## The Derivative, Part 1

In my last post, I showed that the rate of change of any function that plots as a straight line (a linear function) has a constant rate of change and that value is the gradient of the line. However, for a nonlinear function, its rate of change depends on the value of x, that is, where you are on its graph. I also said that a function’s rate of change is called the derivative of the function and that is what I will call it from now on.

Graphically, the derivative at a particular x value is the gradient of the tangent line at that point:

We would like to find an easy way to mathematically find this value as opposed to graphing the function and estimating the tangent line’s gradient at the desired points. Clearly, as seen above, the derivative is another function of x as its value changes depending on what x is. There are several ways to denote the derivative, but we will start with f’(x) (read as “f prime of x”). We would like to find what f’(x) is given a function f(x).

I know that the following derivation of the derivative may look complex and begs the question about how easy it will be to find the derivative, but following this will help solidify your understanding of what a derivative is and the final result will be used many times to find the easy results for various function forms.

We begin by taking the graph of a function and drawing a secant line (a line that connects two points on the graph) and calculate the gradient of that line:

We want to know the gradient of the estimated tangent line which we are using to approximate the tangent line at x. From your study of linear equations, you know that the gradient of a line can be found from any two points on the line. The two points on our estimated tangent line are (x, f(x)) and (x+h, f(x+h)) where h is a small distance away from x. Using these two points, we find the gradient by calculating the rise from the first point to the second point divided by the run between the two points. The rise is the difference between the y coordinates and the run is the difference between the x coordinates (h):

$\text{gradient} \ =\ \frac{f( x+h) -f( x)}{h}$

Now what happens as h gets smaller? The estimate should get closer to the actual value we are seeking. The below graphic from IkamusumeFan [CC BY-SA (https://creativecommons.org/licenses/by-sa/3.0)] illustrates this:

So it appears that we are interested in what our estimated gradient approaches as h approaches zero. This is, in fact, the formal definition of a function’s derivative. Remember my post on limits? Using limit notation then, the definition of the derivative is

$f'( x) \ =\lim _{h\rightarrow 0} \ \frac{f( x+h) -f( x)}{h}$

Notice that if we just substitute zero for h to evaluate the limit, we get the indeterminate form 0/0 as explained in my prior post. So again you may be saying “this doesn’t make finding a derivative easy at all”. At this point, you are correct. But in my next post, I will show how this definition is used to simplify derivatives.

Posted on Categories Math MethodsTags ,

## Rate of Change

We are all familiar with many physical rates of change. The rate of change of distance is called velocity. If distance is being measured in meters and time is being measured in seconds, the rate of change of distance (velocity) is measured in units meters per second (m/sec). Water filling a bucket can be measured in liters. If time is measured in minutes, then the rate of change of the amount of water in the bucket has units liters per minute (ltr/min). Or we cold measure the height of the water in the bucket in centimeters. The rate of change would be in centimeters per minute (cm/min).

Graphically, the rate of change of a function indicates how fast it increases or decreases as you move along the x-axis. From your study of linear equations, the standard form of an equation of a line is y = mx + c, where m is the gradient or slope of the line. For example, the line y = 2x + 5 has a gradient of 2 which means that it rises 2 units for every unit you move to the right along the x-axis. If this was the equation of the distance of a particle moving from some reference point, where the distance (y-axis) was measured in centimeters and the time (x-axis) was measured in seconds, the velocity of the particle would be 2 cm/sec which is the same as the gradient of the graph. However, the gradient (velocity) is constant since the gradient is 2 anywhere on the graph. All linear graphs have a constant gradient (rate of change). What about non-linear graphs?

Look at the graph of a non-linear function below:

You would say that the function is increasing (positive rate of change) up to about x = -0.6, decreases (negative rate of change) between about -0.6 and 0.6, and increases after 0.6. The rate of change is different depending on where you are on the graph. For many physical problems that have been modelled with an equation, we want to know what the rate of change is at different values of x. A very common problem to solve is to find where the rate of change is zero. The solution to this would find the maximum and/or minimum points because these are the points where the rate of change goes from positive to negative or vice versa.

What does this have to do with calculus? The mathematical term for the rate of change of a function is the derivative of the function and finding the derivative of a function will be the first thing I will define in my next post. Finding the derivative of a function is an operation in calculus, and this is usually the first topic developed in a calculus subject.

