# Trigonometry Tutorial 1

During this tutorial you will be asked to perform calculations involving trigonometric functions. You will need a calulator to proceed.

The purpose of this tutorial is to review with you the elementary properties of the trigonometric functions. Facility with this subject is essential to success in all branches of science, and you are strongly urged to review and practice the concepts presented here until they are mastered. Let us consider the right-angle triangle shown in

The angle at \(C\) is a right angle and the angle at \(A\) we will call \(\theta\). The lengths of the sides of the triangle we will denote as \(p\), \(q\) and \(r\). From your elementary geometry, you know several things about this triangle. For example, you know the Pythagorean relation, \(q^2 = p^2 + r^2\). That is, the square of the length of the side opposite the right angle, which we call the hypotenuse, is equal to the sum of the squares of the lengths of the other two sides.

We know other things. For example, we know that if the lengths of the three sides of any triangle \(p\), \(q\) and \(r\) are specified, then the whole triangle is determined, angles included. If you think about this for a moment, you will see it is correct. If I give you three sticks of fixed length and told you to lay them down in a triangle, there's only one triangle which you could make. What we would like to have is a way of relating the angles in the triangle, say \(\theta\), to the lengths of the sides.

It turns out that there's no simple analytic way to do this. Even though the triangle is specified by the lengths of the three sides, there is not a simple formula that will allow you to calculate the angle \(\theta\). We must specify it in some new way.

To do this, we define three ratios of the sides of the triangle.

One ratio we call the **sine** of theta, written \(sin (\theta)\), and it is defined as the ratio of the side opposite \(\theta\) to the hypotenuse, that is \(r/q\).

The **cosine** of \(\theta \), written \(cos (\theta)\), is the side adjacent to \(\theta\) over the hypotenuse, that is, \(p/q\).

This is really enough, but because it simplifies our mathematics later on, we define the **tangent** of \(\theta\), written \(tan(\theta) \), as the ratio of the opposite to the adjacent sides, that is \(r/p\). This is not an independent definition since you can readily see that the tangent of \(\theta\) is equal to the sine of \(\theta\) divided by the cosine of \(\theta\). Verify for yourself that this is correct.

In order to make these functions useful in calculations, we need numerical values of them for the different values of \(\theta\).

All scientific calculators provide this information. The first thing to ensure is that your calculator is set to the anglular measure that you want. Angles are usually measured in either degrees or radians (see tutorial on Dimensional Analysis). The angle \(2^\circ\) is a much different angle than 2 radians since \(180^\circ = \pi radians = 3.1416...\) radians. Make sure that your calculator is set to degrees.

Now suppose that we want the sine of \(24^\circ\). Simply press \(24\) followed by the [sin] key and the display should show the value \(0.4067\). Therefore, the sine of \(24^\circ\) is \(0.4067\). That is, in a triangle like * Panel 1* where \(\theta = 24^\circ\), the ratio of the sides \(r\) to \(q\) is \(0.4067\). Next set your calculator to radians and find the sine of \(0.42\) radians. To do this enter \(0.42\) followed by the [sin] key. You should obtain a value of \(0.4078\). This is nearly the same value as you obtained for the sine of \(24^\circ\). Using the relation above you should confirm that \(24^\circ\) is close to \(0.42\) radians

Obviously, using your calculator to find values of sines is very simple. Now find sine of \(42^\circ 24\) minutes. The sine of \(42^\circ 24\) minutes is \(0.6743\). Did you get this result? If not, remember that \(24\) minutes corresponds to \(24/60\) or \(0.4^\circ\). The total angle is then \(42.4^\circ\)

The determination of cosines and tangents on your calculator is similar. It is now possible for us to solve the simple problem concerning triangles. For example, in * Panel 2*, the length of the hypotenuse is \(3\) cm and the angle \(\theta\) is \(24^\circ\). What is the length of the opposite side \(r\)? The sine of \(24^\circ\) as we saw is \(0.4067\) and it is also, by definition, \(r/3\).

So, sine of \(24^\circ = 0.4067 = r/3\), and therefore, \(r = 3 \times 0.4067 = 1.22cm\).

Conversely, suppose you knew that the opposite side was \(2 cm\) long and the hypotenuse was \(3cm\) long, as in * Panel 3*, what is the angle \(\theta \)? First determine the sine of \(\theta\) .

You should find that the sine of \(\theta\) is \(2/3\), which equals \(0.6667\). Now we need determine what angle has \(0.6667\) as its sine.

If you want your answer to be in degrees, be sure that your calculator is set to degrees. Then enter \(0.6667\) followed by the [INV] key and then the [sin] key. You should obtain a value of \(41.8^\circ\). If your calculator doesn't have a [INV] key, it probably has a [2^{nd}F] key and the inverse sine can be found using it.

One use of these trigonometric functions which is very important is the calculation of components of vectors. In * Panel 4* is shown a vector

**OA**in an xy reference frame. We would like to find the y component of this vector. That is, the projection

**OB**of the vector on the y axis. Obviously,

**OB**=

**CA**and

**CA**/

**OA**= \(sin(\theta)\), so

**CA**=

**OA**\(sin(\theta)\). Similarly, the x-component of

**OA**is

**OC**. And

**OC**/

**OA**= \(cos(\theta)\) so

**OC**=

**OA**\(cos(\theta)\).

There are many relations among the trigonometric functions which are important, but one in particular you will find used quite often. * Panel 1* has been repeated as

*for you. Let us look at the sum \(cos^2 + sin^2\). From the figure, this is \((p/q)^2 + (r/q)^2\), which is \([(p^2 + r^2) / (q^2)]\). The Pythagorean theorem tells us that \(p^2 + r^2 = q^2\) so we have \([(p^2 + r^2) / q^2] = (q^2/q^2) = 1\). Therefore, we have; \(cos^2 + sin^2 = 1\).*

**Panel 5**Our discussion so far has been limited to angles between \(0\) and \(90^\circ\). One can, using the calculator, find the the sine of larger angles (eg \(140^\circ\) ) or negative angles (eg \(-32^\circ\) ) directly. Sometimes, however, it is useful to find the corresponding angle betweeen \(0\) and \(90^\circ\). * Panel 6* will help us here.

In this \(xy\) reference frame, the angle \(\theta\) is clearly between \(90^\circ\) and \(180^\circ\), and clearly, the angle \(\alpha\), which is \(180 - \theta\) ( \(\alpha\) is marked with a double arc) can be dealt with. In this case, we say that the magnitude of sine, cosine, and tangent of \(\theta\) are those of the supplement \(\alpha\) and we only have to examine whether or not they are positive or negative.

We always assume that the hypotenuse \(q\) is positive, but \(r\) and \(p\) have the sign appropriate to their direction with respect to the origin. Clearly, in * Panel 6*, \(r\) is positive and \(p\) is negative. Therefore, we can write

\(r/q = sin(a) = sin(180^\circ - \theta)\)

\(cos(a) = (-p/q) = -cos(a) = -cos(180^\circ - \theta)\)

tangent of \(\alpha = r/(-p) = -tan(a) = - tan(180^\circ - \theta)\)

Notice that only the sin is positive.

For example, what is the sine, cosine and tangent of \(140^\circ\)? The supplement is \(180^\circ - 140^\circ = 40^\circ\). Find the sine, the cosine and the tangent of \(40^\circ\).

\(sin(40^\circ) = 0.6428\)

\(cos(40^\circ) = 0.7660\)

\(tan(40^\circ) = 0.8391\)

Now we know that \(sin(140^\circ) = 0.6428\), \(cos(140^\circ) = -0.7660\), and \(tan(140^\circ) = -0.8391\).