Mathematics » Trigonometry » Defining Ratios In The Cartesian Plane

Summary and Main Ideas

Defining Ratios in the Cartesian Plane

We have defined the trigonometric ratios using right-angled triangles. We can extend these definitions to any angle, noting that the definitions do not rely on the lengths of the sides of the triangle, but on the size of the angle only. So if we plot any point on the Cartesian plane and then draw a line from the origin to that point, we can work out the angle between the \(x\)-axis and that line. We will first look at this for two specific points and then look at the more general case.

Finding an angle for specific points

In the figure below points \(P\) and \(Q\) have been plotted. A line from the origin (\(O\)) to each point is drawn. The dotted lines show how we can construct right-angled triangles for each point. The dotted line must always be drawn to the \(x\)-axis. Now we can find the angles \(A\) and \(B\):

Fact:

We can also extend the definitions of the reciprocals in the same way.

515701398d1ae8da660241375292b5d0.png

From the coordinates of \(P(2;3)\), we can see that \(x=2\) and \(y=3\). Therefore, we know the length of the side opposite \(\hat{A}\) is \(\text{3}\) and the length of the adjacent side is \(\text{2}\). Using: \[\tan\hat{A} = \cfrac{\text{opposite}}{\text{adjacent}} = \cfrac{3}{2}\] we calculate that \(\hat{A} = \text{56.3}°\).

We can also use the theorem of Pythagoras to calculate the hypotenuse of the triangle and then calculate \(\hat{A}\) using: \[\sin\hat{A} = \cfrac{\text{opposite}}{\text{hypotenuse}} \qquad \text{ or } \qquad \cos\hat{A} = \cfrac{\text{adjacent}}{\text{hypotenuse}}\]

Consider point \(Q(-2;3)\). We define \(\hat{B}\) as the angle formed between line \(OQ\) and the positive \(x\)-axis. This is called the standard position of an angle. Angles are always measured from the positive \(x\)-axis in an anti-clockwise direction. Let \(\alpha\) be the angle formed between the line \(OQ\) and the negative \(x\)-axis such that \(\hat{B} + \alpha = 180°\).

From the coordinates of \(Q(-2;3)\), we know the length of the side opposite \(\alpha\) is \(\text{3}\) and the length of the adjacent side is \(\text{2}\). Using: \[\tan\alpha = \cfrac{\text{opposite}}{\text{adjacent}} = \cfrac{3}{2}\] we calculate that \(\alpha = \text{56.3}°\). Therefore \(\hat{B} = 180° – \alpha = \text{123.7}°\).

Similarly, an alternative method is to calculate the hypotenuse using the theorem of Pythagoras and calculate \(\alpha\) using: \[\sin\alpha = \cfrac{\text{opposite}}{\text{hypotenuse}} \qquad \text{ or } \qquad \cos\alpha = \cfrac{\text{adjacent}}{\text{hypotenuse}}\]

Finding any angle

If we were to draw a circle centred on the origin (\(O\)) and passing through the point \(P(x;y)\), then the length from the origin to point \(P\) is the radius of the circle, which we denote \(r\). We denote the angle formed between the line \(OP\) and the \(x\)-axis as \(\theta\).

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We can rewrite all the trigonometric ratios in terms of \(x\), \(y\) and \(r\). The general definitions for the trigonometric ratios are:

\[\begin{array}{cccc} \sin \theta & = \dfrac{y}{r} & \text{ cosec }\theta & = \dfrac{r}{y} \\ \\ \cos \theta & = \dfrac{x}{r} & \sec\theta & = \dfrac{r}{x} \\ \\ \tan \theta & = \dfrac{y}{x} & \cot \theta & = \dfrac{x}{y} \end{array}\]

The CAST diagram

The Cartesian plane is divided into \(\text{4}\) quadrants in an anti-clockwise direction as shown in the diagram below. Notice that \(r\) is always positive but the values of \(x\) and \(y\) change depending on the position of the point in the Cartesian plane. As a result, the trigonometric ratios can be positive or negative. The letters C, A, S and T indicate which of the ratios are positive in each quadrant:

292a9b18391139e3f9470b39ef68703f.png

This diagram is known as the CAST diagram.

We note the following using the general definitions of the trigonometric ratios:

  • Quadrant I

    Both the \(x\) and \(y\) values are positive so all ratios are positive in this quadrant.

  • Quadrant II

    The \(y\) values are positive therefore \(\sin\) and cosec are positive in this quadrant (recall that \(\sin\) and cosec are defined in terms of \(y\) and \(r\)).

  • Quadrant III

    Both the \(x\) and the \(y\) values are negative therefore \(\tan\) and \(\cot\) are positive in this quadrant (recall that \(\tan\) and \(\cot\) are defined in terms of \(x\) and \(y\)).

  • Quadrant IV

    The \(x\) values are positive therefore \(\cos\) and \(\sec\) are positive in this quadrant (recall that \(\cos\) and \(\sec\) are defined in terms of \(x\) and \(r\)).

Important:

The hypotenuse, \(r\), is a length, and is therefore always positive.

The following video provides a summary of the trigonometric ratios in the Cartesian plane.

Special angles in the Cartesian plane

When working in the Cartesian plane we include two other special angles in right-angled triangles: \(\text{0}\)° and \(\text{90}\)°.