Posted on Categories Math MethodsTags ,

## Limits

This post explains what the limit of a function is. The limit operation is used in several areas of maths but I present it here as a pre-requisite to understanding calculus. The limit operation is used to formally define calculus operations.

The concept of a limit is easy to understand but it can be tricky to evaluate. The limit of a function, f(x), is the value the function approaches as x approaches a specified value. Notation-wise:

$\lim _{x\rightarrow a} \ f( x)$

is asking the question “what does f(x) approach as x approaches a?”. Now in many cases, this is a trivial question. The answer is often just f(a). Graphically:

But if the function does not exist at a, or if the function is not continuous at a (you have to lift your pencil off the sheet to continue drawing the graph at a), the answer is not as obvious.

The common values for a, especially as we will eventually see in calculus, are 0 or ±∞. Let’s look at an example:

$\lim _{x\rightarrow \infty } \ \frac{2x+2}{x+1}$

Now limits are nice in that you can take limits of the individual parts of a function and combine the limits of the parts. Looking at the numerator 2x + 2, you can see that it goes to infinity as x goes to infinity. So does the denominator x + 1. But you can’t assume that this reduces to 1. This is an example of an indeterminate form: ∞/∞. Another one is 0/0. When these occur, you usually have to do some algebra on the function to simplify it, then try taking the limit again.

Notice that

$\frac{2x+2}{x+1} \ =\frac{2( x+1)}{x+1} =2$

And the limit of a constant such as 2, is just 2. It doesn’t matter what x is.

There are other ways to arrive at the same answer but algebraically simplifying the function then retaking the limit is the method that will be used in the basic definitions used in calculus.

In my next post, I will introduce the idea of a rate of change of a function. This will be shown to be the derivative of a function.

Posted on Categories Math MethodsTags

## Function Notation Review

I would like to start the journey to explain calculus. This journey actually started with algebra and trigonometry. Calculus is heavily into functions so a review of functions and the associated notation is timely.

Instead of relating a variable with another using the common variables x and y where y is all by itself in the left side (eg y = 3x² -7x + 2), we can replace y with f(x): y = f(x) = 3x² -7x + 2. This notation highlights the fact that x is an independent variable (that is, we are free to choose a valid value of x), but once we do, y or its new notation f(x), depends on the value of x that was chosen. So y is the dependent variable.

This notation also results in simplifying desired operations. Instead of saying “Evaluate y when x = 3″, you can just write f(3). So given a definition of a function, you replace the independent variable with whatever is in the brackets:

f(x) = 3x² -7x + 2
f(3) = 3(3)² -7(3) + 2 = 8

The thing in the brackets does not have to be a number. It can be another variable or even an algebraic expression:

f(x) = 3x² -7x + 2
f(a) = 3a² -7a + 2
f(x+h) = 3(x+h)² -7(x+h) + 2

You can even have functions of functions:

f(x) = 3x² – 5x + 1
g(x) = x² – 7
f[g(x)] = f(x² – 7) = 3(x² – 7)² – 5(x² – 7) + 1
g[f(x)] = g(3x² – 5x + 1) = (3x² – 5x + 1)² – 7

The domain of a function is all the valid values of x that can be used. Many times, the domain of a function (like f(x) and g(x) above) is just any real number. But there are functions where you cannot use just any number. For example, consider

${f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{{x}{-}{2}}$

There is one value of x you cannot use. That value is 2 because that will make the denominator 0, and as you know, this will bring the maths police to your door. So the domain of this function is all real numbers except for 0.

Now consider

${f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{x}{-}{2}}$

Another illegal operation is taking the square root of a negative number. The requirement for this function is that x – 2 has to be 0 or greater. For this to be true, x must be greater than or equal to 2. The phrase ” greater than or equal to” can be replaced by the maths symbol ≥. So the domain of this function is x ≥ 2.

There are other reasons why the domain of a function is restricted, but the most common things to look for is dividing by 0 or taking the square root (or any even root) of a negative number.

I will be using functional notation in the following posts extensively. You will become very familiar and comfortable with it.

Posted on Categories Math MethodsTags

## Trigonometry, Part 7

Another way to analyse trig equations is to use the graph of the trig function involved instead of the unit circle. The following graphs of the basic trig functions were obtained from https://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html.

