Section 7: Trigonometric Functions © 1997 by Karl Hahn

7.2 As the World Turns

Derivatives of Trigonometric Functions

Let's get right back to the train that circles around New Atlantis. When the train is parked at Main Street and 10th Avenue East, it faces due north. Suppose it rolled 0.1 km counterclockwise from that stop. How far north would it be. We already know from our discussion in the last section that it would be sin(0.1) km north of the intersection at Main Street and 10th Avenue East. But look carefully at the diagram. As the train rolls out from the station, isn't it going almost due north. Indeed, during that entire 0.1 km of track isn't it going pretty close to north the whole way?

And so, for each meter of track the train covers in that first 0.1 km, doesn't it go almost a meter north? Another way of saying this would be to say that  sin(0.1)/0.1  is very nearly equal to one. My calculator gives 0.998334166 as the value of that expression carried out to 9 figures. That's pretty close to one.

Still, you have to agree that even in that first 0.1 km, the track does diverge slightly from due north. If you eyeball the diagram carefully you can even see it. But in the first 0.01 km, it diverges from due north even less. If you look at the diagram, in the first tenth of a block, you can't even discern any curvature of the track. It's extremely close to pointing due north the whole way. So if in the first 0.1 km the train was very nearly going one meter north for every meter around the track, in the first 0.01 km, it should be even more nearly going one meter north for every meter around the track. In other words,  sin(0.01)/0.01  ought to be closer to one than  sin(0.1)/0.1  was. And indeed, my calculator gives 0.999983333 for the value of  sin(0.01)/0.01.  That's a whole lot closer to one than  sin(0.1)/0.1  was, and it was pretty close to begin with.

As you try shorter and shorter lengths of track from the station at Main Street and 10th Avenue East, our sense of geometry tells us that the the train will be going closer and closer to due north the entire way. If h is the length of track traveled from the station, then it seems that as h gets shorter and shorter,  sin(h)/h  should get closer and closer to one. To put this in mathematical terms, we have a sense that

           sin(h)
    lim            =  1                                           eq. 7.2-1
    h → 0     h

Between a Rock and a Hard Place

Look at figure 7-7. You can see a small right triangle inscribed in the circle whose interior is shaded yellow. The radius of the circle is one again. The angle at the center is h so the base of this triangle is  cos(h)  and the height is  sin(h).  According to the formula you already know for the area of a triangle (A = (1/2)height × base), the yellow area must be  (1/2)sin(h)cos(h). 

The circumscribed right triangle (that's the big one) that comprises the yellow, the green, and the blue shading, has a base equal to the radius of the circle, which is one, and a height of  tan(h)  (click here to see why this is). Once again, applying the formula for the area of a triangle you get that the combined yellow, green, and blue area is  (1/2)tan(h). 

But what about the area comprising just the yellow and the green shading? It is a pie-slice, and therefore represents the same fraction of the area of the entire circle that the angle, h, is of the entire circumference. In other words, because h is not only the angle in radians, but also the distance around the circumference subtended by that angle, it follows that h is to 2π radians as the area in question is to the area of the circle.

Once again, the radius of the circle is unity, so according the formula for the area of a circle (A = πr²), the area of this circle is exactly π. So if a is the area of the pie-slice, then we have:

h = 2π

a eq. 7.2-2 π

The π's cancel, and hence the yellow-plus-green area must be exactly h/2.

Now observe that the yellow-plus-green-plus-blue area must be greater than the yellow-plus-green area, which in turn must be greater than the yellow area alone. In math symbols, that means

   tan(h)     h     sin(h)cos(h)
           >     >                                                eq. 7.2-3
      2       2          2

      1          h
           >          >  cos(h)                                   eq. 7.2-4
   cos(h)     sin(h)

When you take the limit as h goes to zero, you find that  h/sin(h)  is being squeezed between something just greater than one and something just less than one. And as h goes to zero, both the rock and the hard place close in on exactly one (this technique of trapping a limit between two things that approach each other is, in fact, called the squeeze theorem or the squeeze method). So  h/sin(h)  must be squeezed to that limit as well. And we conclude that

              h
    lim            =  1                                           eq. 7.2-5a
   h → 0   sin(h)

           sin(h)
    lim            =  1                                           eq. 7.2-5b
   h → 0      h

