cos(θ) - 1\)\/θ\), multiply the expression by the "conjugate" of the numerator.

Claim C: (d/d)x sin(x) = 
        cos(x).

Recall that sin(
          a + b) = sin(a) cos(b) + cos(a) sin(b).

(d/d)x sin(x) = 
        (lim_(h -> 0))((sin(x + h) - sin(x))/h).

o = {0, 0}; a = {1, 0}; b = {((Sqrt[2])/2), ((Sqrt[2])/2)}; c = {((Sqrt[2])/2), 0}; 
    d = {1, 1}; e = {1, 1 - 2 (1 - ((Sqrt[2])/2))};
Show[
        Graphics[{Circle[{0, 0}, 1, {0, (π/4)}], 
            Line[{a, o, b}], {GrayLevel[ .5], Line[{b, e, a, d, b}]}, 
            Text["O", {0, 0}, {0, 2}], Text["A", a, {0, 2}], 
            Text["B", 
              b, {1, -1}],   
              Text["D", d, {-2, 0}], Text["E", e, {-2, 0}], 
            Text["θ", {0, 0}, {-4, -1}]}], 
        AspectRatio -> 1, DefaultFont -> {"Times", 12}, 
        TextStyle -> {FontSlant -> "Oblique"}, 
        PlotRange -> All];