Let’s talk about the sine function, the first graph, though the following discussion can be generalised to the other two.

There are key points on this graph: the peaks and where it crosses the x-axis. The values of x where the sine is zero are 0, ±𝜋, ±2𝜋, ±3𝜋, … . The positive and negative peaks occur at ±𝜋/2, ±3𝜋/2, ±5𝜋/2, … . Also note that the sine goes from -1 to 1, a total distance of 2. The amplitude is defined as half of this distance, so the amplitude of the basic sine wave is 1.

Another aspect to notice is that the curve repeats every 2𝜋 radians. That means that given any x value, adding or subtracting any multiple of 2𝜋 to x will result in the same sine value. This is the definition of a function’s period. The basic sine function has a period of 2𝜋. Mathematically,

sin(x +2n𝜋) = sin(x), where n is any integer.

So let’s look at the solution to sin x = 1/2, 0 ≤ x ≤ 2𝜋. Instead of the unit circle, the sine graph can be used instead:

Noticing that 𝜋/6 has a sine of 1/2, we can use the symmetry of the sine graph to find the other angle with the same sine, 𝜋 – 𝜋/6 = 5𝜋/6. These are the two answers for the given domain.

What if the given domain was -2𝜋 ≤ x ≤ 2𝜋? Because the period of the basic sine wave is 2𝜋, you can add or subtract multiples of 2𝜋 to the basic angles found to get the answers in the required interval. In this case, subtract 2𝜋 from each of the basic angles previously found to get:

𝜋/6 – 2𝜋 = -11𝜋/6
5𝜋/6 – 2𝜋 = -7𝜋/6

If the problem at hand is more complex like 2sin(2x +𝜋) -1 which I solved in my last post, you solve it as before except you use the sine graph instead of the unit circle to find the basic angles, then proceed as before.

Posted on Categories Math MethodsTags

## Trigonometry, Part 6

Remember what the term argument means? The argument of sin (2x + 7) is 2x + 7, that is an argument of function is the thing the function is operating on. In this case, the sine function is operating on 2x + 7.

So when the argument of a trig function is other than x, the basic method of solving an equation with this function is basically the same as what we did in the last post but an extra bit of algebra is required.

For example, solve 2sin(2x +𝜋) -1 = 0 for 0 ≤ x ≤ 2𝜋. The argument of the sine is 2x + 𝜋. We want to first, solve the equation just as we did before, but in terms of the argument 2x + 𝜋:

2sin(2x +𝜋) -1 = 0 ⟹ sin(2x +𝜋) = 1/2

From our table of common values, we see that the angle 𝜋/6 has a sine of 1/2. So the basic starting point to find all solutions is 2x +𝜋 = 𝜋/6. To correctly limit the number of solutions, let’s convert the given domain 0 ≤ x ≤ 2𝜋 to be in terms of 2x +𝜋. To do this, algebraically change the inequality so that the middle expression is 2x +𝜋:

0 ≤ x ≤ 2𝜋 ⟹ 0 ≤ 2x ≤ 4𝜋 ⟹ 𝜋 ≤ 2x + 𝜋 ≤ 5𝜋

So we need to find all the angles between 𝜋 and 5𝜋 that have a sine of 1/2. Again, a unit circle diagram will help. I will start with 𝜋/6, but realise that 𝜋/6 is not in the required modified domain:

So starting with the basic angle of 𝜋/6, I add multiples of 2𝜋 until we are outside the given domain. For 𝜋/6, adding 2𝜋 gives 13𝜋/6. Adding 2𝜋 again gives 25𝜋/6. Adding another 2𝜋 gives 37𝜋/6, but 37𝜋/6 is greater than 5𝜋 so it is not a valid solution.

So far, the intermediate solutions are 13𝜋/6 and 25𝜋/6. The other basic angle found from the unit circle is 5𝜋/6. Again, this angle is not in our modified domain so we need to add multiples of 2𝜋 to find the angles within 𝜋 and 5𝜋. So just like we did for 𝜋/6, adding 2𝜋 to 5𝜋/6 gives 17𝜋/6. Adding another 2𝜋 gives 29𝜋/6. Adding another 2𝜋 puts us outside the domain.

