Trigonometric Functions of Any Angle

To define the trigonometric functions of any angle - including angles less than 0º or greater than 360º - we need a more general definition of an angle. We say that an angle is formed by rotating a ray \(\overrightarrow{OA}\) about the endpoint O (called the vertex), so that the ray is in a new position, denoted by the ray \(\overrightarrow{OB}\). The ray \(\overrightarrow{OA}\) is called the initial side of the angle, and \(\overrightarrow{OB}\) is the terminal side of the angle (see Figure 1).

We denote the angle formed by this rotation as \(\angle\)AOB, or simply \(\angle\)O, or even just O. If the rotation is counter-clockwise then we say that the angle is positive, and the angle is negative if the rotation is clockwise (see Figure 2). One full counter-clockwise rotation of \(\overrightarrow{OA}\) back onto itself (called a revolution), so that the terminal side coincides with the initial side, is an angle of 360º; in the clockwise direction this would be −360º. Not rotating \(\overrightarrow{OA}\) constitutes an angle of 0º. More than one full rotation creates an angle greater than 360º. For example, notice that 30º and 390º have the same terminal side in Figure 3, since 30+360 = 390.

We can now define the trigonometric functions of any angle in terms of Cartesian coordinates. Recall that the xy-coordinate plane consists of points denoted by pairs (x, y) of real numbers. The first number, x, is the point’s x coordinate, and the second number, y, is its y coordinate. The x and y coordinates are measured by their positions along the x-axis and y-axis, respectively, which determine the point’s position in the plane. This divides the xy-coordinate plane into four quadrants (denoted by QI, QII, QIII, QIV), based on the signs of x and y (see Figure 4-6).

Now let θ be any angle. We say that θ is in standard position if its initial side is the positive x-axis and its vertex is the origin (0,0). Pick any point (x, y) on the terminal side of θ a distance r > 0 from the origin (see Figure 6). (Note that r = \(\sqrt{x^2 + y^2}\) Why?) We then define the trigonometric functions of θ as follows:

As in the acute case, by the use of similar triangles these definitions are well-defined (i.e. they do not depend on which point (x, y) we choose on the terminal side of θ). Also, notice that |sin θ| ≤ 1 and |cos θ| ≤ 1, since | y| ≤r and |x| ≤ r in the above definitions.

Notice that in the case of an acute angle these definitions are equivalent to our earlier definitions in terms of right triangles: draw a right triangle with angle θ such that x = adjacent side, y = opposite side, and r = hypotenuse. For example, this would give us sin θ = \(\frac{y} {r} = \frac {opposite} {hypotenuse} \) and cos θ = \(\frac{x}{r}\) = \(\frac{adjacent}{hypotenuse}\) , just as before (see Figure 7).

In Figure 8 we see in which quadrants or on which axes the terminal side of an angle 0º ≤ θ < 360º may fall. From Figure 4 and formulas (1.2) and (1.3), we see that we can get negative values for a trigonometric function. For example, sin θ < 0 when y < 0. Figure 9 summarizes the signs (positive or negative) for the trigonometric functions based on the angle’s quadrant:

Example 1

Find the exact values of all six trigonometric functions of 120º.

Solution: We know 120º = 180º −60º. As we realize, we can use the point (−1,\(\sqrt{3}\)) on the terminal side of the angle 120º in QII, since a basic right triangle with a 60º angle has adjacent side of length 1, opposite side of length \(\sqrt{3}\), and hypotenuse of length 2, as in the figure above. Drawing that triangle in QII so that the hypotenuse is on the terminal side of 120º makes r = 2, x = −1, and y = \(\sqrt{3}\). Hence:

$$sin 120^\circ= \frac{y}{r} = \frac{\sqrt{3}}{2}$$	$$cos 120^\circ= \frac{x}{r} = \frac{-1}{2}$$	$$tan 120^\circ= \frac{y}{x} = \frac{\sqrt{3}}{-1}$$
$$csc 120^\circ= \frac{r}{y} = \frac{2}{\sqrt{3}}$$	$$sec 120^\circ= \frac{r}{x} = = \frac{2}{-1}$$	$$cot 120^\circ= \frac{x}{y} = \frac{-1}{\sqrt{3}}$$

The following table summarizes the values of the trigonometric functions of angles between 0º and 360º which are integer multiples of 30º or 45º:

Since 360º represents one full revolution, the trigonometric function values repeat every 360º. For example, sin 360º = sin 0º, cos 390º = cos 30º, tan 540º = tan 180º, sin (−45º) = sin 315º, etc. In general, if two angles differ by an integer multiple of 360º then each trigonometric function will have equal values at both angles. Angles such as these, which have the same initial and terminal sides, are called coterminal.

As you may have previously studied, the values of trigonometric functions of an angle θ larger than 90º are found by using a certain acute angle as part of a right triangle. That acute angle has a special name: if θ is a nonacute angle then we say that the reference angle for θ is the acute angle formed by the terminal side of θ and either the positive or negative x-axis. For example, 60º is the reference angle for the nonacute angle θ = 120º; and in a similar way, 45º is the reference angle for θ = 225º; and 30º is the reference angle for θ = 330º.

Example 6

Suppose that cos θ = \(\frac{−4}{5}\) . Find the exact values of sin θ and tan θ.

Solution: Draw a right triangle and interpret cos θ as the ratio adjacent hypotenuse of two of its sides. Since cos θ = −4 5 , we can use 4 as the length of the adjacent side and 5 as the length of the hypotenuse. By the Pythagorean Theorem, the length of the opposite side must then be 3. Since cos θ is negative, we know from the Figure 9 that θ must be in either QII or QIII. Thus, we have two possibilities, as shown in Figure 1.4.8 below:

When θ is in QII, we see from Figure 1.4.8(a) that the point (−4,3) is on the terminal side of θ, and so we have x = −4, y = 3, and r = 5. Thus, sin θ = \(\frac{y}{r}\) =\(\frac{3}{5}\) and tan θ = \(\frac{y}{x}\) = \(\frac{3}{−4}\) . When θ is in QIII, we see from Figure 1.4.8(b) that the point (−4,−3) is on the terminal side of θ, and so we have x = −4, y = −3, and r = 5. Thus, sin θ = \(\frac{y}{r}\) = \(\frac{−3}{5}\) and tan θ = \(\frac{y}{x}\) = \(\frac{−3}{−4}\) = \(\frac{3}{4}\). Thus, either sin θ = \(\frac{3}{5}\) and tan θ = −\(\frac{3}{4}\) or sin θ = \(\frac{−3}{5}\) and tan θ = \(\frac{3}{4}\).

Since reciprocals have the same sign, csc θ and sin θ have the same sign, sec θ and cos θ have the same sign, and cot θ and tan θ have the same sign. So it suffices to remember the signs of sin θ, cos θ, and tan θ: