6.2 Trigonometric functions of angles 6. The Trigonometric Functions 6.2 Trigonometric functions of angles Copyright © Cengage Learning. All rights reserved.

Trigonometric Functions of Angles A triangle is a right triangle if one of its angles is a right angle. If  is any acute angle, we may consider a right triangle having  as one of its angles, as in Figure 1, where the symbol specifies the 90° angle. Six ratios can be obtained using the lengths a, b, and c of the sides of the triangle: Figure 1

Trigonometric Functions of Angles We can show that these ratios depend only on , and not on the size of the triangle, as indicated in Figure 2. Since the two triangles have equal angles, they are similar, and therefore ratios of corresponding sides are proportional. Figure 2

Trigonometric Functions of Angles For example, Thus, for each , the six ratios are uniquely determined and hence are functions of . They are called the trigonometric functions and are designated as the sine, cosine, tangent, cotangent, secant, and cosecant functions, abbreviated sin, cos, tan, cot, sec, and csc, respectively.

Trigonometric Functions of Angles The symbol sin( ), or sin , is used for the ratio b/c, which the sine function associates with . Values of the other five functions are denoted in similar fashion. To summarize, if  is the acute angle of the right triangle in Figure 1, then, by definition,

Trigonometric Functions of Angles The domain of each of the six trigonometric functions is the set of all acute angles. Later in this section we will extend the domains to larger sets of angles, and in the next section, to real numbers. If  is the angle in Figure 1, we refer to the sides of the triangle of lengths a, b, and c as the adjacent side, opposite side, and hypotenuse, respectively.

Trigonometric Functions of Angles We shall use adj, opp, and hyp to denote the lengths of the sides. We may then represent the triangle as in Figure 3. Figure 3

Trigonometric Functions of Angles With this notation, the trigonometric functions may be expressed as follows.

Trigonometric Functions of Angles Note that since sin  and csc  are reciprocals of each other, giving us the two identities in the left-hand column of the next box. Similarly, cos  and sec  are reciprocals of each other, as are tan  and cot .

Example 1 – Finding trigonometric function values If  is an acute angle and cos  = , find the values of the trigonometric functions of . Solution: We begin by sketching a right triangle having an acute angle  with adj = 3 and hyp = 4, as shown in Figure 4, and proceed as follows: Figure 4

Example 1 – Solution 32 + (opp)2 = 42 (opp)2 = 16 – 9 = 7 opp = cont’d Pythagorean theorem isolate (opp)2 take the square root

Example 1 – Solution cont’d Applying the definition of the trigonometric functions of an acute angle of a right triangle, we obtain the following:

Example 2 – Finding trigonometric function values of 60°, 30°, and 45° Find the values of the trigonometric functions that correspond to  : (a)  = 60° (b)  = 30° (c)  = 45° Solution: Consider an equilateral triangle with sides of length 2. The median from one vertex to the opposite side bisects the angle at that vertex, as illustrated by the dashes in Figure 5. Figure 5

Example 2 – Solution cont’d By the Pythagorean theorem, the side opposite 60° in the shaded right triangle has length . Using the formulas for the trigonometric functions of an acute angle of a right triangle, we obtain the values corresponding to 60° and 30° as follows: (a)

Example 2 – Solution cont’d (b)

Example 2 – Solution cont’d (c) To find the values for  = 45°, we may consider an isosceles right triangle whose two equal sides have length 1, as illustrated in Figure 6. By the Pythagorean theorem, the length of the hypotenuse is . Figure 6

Example 2 – Solution cont’d Hence, the values corresponding to 45° are as follows:

Trigonometric Functions of Angles Because of the importance of these special values, it is a good idea either to memorize the table or to learn to find the values quickly by using triangles. Special Values of the Trigonometric Functions

Trigonometric Functions of Angles The next example illustrates a practical use for trigonometric functions of acute angles.

Example 3 – Finding the height of a flagpole A surveyor observes that at a point A, located on level ground a distance 25.0 feet from the base B of a flagpole, the angle between the ground and the top of the pole is 30°. Approximate the height h of the pole to the nearest tenth of a foot. Solution: Referring to Figure 7, we see that we want to relate the opposite side and the adjacent side, h and 25, respectively, to the 30° angle. Figure 7

Example 3 – Solution cont’d This suggests that we use a trigonometric function involving those two sides—namely, tan or cot. It is usually easier to solve the problem if we select the function for which the variable is in the numerator. Hence, we have or, equivalently, h = 25 tan 30. We use the value of tan 30° from Example 2 to find h:

Trigonometric Functions of Angles Let us next list all the fundamental identities. These identities are true for every acute angle , and  may take on various forms. We shall see later that these identities are also true for other angles and for real numbers.

Trigonometric Functions of Angles We can use the fundamental identities to express each trigonometric function in terms of any other trigonometric function. Two illustrations are given in the next example.