So we have four intermediate solutions:

2x + 𝜋 = 13𝜋/6
2x + 𝜋 = 25𝜋/6
2x + 𝜋 = 17𝜋/6
2x + 𝜋 = 29𝜋/6

To find the final solutions, we need to solve each of these equations for x:

2x + 𝜋 = 13𝜋/6 ⟹ x = 7𝜋/12
2x + 𝜋 = 25𝜋/6 ⟹ x = 19𝜋/12
2x + 𝜋 = 17𝜋/6 ⟹ x = 11𝜋/12
2x + 𝜋 = 29𝜋/6 ⟹ x = 23𝜋/12

A generic method for solving many trig equations is:

1. Find the basic angles that solve the equation in terms of the argument of the trig function in the equation.
2. Modify the given domain to be in terms of the argument.
3. Add or subtract multiples of 2𝜋 to the basic angles to find all angles within the modified domain. Keep in mind that the basic angles themselves may or may not be in the domain.
4. Set the argument equal to all solutions and solve for the variable used in the problem (usually 𝜃 or x)

There is another way to analyse trig equations which I will show on my next post.

Posted on Categories Math MethodsTags

## Trigonometry, Part 5

Now let’s turn to solving equations with trig functions. Before we get into it though, let me introduce the inverse trig functions.

Just like square and square root, logs and exponentials, add and subtract, etc are inverse operations, there are corresponding inverse functions for the trig functions. These functions are identified usually by a -1 exponent. For example, sin-1(0.25) is asking the question: what angle has a sine of 0.25? Since sin-1x is an inverse function, then sin-1(sin x) = sin(sin-1 x) = x. That is, they undo each other. The same notation is used for the inverse cosine and tangent: cos-1x, tan-1x. Some older calculators may use the notation arcsin(arcsine), arccos(arccosine), and arctan(arc tangent), or they may further abbreviate these as asin, acos, atan. However, these are the same things as the corresponding inverse functions.

Solving trig equations, for the most part, use the same skills you already know for other types of algebraic equations. However, you need to be aware of the cyclic nature of trig functions and the varying signs of these functions in different quadrants, in order to get the correct and complete answers. For example, solve

√2̅cos(x) + 1 = 0, 0 ≤ x ≤ 2𝜋

First notice that frequently, a domain of the equation is specified. This is the 0 ≤ x ≤ 2𝜋 part. In most trig equations, where the unknown is in the trig expression, a domain needs to be specified or there will be an infinite number of solutions. By the way, the expression that is being acted upon by the trig function is called the argument of that function. For example the argument of sin(x² + 7) is x² + 7. In our example, the argument of the cosine is simply x.

So to solve this, you use your algebra skills to get,

$\begin{equation*} \text{cos}( x) =-\frac{1}{\sqrt{2}} \end{equation*}$

So we can take the inverse cosine of both side and use our calculator to find x, and you would find that x =3𝜋/4. However, let’s look at the equation before we take the inverse cosine. You may be asked to solve this without a calculator. You should notice that the cos(x) is equal to an entry in the table of common values I presented in my last post (except for the minus sign). It appears that our answers are associated with the angle 𝜋/4.

Now we have two trig identities that will solve this for us, namely

cos(𝜋 – 𝜃) = -cos𝜃
cos(𝜋 + 𝜃) = -cos𝜃

So, as we want the negative of cos(𝜋/4), then the two angles that will solve this equation within the domain 0 ≤ x ≤ 2𝜋 are 𝜋 – 𝜋/4 = 3𝜋/4 and 𝜋 + 𝜋/4 = 5𝜋/4. Notice that your calculator only gave one of these answers.

Instead of using the trig identities directly, I prefer to use the unit circle to guide me to all correct answers. I highly recommend drawing a unit circle for a particular problem to help guide you to the correct solutions. For this problem:

For practice, draw the unit circle for the same problem, with the provided domain as -𝜋 ≤ x ≤ 𝜋. You should be able to see that the two answers would be -3𝜋/4 and 3𝜋/4.

I will do more examples in my next post.

