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Math 006 - Introduction to Calculus


Introduction to Trigonometry for Lecture 9

This page has two kinds of things - easy and difficult! (1) more basic stuff about trignonmetry (2) examples of derivatives of trigonometric functions (3) examples of inverses of trigonometric function

What is trigonometry all about?

Trigonometry is about angles. What is an angle? Here are some angles:

Where do they come from? Well, there are lots of problems to solve involving angles:

How to Measure angles?

How can we work with angles, how can we measure them?

Here are three angles I want to measure:

These angles should be familiar to you already, so they are a simple example.

Old fashioned way of measureing angles: DEGREES

Use a protractor, divided into 360 degrees, to measure your angle in degrees:

In terms of degrees, these angles are 45, 60 and 90.

Measuring angles by arc length: RADIANS

We can make these angles correspond to points on a circle of radius 1:

Every angle corresponds to a point on the edge of the circle; every point on the edge of the circle corresponds to an angle. So we can use points on the circle to measure the size of the angle:

How do you measure where these points are? You just measure the distance from the black point to the point your're interested in:

We call these units of measurement radians. In terms of radians, the three angles are pi/2, pi/3, pi/4

Measuring angles by ratios in right angled trigangles: TRIGONOMETRY

We can put an angle in a triangle, where one of the other angles is a right angle:

Actually, we can't quite, as the sqaure shows, but this is still a good idea. an angle can only go into one shape of right angle triangle, (the scale can be changed, but the basic shape stays the same) so using a right angle triangle is another way to compare angles. But how do we give a number to measure a right angled triangle? Three ways, sine and cosine and tangent, which come from ratios of the side lengths:


Examples of finding derivatives of trigonometric functions

Make sure you can do the following kinds of derivatives in case you are asked questions like these on the exam!

Find the derivative of

For all these examples the main point to remember is that you have a function of a function, so you can find the derivative by using the chain rule.

Solutions

Remember that sinn(x) means (sin(x))n.

1. sin(x4 + 7x + 2)

Use chain rule, and get:

cos((x4 + 7x + 2)*(4x3 + 7))

2. sin(cos(sin(x)))

Use chain rule, get:

cos(cos(sin(x))) * (-sin(sin(x)) * cos(x))

3. sin10(x)

Use chain rule and power rule, to get:

10(sin9(x))cos(x)

4. sin6(cos(x))

Use the chain rule again

6sin5(cos(x))*cos(cos(x))*(-sin(x))

Note, when you're thinking about using the chain rule, you have to think like you're peeling an onion, and ask "what is the last layer I put on?" - that will be the first thing to differentiate, and so on. In this case, the outer most thing is the power 6; before that is sin, and before that is cos.

5. sin6(sin(x)+x+2)

Chain rule again

6 sin5(sin(x)+x+2)*(cos(x) + 1)

6. sin3(x)*cos(7x2+10x)

Watch out here, chain rule and product rule. We get:

((3 sin2(x))*cos(7x2 + 10x) + sin3(x)*((-sin(7x2 + 10x)*(14x + 10)

Since there is not much explaination given for any of the above, I am going to draw a diagram to explain better what is going on, in case you need extra details of how I got the answers.


Example of finding the maximum and minimum of a function involving trigonometry.

This is example 4(b) page 314 of the text book. There are several approaches to doing this problem, and you can use which ever you find easiest to follow.

Note: You don't really need to learn this solution for the exam. You'll be fine on trigonometry as long as you can do the derivatives as in the above section. I am writing out the solution here because this was on the assignment, so if you had problems you may like to know how it was done.

Question: Find the critical numbers, intervals of increase and decrease, and local maximum and minimum of

f(x) = cos(x) - sin(x)
on the interval [-pi,pi]

Answer: The question doesn't say anything about drawing a graph, but drawing a rough graph might help us to understand what kind of answer to expect:

(You can draw this function by making a table of values and plotting points)

From this diagram, it looks like there is one maximum and one minimum value in the interval from -pi to pi. So hopefully this is the result we'll get.

To find the critical values, first differentiate:

f '(x) = -sin(x) - cos(x)

For the critical values, f '(x) = 0:

-sin(x) - cos(x) = 0

So rearranging this a bit:

sin(x) = -cos(x)

The question is, how can we solve this to find x?