Example 4 – Using fundamental identities Let  be an acute angle. (a) Express sin  in terms of cos . (b) Express tan  in terms of sin . Solution: (a) We may proceed as follows: sin2  + cos2  = 1 sin2  = 1 – cos2  sin  = Pythagorean identity isolate sin2  take the square root sin  > 0 for acute angles

Example 4 – Solution (b) If we begin with the fundamental identity cont’d (b) If we begin with the fundamental identity then all that remains is to express cos  in terms of sin . We can do this by solving sin2  + cos2  = 1 for cos , obtaining Hence,

Trigonometric Functions of Angles Fundamental identities are often used to simplify expressions involving trigonometric functions, as illustrated in the next example.

Example 5 – Showing that an equation is an identity Show that the following equation is an identity by transforming the left-hand side into the right-hand side: (sec  + tan  ) (1 – sin  ) = cos  Solution: We begin with the left-hand side and proceed as follows: reciprocal and tangent identities add fractions

Example 5 – Solution cont’d multiply sin2  + cos2  = 1 cancel cos 

Trigonometric Functions of Angles Since many applied problems involve angles that are not acute, it is necessary to extend the definition of the trigonometric functions. We make this extension by using the standard position of an angle  on a rectangular coordinate system.

Trigonometric Functions of Angles If  is acute, we have the situation illustrated in Figure 11, where we have chosen a point P(x, y) on the terminal side of  and where d(O, P) = r = . Figure 11

Trigonometric Functions of Angles Referring to triangle OQP, we have We shall define the six trigonometric functions so that their values agree with those given previously whenever the angle is acute. It is understood that if a zero denominator occurs, then the corresponding function value is undefined.

Trigonometric Functions of Angles Our discussion of domains is summarized in the following table, where n denotes any integer.

Trigonometric Functions of Angles For any point P(x, y) in the preceding definition, | x |  r and | y |  r or, equivalently | x / r |  1 , and | y / r |  1. Thus, | sin  |  1, | cos  |  1, | csc  |  1, and | sec  |  1, for every  in the domains of these functions.

Example 8 – Finding trigonometric function values of an angle in standard position An angle  is in standard position, and its terminal side lies in quadrant III on the line y = 3x. Find the values of the trigonometric functions of . Solution: The graph of y = 3x is sketched in Figure 14, together with the initial and terminal sides of . Figure 14

Example 8 – Solution cont’d Since the terminal side of  is in quadrant III, we begin by choosing a convenient negative value of x, say x = –1. Substituting in y = 3x gives us y = 3(–1) = –3, and hence P(–1, –3) is on the terminal side. Applying the definition of the trigonometric functions of any angle with x = –1, y = –3, and gives us

Trigonometric Functions of Angles The definition of the trigonometric functions of any angle may be applied if  is a quadrantal angle. The procedure is illustrated by the next example.

If  = 3 /2, find the values of the trigonometric functions of . Example 9 – Finding trigonometric function values of a quadrantal angle If  = 3 /2, find the values of the trigonometric functions of . Solution: Note that 3 /2 = 270. If  is placed in standard position, the terminal side of  coincides with the negative y-axis, as shown in Figure 15. Figure 15

Example 9 – Solution cont’d To apply the definition of the trigonometric functions of any angle, we may choose any point P on the terminal side of . For simplicity, we use P(0, –1). In this case, x = 0, y = –1, r = 1, and hence

Example 9 – Solution cont’d The tangent and secant functions are undefined, since the meaningless expressions tan  = (–1)/0 and sec  = 1/0 occur when we substitute in the appropriate formulas.

Trigonometric Functions of Angles Let us determine the signs associated with values of the trigonometric functions. If  is in quadrant II and P(x, y) is a point on the terminal side, then x is negative and y is positive. Hence, sin  = y / r and csc  = r / y are positive, and the other four trigonometric functions, which all involve x, are negative.

Trigonometric Functions of Angles Checking the remaining quadrants in a similar fashion, we obtain the following table. Signs of the Trigonometric Functions

Trigonometric Functions of Angles The diagram in Figure 16 may be useful for remembering quadrants in which trigonometric functions are positive. If a function is not listed (such as cos in quadrant II), then that function is negative. Positive trigonometric functions Figure 16

Example 10 – Finding the quadrant containing an angle Find the quadrant  containing if both cos  > 0 and sin  < 0. Solution: Referring to the table of signs or Figure 16, we see that cos  > 0 (cosine is positive) if  is in quadrant I or IV and t that sin  < 0 (sine is negative) if  is in quadrant III or IV. Hence, for both conditions to be satisfied , must be in quadrant IV.

Example 11 – Finding values of trigonometric functions from prescribed conditions If sin  = and tan  < 0, use fundamental identities to find the values of the other five trigonometric functions. Solution: Since sin  = > 0 (positive) and tan  < 0 (negative),  is in quadrant II. Using the relationship sin2  + cos2  = 1 and the fact that cos  is negative in quadrant II, we have

Example 11 – Solution Next we use the tangent identity to obtain cont’d Next we use the tangent identity to obtain Finally, using the reciprocal identities gives us

Example 12 – Using fundamental identities Rewrite in nonradical form without using absolute values for  <  < 2. Solution: cos2  + sin2  = 1 1 + cot2  = csc2 

Example 12 – Solution cont’d Since  <  < 2, we know that  is in quadrant III or IV. Thus, csc  is negative, and by the definition of absolute value, we have | csc  | = –csc .