           1 - cos(h)
    lim                =  0                                       eq. 7.2-5c
   h → 0       h

This follows from the identity given in eq. 7.1b-8b. This is because

1 - cos(h) lim = h → 0 h

2 sin2(h/2) lim = h → 0 h

lim sin(h/2) h → 0

sin(h/2) (h/2)

As we've already seen, the right-hand factor (in the big parentheses) goes to 1 as h/2 goes to zero. But the other factor, just to its left,  sin(h/2),  goes to zero. And the limit of the product is the product of the limits.

So How Fast is the Train Going North?

Recall that the train travels around the track at 1 km per minute. Suppose nobody wants to get on or off the train at Main Street and 10th Avenue East. So the train doesn't bother to stop there -- it rolls right through the station at 1 km per minute. So what is its northward speed at that exact moment? Since the train is pointed due north at that moment, our intuition tells us that its northward speed must also be 1 km per minute as well. But now think of it in calculus terms. Remember that speed is the derivative of position with respect to time. If  x(t)  is the trains east-west position as a function of time and  y(t)  is its north-south position as a function of time, then the northward speed when the train rolls through the station will be  y'(t)  at the moment, t, that it is at the station.

Let's call the moment that the train passes the station,  t = 0.  Then the train's northward speed at that moment is  y'(0).  And according to our definition of the derivative,

                  y(h) - y(0)
   y'(0) =  lim                                                   eq. 7.2-6
           h → 0       h

             dsin(t)            sin(t+h) - sin(t)
   y'(t)  =           =   lim                                     eq. 7.2-7
               dt        h → 0          h

                    sin(h)
   y'(0)  =   lim           =  1                                  eq. 7.2-8
             h → 0     h

                    sin(t)cos(h) + sin(h)cos(t) - sin(t)
   y'(t)  =   lim                                                 eq. 7.2-9a
             h → 0                   h

                    sin(t)cos(h)   sin(h)cos(t)   sin(t)
   y'(t)  =   lim                +              -                 eq. 7.2-9b
             h → 0       h              h           h

                     sin(h)cos(t)
   y'(t)  =   lim                                                 eq. 7.2-9c
             h → 0        h

   y'(t)  =  cos(t)                                               eq. 7.2-9d

That's it. The derivative of the sine is the cosine. Look carefully at the graph of sine and cosine and try to see how the slope of the sine trace (in blue) is always exactly equal to the cosine trace (in green).

And what about the train's east-west speed at any time, t? Remember that the east-west position of the train is given by  x(t) = cos(t).  So

                    cos(t+h) - cos(t)
   x'(t)  =   lim                                                 eq. 7.2-10a
             h → 0          h

                    cos(t)cos(h) - sin(t)sin(h) - cos(t)
   x'(t)  =   lim                                                 eq. 7.2-10b
             h → 0                  h

                    -sin(t)sin(h)
   x'(t)  =   lim                                                 eq. 7.2-10c
             x → 0        h

   x'(t) = -sin(t)                                                eq. 7.2-10d

And you're left with the derivative of the cosine is minus the sine. I urge you to study the graph in figure 7-3 again and see how the sine trace (in blue) is always the opposite of the slope of the cosine trace (in green).

Derivative of tan, cot, sec, and csc

In the trig identity section, you learned that these four functions were merely quotients of the sine and cosine functions. So to find the derivatives of tan, cot, sec, and csc, it would make sense to use the quotient rule. Let's begin by finding the derivative of  f(x) = tan(x). 

                       sin(x)
   f(x)  =  tan(x)  =                                             eq. 7.2-11a
                       cos(x)

             cos(x)cos(x) - sin(x)(-sin(x))
   f'(x)  =                                                       eq. 7.2-11b
                        cos2(x)

             cos2(x) + sin2(x)
   f'(x)  =                                                       eq. 7.2-11c
                  cos2(x)

