The derivative of the cosine via the definition of the derivative

In this problem, we’ll prove that

using the definition of the derivative. Here’s the outline:

• Write down the difference quotient for the cosine.
• Find an identity to expand out the \cos(x+h) and then do so in your difference quotient.
• Do a little algebra to write your expanded difference quotient in the form
  F_1(x)G_1(h) + F_2(x)G_2(h).
• Evaluate the limits as h\to0 of G_1(h) and G_2(h) (they should be familiar from our work with the sine function).
• Putting it all together, conclude that

Step 1: d/dxcos(x) = lim_(h->0) (cos(x+h) -cos(x) )/h

Converting the derivative to the difference quotient

Step 2: cos(x+h) = cos(x)cos(h) - sin(x)sin(h)

Using an identity to expand out cos(x+h)

Step 3: lim_(h->0) (cos(x+h) -cos(x) )/h= lim_(h->0) ((cos(x)cos(h) - sin(x)sin(h)) -cos(x) )/h

replacing cos(x+h) with the expanded form

Step 4: lim_(x->h) ((cos(x)cos(h) - sin(x)sin(h)) -cos(x) )/h =lim_(x->h) (cos(x)cos(h) - cos(x) - sin(x)sin(h) )/h

Using Commutative Property To Rearrange The Equation

Step 5: = (lim_(x->h) (cos(x)cos(h) - cos(x)) / h) - (lim_(x->h) (sin(x)sin(h))/h)

Limit of the Sum is equal to the sum of the limits

Step 6: = (cos(x) lim_(x->h) (1*cos(h) - 1) / h) - (lim_(x->h) (sin(x)sin(h))/h)

Factoring out a cos(x) from the first half of the equation

Step 7: = (cos(x) lim_(x->h) (1*cos(h) - 1) / h) - (sin(x) lim_(x->h) (1*sin(h))/h)

Factoring out a sin(x) from the second half of the equation

Step 8: = cos(x) * 0 - sin(x) *1

Using the squeeze theorem, we know for a fact that lim_(theta->0) (cos(theta) - 1) / theta = 0 and that lim_(theta->0) (sin(theta)) / theta = 1 we can simplify some of our terms

Step 9: cos(x) * 0 - sin(x) *1 = -sin(x)

sprinkle in some simplification …

Step 10: d/dx cos(x) = -sin(x)

We can finally determine that the derivative of cos(x) is indeed equal to -sin(x)

Some help was received to understand how the squeeze theorem applies to the lim_(theta->0) (cos(theta) - 1) / theta