Posted on Categories Math MethodsTags

## Trigonometry, Part 4

I will now do some examples of using the trig identities covered in the previous posts. But before I do, I want to show you a table that gives the values of the main trig functions for common angles:

Now let’s do some examples:

1. If cos(𝜃) = 0.8829 and 𝜃 is in the first quadrant, find cos(3𝜋/2 – 𝜃).

According to the identity developed before, cos(3𝜋/2 – 𝜃) = -sin𝜃. But what is sin(𝜃)? To find this, we need the Pythagorean identity, sin²(𝜃) + cos²(𝜃) = 1:

$\begin{equation*} \text{sin}^{2} \theta +\text{cos}^{2} \theta \ =1\ \ \ \Longrightarrow \ \ \ \text{sin}( \theta ) =\sqrt{1-\text{cos}^{2} \theta } \end{equation*}$ $\begin{equation*} \text{sin}( \theta ) =\sqrt{1-0.8829^{2}} =0.4696 \end{equation*}$

Therefore, cos(3𝜋/2 – 𝜃) = -sin𝜃 = -0.4696

2. If sin(𝜃) = 0.1736, and 𝜃 is in the first quadrant, find tan(3𝜋/2 + 𝜃).

According to the identity developed before, tan(3𝜋/2 + 𝜃) = -1/tan𝜃 = -cos𝜃/sin𝜃. Again, we need the Pythagorean identity to find cos𝜃:

$\begin{equation*} \text{cos}( \theta ) =\sqrt{1-0.1736^{2}} =0.9848 \end{equation*}$

Therefore, tan(3𝜋/2 + 𝜃) = -0.9848/0.1736 = -5.6729.

3. Find the exact value of cos(5𝜋/6) (without a calculator).

The clues here are that I have been developing trig identities and I just gave you a table of exact values of trig functions for common angles. Perhaps we can equate 5𝜋/6 in terms of a common angle. Notice that 6𝜋/6 – 𝜋/6 = 5𝜋/6. That is 5𝜋/6 = 𝜋 – 𝜋/6. We have a trig identity for this: cos(𝜋 – 𝜃) = -cos𝜃. In our case 𝜃 = 𝜋/6 and cos(𝜋/6) = √3̅/2 so cos(5𝜋/6) = -√3̅/2.

By the way, you can find a decimal approximation to √3̅/2, but this will just be an approximation no matter how many decimal places you include since √3̅/2 is an irrational number and has non-repeating decimals. So the exact value is √3̅/2 since we have agreed that √3̅ is the notation for the exact square root of 3.

4. If sin𝜃 = 4/5 and 𝜋/2 < 𝜃 < 𝜋, find the exact value of cos𝜃

The part 𝜋/2 < 𝜃 < 𝜋 means that the angle is in the second quadrant which means the cosine is negative. As before, we can find the cosine via the Pythagorean identity:

$\begin{equation*} \text{cos}( \theta ) =\sqrt{1-(4/5)^{2}} =3/5 \end{equation*}$

But as the angle is in the second quadrant, cos𝜃 = – 3/5.

Posted on Categories Math MethodsTags

## Trigonometry, Part 3

The identities shown in my last post showed how measuring 𝜃 from the negative x-axis affected its trig functions. Today’s post shows how measuring 𝜃 from the y-axis affects its trig functions.

Consider the diagram below:

The angle 𝜃 is shown measured from both the x and the y axes. The angle 𝜃 measured from the y-axis is 𝜋/2 – 𝜃 when conventionally measured from the positive x-axis. Notice that the right triangles formed from both measurements are identical – just in different orientations. The x coordinate of the 𝜋/2 – 𝜃 angle on the unit circle, cos(𝜋/2 – 𝜃), is the same as the y coordinate of the 𝜃 angle, sin𝜃. Similarly the y coordinate of the 𝜋/2 – 𝜃 angle on the unit circle, sin(𝜋/2 – 𝜃), is the same as the x coordinate of the 𝜃 angle, cos𝜃. This illustrates the following identities:

cos(𝜋/2 – 𝜃) = sin𝜃
sin(𝜋/2 – 𝜃) = cos𝜃
tan(𝜋/2 – 𝜃) = sin(𝜋/2 – 𝜃)/cos(𝜋/2 – 𝜃) = cos𝜃/sin𝜃 = 1/tan𝜃 = cot𝜃

I’ve introduced another trig function here, the cotangent, abbreviated cot. The cotangent is the reciprocal of the tangent.

When solving equations involving trig functions (which we will eventually do), these and the following identities can be used to convert sines to cosines and vice versa.