There are (at least) four methods you might use, and I can't say one is over all best. Here they are:

These methods are given below. Read them, or skip to the continuation of this question


Solving sin(x) = -cos(x)


Using tan(x)

If

sin(x) = -cos(x)

Then as long as cos(x) is not zero (and it can't be, since if cos(x) = 0, then sin(x) is 1 or -1, so they are not equal), so we get:

sin(x)/cos(x) = -1

But sin(x)/cos(x) = tan(x), so we get

tan(x) = -1

now you can just use your calculator to find

x = tan-1(-1)

The only problem is that the calclator only tells you one possible solution, but there will be more than one solution. You need to know the rule for getting from one value of tan-1(a) to another. We better look at the graph of tan(x) to see what this rule is. (Note, I'm not expecting you to know this rule, or the rule for getting the various values of sin-1(a) and cos-1(a) for the exam, but you might need this for the assignment.)

If tan(x) = a then also
tan(x - 2pi) = tan(x - pi) = tan(x + pi) = tan(x + 2pi) = tan(x + 3pi) = tan(x + 4pi) = a
and for any whole number k, (positive or negative),
tan(x + k*pi) = a

So if your calculator tells you that one answer to tan-1(-1) is -45 degrees, which is the same as -pi/4 radians, then, in radians, you know that other solutions are -pi/4 + pi = 3*pi/4, -pi/4 + 2pi = 7pi/4, etc. If you add too much or take away too much you'll get outside the interval [-pi,pi], so the only critical values in this interval are -pi/4 and 3pi/4.


Using graphs of sin and cos

We want to find where

sin(x) = -cos(x)

If you draw sin and -cos, on the same graph, then the points where the two curves cross are where sin(x) = -cos(x):

Just looking at the graph will give you the answer that x = -pi/4 or 3pi/4.


Using triangles

We can go back to the original definition of cos and sin in terms of the triangle:

Examine this diagram a bit more to see where cos(x) = -sin(x):

We see from the diagram that the solutions are x = -pi/4 or 3pi/4.


Using pythagoras

We know that we have

sin2(x) + cos2 = 1

This is true whatever x is. For the x we have,

sin(x) = -cos(x)
If we square both sides, we get
sin2(x) = cos2(x)

Then if we substitute one of these equations in the other, we get

sin2(x) + sin2 = 1

2*sin2(x) = 1

sin2(x) = 1/2

sin(x) = (1/2)1/2

x = +/- sin-1((1/2)1/2)

Then you can use your calculator to find what x might be. The calculator might say that x = +/- 45 degrees and then you add multiples of 2pi etc, to get the other solutions, which will be +/- 3pi/4 in this interval.

The only problem with this method is that when you use squares, you usually double the number of solutions, so not all the solutions you get are really solutions - but you can easily use your calculator to check which two of the four solutions really satisfy the original equations, which turn out to be -pi/4 and 3pi/4.


Critical values

From all of the above, we now know that

The critical values of cos(x) - sin(x) in the interval [-pi,pi] are
-pi/4 and 3pi/4

The next thing to do is say where the function is increasing or decreasing - that is, where is the derivative positive or negative.

There are several approaches here too. The easiest is to say that between being zero, the derivative it is either positive or negative, so you just need to check one value between the critical numbers. Other approaches can also be made, following on in the same kind of way as for each of the above methods.

For instance, where is f'(x) > 0 ? (this will be where the slope of the tangent is positive, and so the function is increasing.)

At f '(x) > 0,

f '(x) = -sin(x) - cos(x) > 0

Which means that

when f(x) is increasing, - cos(x) > sin(x)

One way to see when this happens is to look at the graph of -cos(x) and sin(x):

From this graph, we see that we have the following table (fill in the end points and the critical values first, and then use this to see what the intervals are):

interval f' (x)
-pi -
[-pi,-pi/4] - Increaseing
-pi/4 0
[-pi/4,3pi/4] + Decreasing
3pi/4 0
[3pi/4,pi] - Increaseing
pi -

Which of the critical values is maximum and which is minimum? (Of course, if we have a good sketch graph, we already know...)

If the slope goes + 0 -, then the critical value is maximum, and if the slope goes - 0 + , then it's minimum. We can fund the values by finding f(x) at these points. So:

At -pi/4 there is a maximum, of value 2*21/2
At 3pi/4 there is a minimum, of value -2*21/2



Please send corrections to spelling mistakes, and any other comments, to me.



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