                1
   f'(x)  =           =  sec2(x)                                  eq. 7.2-11d
             cos2(x)


             cos2(x)   sin2(x)
   f'(x)  =          +          =  1 + tan2(x)                    eq. 7.2-11e
             cos2(x)   cos2(x)

The two forms (equation 7.2-11d and 7.2-11e) are identities of each other, and therefore entirely equivalent. So the derivative of the tangent is the secant squared. But it is also true that the derivative of the tangent is one plus the tangent squared. The latter is a very important property of the tangent function (that is to take its derivative you simply square it and add 1), and we will make use of it later on.

If you recall that  cot(x)  is simply the reciprocal of the tangent, you should be able to apply the same attack on it as we did above on the tangent function to show that the derivative of the cotangent is minus the cosecant squared, and that equivalently the derivative of the cotangent is minus one minus the cotangent squared. I urge you to take out pencil and paper right now and do this as an exercise.

Finding the derivative of secant and cosecant is even easier than doing so for tangent and cotangent. You can use the quotient rule if you like, or you can observe that the derivative of  f(x) = 1/x  is  f'(x) = -1/x².  Since secant and cosecant are reciprocals of cosine and sine respectively, you need only apply the chain rule and the above derivative to find their derivatives. For example, if  f(x) = 1/x  and  g(x) = cos(x)  and  h(x) = f(g(x)) = 1/cos(x),  then the chain rule says

                                  -1
   h'(x)  =  f'(g(x)) g'(x)  =          (-sin(x))  =  tan(x) sec(x)
                                cos2(x)
                                                                  eq. 7.2-12

So the derivative of the secant is the tangent times the secant. Once again, as an exercise, I urge you right now to use the same attack to show that the derivative of the cosecant is minus the cotangent times the cosecant.

   y  =  arcsin(x)                                                eq. 7.2-13

   dy           arcsin(x + h) - arcsin(x)
       =  lim                                                     eq. 7.2-14
   dx    h → 0              h

   sin(y)  =  x                                                   eq. 7.2-15

   cos(y) y'  =  1                                                eq. 7.2-16

   sin2(y) + cos2(y)  =  1                                        eq. 7.2-17

               ___________
   cos(y)  =  √1 - sin2(y)                                        eq. 7.2-18

               ______
   cos(y)  =  √1 - x2                                              eq. 7.2-19

    ______
   √1 - x2 y'  =  1                                               eq. 7.2-20

             1
   y'  =                                                          eq. 7.2-21
          √1 - x2

And that is the derivative of  arcsin(x).  Go over it several times, since they might pop this derivation on you on the exam.

Interesting, isn't it, how a simple algebraic function made only from squares and square roots can be the derivative of something as abstract as an inverse trig function. What's happening here is that this derivative is telling us something about the relationship of  sin(x)  to its derivative. Think about it. Taking the derivative of  sin(x)  is equivalent to squaring it, subtracting the result from 1, then taking the square root of that. Of course, that doesn't work for any function. But for  sin(x)  it does. It's a special relationship that  sin(x)  enjoys with its own derivative. The concept of a function's derivative being related to the function itself according to some algebraic rule is an important one that you will be coming back to when you get to more advanced material. For now, just make a mental note of it and move on. And if the thought tickles your fancy, all the better.

Exercises

1) Use what you learned above about a) the derivative of the tangent function, and b) taking the derivative of an inverse function to determine the derivative of  f(x) = arctan(x).  Hint: In this instance, the more useful of the two forms for the derivative of the tangent function is the "one plus tangent squared" form. Click here to view solution.

2) Use the product rule to determine the derivative of  f(x) = sin(x)cos(x).  Now observe the trig identity,  sin(x)cos(x) = (1/2)sin(2x).  Show that if you take the derivative of the right-hand side of this identity, it is consistent with the derivative you get using the left-hand side of the identity. Click here to view solution.