Now let’s measure 𝜃 from the other side of the y-axis. This gives us the conventional angle 𝜋/2 + 𝜃:

The only change here is that the x coordinate on the unit circle is now negative. So the resulting identities are:

cos(𝜋/2 + 𝜃) = -sin𝜃
sin(𝜋/2 + 𝜃) = cos𝜃
tan(𝜋/2 + 𝜃) = sin(𝜋/2 + 𝜃)/cos(𝜋/2 + 𝜃) = cos𝜃/-sin𝜃 = -1/tan𝜃 = -cot𝜃

Now the angle to the negative y-axis is 3𝜋/2 (270°). An angle measured to the left from the negative y-axis is the conventional angle 3𝜋/2 – 𝜃. The coordinates of this angle on the unit circle are swapped and negative of the angle 𝜃 in the first quadrant. So the picture looks like this:

And the corresponding identities are:

cos(3𝜋/2 – 𝜃) = -sin𝜃
sin(3𝜋/2 – 𝜃) = -cos𝜃
tan(3𝜋/2 – 𝜃) = sin(3𝜋/2 – 𝜃)/cos(3𝜋/2 – 𝜃) = -cos𝜃/-sin𝜃 = 1/tan𝜃 = cot𝜃

I will leave it as an exercise for you to show that for an angle measured to the right of the negative y-axis, the corresponding identities are:

cos(3𝜋/2 + 𝜃) = sin𝜃
sin(3𝜋/2 + 𝜃) = -cos𝜃
tan(3𝜋/2 + 𝜃) = sin(3𝜋/2 + 𝜃)/cos(3𝜋/2 + 𝜃) = -cos𝜃/sin𝜃 = -1/tan𝜃 = -cot𝜃

Next time, I will use the identities shown so far in some example problems.

Posted on Categories Math MethodsTags

## Trigonometry, Part 2

Now let’s use the unit circle to see some of the common trig identities. These identities (rules) will be used in future posts.

Let’s assume we have an acute angle 𝜃. An acute angle is one that is between 0 and 𝜋/2 (or 0 to 90°). The following identities are valid for any angle, not just acute ones – it is just easier to see the logic in the diagram if we assume this.

The following picture shows the relationship between an angle 𝜃 in the first quadrant, and an angle in the second quadrant which is symmetric with 𝜃:

You can see that to measure this symmetric angle from the postive x-axis, you just subtract it from 𝜋. The coordinates of the intersected point on the unit circle are negative for the x coordinate but the same y coordinate as the original angle 𝜃. So the following identities are evident from this picture:

cos(𝜋 – 𝜃) = -cos𝜃
sin(𝜋 – 𝜃) = sin𝜃
tan(𝜋 – 𝜃) = -cos𝜃/sin𝜃 = -tan𝜃

Again, these are true for any angle, not just acute ones.

As an example, let 𝜃 = 𝜋/3, (60°). The following is true for 𝜋/3:

cos(𝜋/3) = 1/2
sin(𝜋/3) = √3̅/2
tan(𝜋/3) = √3̅

Now 𝜋 – 𝜋/3 = 2𝜋/3. So using these identities, we know that

cos(2𝜋/3) = -1/2
sin(2𝜋/3) = √3̅/2
tan(2𝜋/3) = -√3̅

Now let’s look at a symmetric angle in the third quadrant. To measure this angle from the positive x-axis, you add it to 𝜋. The corresponding coordinates of the intersected point on the unit circle are both the negative of the coordinates for 𝜃. So the following identities are shown in this picture:

So these identities are

cos(𝜋 + 𝜃) = -cos𝜃
sin(𝜋 + 𝜃) = -sin𝜃
tan(𝜋 + 𝜃) = -cos𝜃/-sin𝜃 = tan𝜃

Using our same example, 𝜋 + 𝜋/3 = 4𝜋/3. Using these identities:

cos(4𝜋/3) = -1/2
sin(4𝜋/3) = -√3̅/2
tan(4𝜋/3) = √3̅

As was mentioned before, angles measured clockwise from the positive x-axis are negative. So the following trig identities are shown in the figure above:

cos(-𝜃) = cos𝜃
sin(-𝜃) = -sin𝜃
tan(-𝜃) = cos𝜃/-sin𝜃 = -tan𝜃

So,

cos(-𝜋/3) = 1/2
sin(-𝜋/3) = -√3̅/2
tan(-𝜋/3) = -√3̅

There are a couple more identities I would like to show but I’ll save that for next time.

Posted on Categories Math MethodsTags