Notice that when \(\theta = 0°\) the length of the opposite side is equal to \(\text{0}\) and the length of the adjacent side is equal to the length of the hypotenuse. Therefore:

\begin{align*} \sin 0° & = \cfrac{\text{opposite}}{\text{hypotenuse}} = \cfrac{0}{\text{hypotenuse}} = 0 \\ \cos 0° & = \cfrac{\text{adjacent}}{\text{hypotenuse}} = \cfrac{\text{hypotenuse}}{\text{hypotenuse}} = 1 \\ \tan 0° & = \cfrac{\text{opposite}}{\text{adjacent}}=\cfrac{0}{\text{adjacent}} = 0 \end{align*}

When \(\theta =90°\) the length of the adjacent side is equal to \(\text{0}\) and the length of the opposite side is equal to the length of the hypotenuse. Therefore:

\begin{align*} \sin 90° & = \cfrac{\text{opposite}}{\text{hypotenuse}} = \cfrac{\text{hypotenuse}}{\text{hypotenuse}} = 1 \\ \cos 90° & = \cfrac{\text{adjacent}}{\text{hypotenuse}} = \cfrac{0}{\text{hypotenuse}} = 0 \\ \tan 90° & = \cfrac{\text{opposite}}{\text{adjacent}} = \cfrac{\text{opposite}}{0} = \text{undefined} \end{align*}

Now we can extend our knowledge of special angles.

\(\theta\)

\(\text{0}\)°

\(\text{30}\)°

\(\text{45}\)°

\(\text{60}\)°

\(\text{90}\)°

\(\sin\theta\)

0

\(\dfrac{1}{2}\)

\(\dfrac{1}{\sqrt{2}}\)

\(\dfrac{\sqrt{3}}{2}\)

1

\(\cos\theta\)

1

\(\dfrac{\sqrt{3}}{2}\)

\(\dfrac{1}{\sqrt{2}}\)

\(\dfrac{1}{2}\)

0

\(\tan\theta\)

0

\(\dfrac{1}{\sqrt{3}}\)

1

\(\sqrt{3}\)

undefined

Example

Question

\(P(-3;4)\) is a point on the Cartesian plane with origin \(O\). \(\theta\) is the angle between \(OP\) and the positive \(x\)-axis. Without using a calculator, determine the value of:

  1. \(\cos\theta\)

  2. \(3\tan\theta\)

  3. \(\cfrac{1}{2}\text{cosec }\theta\)

Sketch point \(P\) in the Cartesian plane and label the angle \(\theta\)

c17e444faaf40315d0499f105c9e051e.png

Use the theorem of Pythagoras to calculate \(r\)

\begin{align*} {r}^{2} & = {x}^{2} + {y}^{2} \\ & = {(-3)}^{2} + {(4)}^{2} \\ & = 25 \\ \therefore r & = 5 \end{align*}

Note \(r\) is positive as it is the radius of the circle.

Substitute values for \(x\), \(y\) and \(r\) into the required ratios

We note that \(x = -3\), \(y = 4\) and \(r = 5\).

  1. \(\cos\theta = \dfrac{x}{r} = \dfrac{-3}{5}\)

  2. \(3\tan\theta = 3(\dfrac{y}{x}) = 3(\dfrac{4}{-3}) = -4\)

  3. \(\dfrac{1}{2}\text{cosec }\theta = \dfrac{1}{2}(\dfrac{r}{y}) = \cfrac{1}{2}(\dfrac{5}{4}) = \dfrac{5}{8}\)

Example

Question

\(X\hat{O}K=\theta\) is an angle in the third quadrant where \(X\) is a point on the positive \(x\)-axis and \(K\) is the point \((-5;y)\). \(OK\) is \(\text{13}\) \(\text{units}\).

  1. Determine, without using a calculator, the value of \(y\).

  2. Prove that \({\tan}^{2}\theta + 1 = {\sec}^{2}\theta\) without using a calculator.

Sketch point \(K\) in the Cartesian plane and label the angle \(\theta\)

f5d17ce93a3dc2c76dea94ed52d09cdc.png

Use the theorem of Pythagoras to calculate \(y\)

\begin{align*} {r}^{2} & = {x}^{2} + {y}^{2} \\ {y}^{2} & = {r}^{2} – {x}^{2} \\ & = {(13)}^{2} – {(-5)}^{2} \\ & = 169 – 25 \\ & = 144 \\ y & = ±12 \end{align*}

Given that \(\theta\) lies in the third quadrant, \(y\) must be negative.

\[\therefore y = -12\]

Substitute values for \(x\), \(y\) and \(r\) and simplify

\(x = -5\), \(y = -12\) and \(r = 13\).

LHS

\begin{align*} {\tan}^{2}\theta + 1 & = {\left(\cfrac{y}{x}\right)}^{2} + 1 \\ & = {\left(\cfrac{-12}{-5}\right)}^{2} + 1 \\ & = \left(\cfrac{144}{25}\right) + 1 \\ & = \cfrac{144 + 25}{25} \\ & = \cfrac{169}{25} \end{align*}

RHS

\begin{align*} {\sec}^{2}\theta & = {\left(\cfrac{r}{x}\right)}^{2} \\ & = {\left(\cfrac{13}{-5}\right)}^{2} \\ & = \cfrac{169}{25} \end{align*}

Therefore the LHS = RHS and we have proved that \({\tan}^{2}\theta + 1 = {\sec}^{2}\theta\).

Fact:

Whenever you have to solve trigonometric problems without a calculator, it can be very helpful to make a sketch.

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