3) Use the product rule and the chain rule to find the derivative of

   f(x)  =  x sin( 1/x )

4) Use implicit differentiation on each of the following (where y is a function of x in each case):

a)   sin(y) = cos(x2)

b)   y tan(x) = x

c)   sin(xy)cos(x) = y2

d)   tan2(y) = cos(2x)

          1 - cos(h)
    lim             
   h → 0      h2

6) Use figure 7-7 (shown here again for your convenience) to prove that for positive h in the first quadrant, it is always true that  sin(h) < h.  Hint: Construct another line segment from the point (cos(h),sin(h)) to the point  (1,0)  and see what the area of the resulting triangle is compared to the area of the pie-slice. Once you have proved that, use the inequality in a delta-epsilon proof that the function,  f(x) = sin(x),  is continuous at the point,  x = 0.  Take your best shot at this, and then click here.

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Proof that  sin(x)  is Continous Everywhere

In the last exercise above you proved that  f(x) = sin(x)  is continuous at  x = 0.  It would be better to know that  f(x) = sin(x)  is continuous for all x. And since there is an outside possibility that your instructor might spring something like this on an exam, here is a delta-epsilon proof of that, taken step by step. This is likely the hardest delta-epsilon proof that you will encounter in your first year of calculus, so if you go over this until you understand it, you will be ready for any delta-epsilon proof you are likely see on any exam.

The outline of the proof is that first we set up the delta-epsilon contract that we have to meet for

    lim    sin(x)  =  sin(c)                                      eq. 7.2-22
   x → c

to be true for all x. Recall that this is the limit that defines continuity.

We will make a change of variables that puts the δ term to work invisibly inside the contract expression. You will see that once we set up the delta-epsilon contract in this way, there will be a trig identity that we can apply right away. We will then show that each of the terms in the resulting expression is less than some very simple expression. By combining the simple expressions, we will have an easy way to quantify the δ in terms of ε so that the contract must be met.

But first, some more basic inequalities concerning sine and cosine. You recall that for  -π/2 < h < π/2  (but h not equal zero) we showed (in problem 6) that it is always the case that   |sin(h)| < |h|.  Observe also (using figure 7-3 if you need to) that it is always the case that   |sin(h)| ≤ 1,  and that   |cos(h)| ≤ 1,  no matter what h is. The following inequalities are immediate consequences of this (again assuming that h is not equal zero):

   cos2(h)  ≤  |cos(h)|                                           eq. 7.2-23a


   1 - |cos(h)|  ≤  1 - cos2(h)  =  sin2(h)  <  h2                eq. 7.2-23b

or more succinctly,

   1 - |cos(h)|  <  h2                                            eq. 7.2-23c

Also please recall from algebra that for any a and b, it is always true that  |a ± b| ≤ |a| + |b|.  We will be using all these inequalities very shortly.

One last note. This may seem obvious, but it bears repeating before getting into the meat of this proof. And that is that the relation, <, is transitive. So is the relation, ≤. That means, for example, that if  a < b  and  b < c,  then  a < c.  Likewise, if  a ≤ b  and  b ≤ c,  then  a ≤ c.  We will be using this over and over in the proof that follows to show that the contract is still met even when we replace the original expressions with new ones that are greater.

Step 1: Write the contract. To prove the limit shown in equation 7.2-22, the contract is that whatever ε > 0 I might name, you need to be able to name a δ > 0 so that

   |sin(x) - sin(c)|  <  ε                                        eq. 7.2-24

whenever  |x - c| ≤ δ.  In other words, no matter how close I demand that  sin(x)  be to  sin(c),  you can find an x close enough to c to make it happen. The degree of closeness of x to c is expressed by the δ. How close I am demanding  sin(x)  to be to  sin(c)  is expressed by the ε.

