Derivatives of Trigonometric Functions

We will next take a look at trigonometric functions, and how we can find derivatives for them. I will first present a table of derivatives, and then we will work on proving each of them.

In total, there are six derivatives that we will focus on.

  1. \frac{dy}{dx}(sin(x)) = cos(x)
  2. \frac{dy}{dx}(cos(x)) = -sin(x)
  3. \frac{dy}{dx}(tan(x)) = sec^2(x)
  4. \frac{dy}{dx}(csc(x)) = -csc(x)cot(x)
  5. \frac{dy}{dx}(sec(x)) = sec(x)tan(x)
  6. \frac{dy}{dx}(cot(x)) = csc^2(x)

The first two derivatives for sin(x) and cos(x) are the most important to prove. Since the other functions can be rewritten in terms of sin(x) and cos(x), once we have those derivatives the others will fall into place.

Proof: \frac{dy}{dx}(sin(x)) = cos(x)

Let’s start with the definition of a derivative, and see if we can reach the conclusion that \frac{dy}{dx}(sin(x)) = cos(x)

lim_{h \to 0} \frac{sin(x+h) - sin(x)}{h}

In order to progress here, we will need to use a trigonometric identity. The identity is that sin(x + h) = sin(x)cos(h) + cos(x)sin(h). Therefore:

lim_{h \to 0} \frac{sin(x+h) - sin(x)}{h}
=lim_{h \to 0} \frac{sin(x)cos(h) + cos(x)sin(h) - sin(x)}{h}
=lim_{h \to 0} sin(x)(\frac{cos(h) - 1}{h}) + cos(x)(\frac{sin(h)}{h})
=lim_{h \to 0} sin(x) * lim_{h \to 0} \frac{cos(h) - 1}{h} + lim_{h \to 0} cos(x) * lim_{h \to 0} \frac{sin(h)}{h}

At this point, we are stuck again. We don’t seem to have a good way to evaluate lim_{h \to 0} \frac{cos(h) - 1}{h} or lim_{h \to 0} \frac{sin(h)}{h}.

To start, we will take a look at lim_{h \to 0} \frac{sin(h)}{h}. This can be proven with L’Hoptial’s rule rather easily once we learn it, however with the tools we currently have, the proof is more complete. We will instead attempt to prove this limit geometrically.

Suppose we have a unit circle, and we have two triangles, one that sits within the unit circle, and one that sits outside it, like the diagram below.

First, let’s figure out the area of the triangle made with the blue, orange, and black line, highlighted in red below.

The area of the triangle is \frac{1}{2} * b * h = \frac{1}{2} * 1 * sin(x) = \frac{sin(x)}{2}

Next, let’s take a look at the area up to the actual arc of the triangle, shown below in red.

The area of the arc is equal to \frac{x}{2\pi} * \pi * r^2 = \frac{x}{2}, since our radius is 1.

Finally, let’s take the area of the larger blue triangle. This is \frac{1}{2} * 1 * tan(x) = \frac{1}{2}tan(x).

What we learned from this is that the first triangle is the smallest, the arc is in the middle, and the final triangle is the largest. We can write this as an inequality.

\frac{1}{2}sin(x) < \frac{x}{2} < \frac{1}{2}tan(x)
\Longrightarrow sin(x) < x < tan(x)
\Longrightarrow 1 < \frac{x}{sin(x)} < \frac{tan(x)}{sin(x)}
\Longrightarrow 1 < \frac{x}{sin(x)} < \frac{1}{cos(x)}
\Longrightarrow 1 > \frac{sin(x)}{x} > cos(x)

What we’ve done here is manipulated our inequality in such a way that we place \frac{sin(x)}{x} between an upper and lower bound. The reason why this is useful is because it tells us information about where \frac{sin(x)}{x} might be.

Let’s graph the three functions to see what they look like at x = 0.

You can see that all three functions look to be equal to 1, at x = 0. If we evaluate our inequality at x = 0, we get exactly that.

1 > \frac{sin(x)}{x} > cos(0) \Longrightarrow 1 > \frac{sin(x)}{x} > 1

You can see from this that as the functions that bound \frac{sin(x)}{x} approach x = 0, they approach y = 1. The only way this can be true is if \frac{sin(x)}{x} also approaches 1.

Therefore, we can conclude that lim_{h \to 0} \frac{sin(h)}{h} = 1. This solves our first limit problem.

The second limit problem can luckily be rearranged in a way that is easy to solve given the first limit solution.

lim_{h \to 0} \frac{cos(h) - 1}{h}
= lim_{h \to 0} \frac{cos(h) - 1}{h} * \frac{cos(h) + 1}{cos(h) + 1}
= lim_{h \to 0} \frac{cos^2(h) - 1}{h(cos(h)+1)}
= lim_{h \to 0} \frac{-sin^2(h)}{h(cos(h)+1)}
= - lim_{h \to 0} \frac{sin(h)}{h} * lim_{h \to 0} \frac{sin(h)}{cos(h) + 1}
= -1 * \frac{0}{1+1} = 0

Once we have this, we can solve our equation for sin(x).

=lim_{h \to 0} sin(x) * lim_{h \to 0} \frac{cos(h) - 1}{h} + lim_{h \to 0} cos(x) * lim_{h \to 0} \frac{sin(h)}{h}
=sin(x) * 0 + cos(x) * 1 = cos(x)

Therefore, we have proven that \frac{dy}{dx}(sin(x)) = cos(x)

We can basically do the same proof to show that \frac{dy}{dx}(cos(x)) = -sin(x). I won’t show this proof due to the similarity.

The next proof we will look at is \frac{dy}{dx}(tan(x)) = sec^2(x)

This proof is rather simple, due to the fact that tan(x) = \frac{sin(x)}{cos(x)}

\frac{dy}{dx}(\frac{sin(x)}{cos(x)}
= \frac{cos(x)cos(x) + sin(x)sin(x)}{cos^2(x)}

From here, we can use the identity cos^2(x) + sin^2(x) = 1

= \frac{1}{cos^2(x)}
= sec^2(x)

Similar to this proof, the inverse trigonometric functions can be proved using either the reciprocal rule, or the quotient rule.

Proof: \frac{dy}{dx}(csc(x)) = -csc(x)cot(x)

\frac{dy}{dx}(\frac{1}{sin(x)}
= \frac{-cos(x)}{sin^2(x)}
= \frac{-cos(x)}{sin(x)} * \frac{1}{sin(x)}
= -csc(x)cot(x)

The remaining inverse trigonometric proofs are similar to the proof of csc(x).

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