Step 2: A change of variable. Observe that if  |x - c| ≤ δ,  then there is some h, where  0 < h ≤ δ,  that makes is so that  x = c ± h.  So we can substitute  c ± h  into eq. 7.2-24 wherever we see x.

   |sin(c ± h) - sin(c)|  <  ε                                    eq. 7.2-25

Step 3: Apply the identity for the sine of a sum (or difference). You remember,  sin(a + b) = sin(a)cos(b) + sin(b)cos(a)  and sin(a - b) = sin(a)cos(b) - sin(b)cos(a). It doesn't take much of a stretch to see that you can combine these two rules into  sin(a ± b) = sin(a)cos(b) ± sin(b)cos(a).  So we apply that last form to eq. 7.2-25:

   |sin(c)cos(h) ± sin(h)cos(c) - sin(c)|  <  ε                   eq. 7.2-26a

You can combine the first and third terms that are inside the absolute value sign to get

   |sin(c)(cos(h) - 1) ± sin(h)cos(c)|  <  ε                      eq. 7.2-26b

Step 4: Use the absolute value inequality. If  |a ± b| ≤ |a| + |b|,  and we can choose a and b in such a way that we guarantee  |a| + |b| < ε,  then that would force  |a ± b| ≤ |a| + |b| < ε.  In this case,  a = sin(c)(cos(h) - 1)  and  b = sin(h)cos(c).  Hence if we can choose h so that

   |sin(c)(cos(h) - 1)| + |sin(h)cos(c)|  <  ε                    eq. 7.2-27a

that would guarantee our contract. Notice that  |sin(c)(cos(h) - 1)| = |sin(c)(1 - cos(h))|.  So

   |sin(c)(1 - cos(h))| + |sin(h)cos(c)|  <  ε                    eq. 7.2-27b

is the same thing.

Step 5: Restrict the range of h. This is the same as restricting the range of δ. But the contract allows us to make δ as close to zero as we like. So there is no problem restricting it to  0 < δ < 1.  Which, of course, restricts h to a range of  0 < h < 1.  Why would we want to do that? So that we keep things like  cos(h)  within a range where its behavior is simple. In this case we are interested in keeping  cos(h)  positive. You will see why momentarily.

Step 6: Substitute simpler terms that are known to be greater. Once again, if something greater than our contract expression is less than ε, then that forces the contract expression to be less than ε as well (because of transitivity). So, for example, we know that if  0 < cos(h) ≤ 1,  then  cos(h) = |cos(h)|.  And we know from the discussion above that  1 - |cos(h)| ≤ h².  Consequently wherever we see  1 - cos(h),  we can replace it with h² and rest assured that replacement is at least as large as what it replaced. Hence our contract remains satisfied if

   |sin(c)h2| + |sin(h)cos(c)|  <  ε                              eq. 7.2-28a

Likewise, since  |sin(h)| < h and h  is known to be positive, we can replace  sin(h)  with h, knowing that it won't make the expression any smaller. Hence

   |sin(c)h2| + |h cos(c)|  <  ε                                  eq. 7.2-28b

still guarantees the contract. We also know that no matter what c is,  |sin(c)| ≤ 1,  and  |cos(c)| ≤ 1.  So wherever we see them, we can replace them with 1 and still guarantee the contract:

   |h2| + |h|  <  ε                                               eq. 7.2-28c

And since h is positive, this becomes simply

   h2 + h  <  ε                                                   eq. 7.2-28d

At this point, we could just go and solve the quadratic inequality above. But there is an easier way. Remember that  h < 1.  Consequently  h² < h  (remember that this works because we know h to be positive). So by the same thinking as was responsible for all our other replacements above, we have

   h + h  =  2h  <  ε                                             eq. 7.2-28e

or equivalently

   h  <  ε/2                                                      eq. 7.2-28f

If you can choose h so that it satisfies this inequality, this still guarantees the contract.

Step 7: Bring back the δ. Since h ≤ δ, you can replace h with δ, and the inequality still guarantees the contract. So in the end, whatever ε I choose, you need only give me a δ that is less than half of ε and no more than 1 (remember we restricted the range of h before), and the contract is met.

And that is the proof that  sin(x)  is continuous. Admittedly it's a lot to swallow all at once. But go back and reread the outline of the proof up at the beginning. Then see how each step works into something in the outline. The outline tells you through which towns the proof passes. The detailed steps tell you where to make your right and left turns so that you end up unerringly at your destination. If you think about complicated proofs that way, you will never get lost.

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