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Copyright © Cengage Learning. All rights reserved. Trigonometric Functions: Right Triangle Approach.

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Presentation on theme: "Copyright © Cengage Learning. All rights reserved. Trigonometric Functions: Right Triangle Approach."— Presentation transcript:

1 Copyright © Cengage Learning. All rights reserved. Trigonometric Functions: Right Triangle Approach

2 Copyright © Cengage Learning. All rights reserved. 6.3 Trigonometric Functions of Angles

3 3 Objectives ► Trigonometric Functions of Angles ► Evaluating Trigonometric Functions at Any Angle ► Trigonometric Identities ► Areas of Triangles

4 4 Trigonometric Functions of Angles

5 5 Trigonometric Functions of Angles Let POQ be a right triangle with acute angle  as shown. Place  in standard position. (b)(a)

6 6 Trigonometric Functions of Angles Then P = P (x, y) is a point on the terminal side of . In triangle POQ, the opposite side has length y and the adjacent side has length x. Using the Pythagorean Theorem, we see that the hypotenuse has length r = So The other trigonometric ratios can be found in the same way. These observations allow us to extend the trigonometric ratios to any angle.

7 7 Trigonometric Functions of Angles We define the trigonometric functions of angles as follows.

8 8 Trigonometric Functions of Angles The angles for which the trigonometric functions may be undefined are the angles for which either the x- or y-coordinate of a point on the terminal side of the angle is 0. These are quadrantal angles—angles that are coterminal with the coordinate axes.

9 9 Trigonometric Functions of Angles It is a crucial fact that the values of the trigonometric functions do not depend on the choice of the point P(x, y). This is because if P (x, y ) is any other point on the terminal side, as in Figure 3, then triangles POQ and P OQ are similar. Figure 3

10 10 Evaluating Trigonometric Functions at Any Angles

11 11 Evaluating Trigonometric Functions at Any Angle From the definition we see that the values of the trigonometric functions are all positive if the angle  has its terminal side in Quadrant I. This is because x and y are positive in this quadrant. [Of course, r is always positive, since it is simply the distance from the origin to the point P (x, y).] If the terminal side of  is in Quadrant II, however, then x is negative and y is positive. Thus, in Quadrant II the functions sin  and csc  are positive, and all the other trigonometric functions have negative values.

12 12 Example – Finding Trigonometric Functions of Angles Find (a) cos 135  (b) tan 390 .

13 13 Evaluating Trigonometric Functions at Any Angle

14 14 Evaluating Trigonometric Functions at Any Angle The figure shows that to find a reference angle it’s useful to know the quadrant in which the terminal side of the angle  lies. The reference angle for an angle 

15 15 Example – Finding Reference Angles Find the reference angle for (a) (b)  = 870°

16 16 Evaluating Trigonometric Functions at Any Angle

17 17 Example – Using the Reference Angle to Evaluate Trigonometric Functions Find (a) and (b)

18 18 Trigonometric Identities

19 19 Trigonometric Identities The trigonometric functions of angles are related to each other through several important equations called trigonometric identities. We’ve already encountered the reciprocal identities. These identities continue to hold for any angle , provided that both sides of the equation are defined.

20 20 Trigonometric Identities The Pythagorean identities are a consequence of the Pythagorean Theorem.

21 21 Example – Expressing One Trigonometric Function in Terms of Another (a)Express sin  in terms of cos . (b) Express tan  in terms of sin , where  is in Quadrant II.

22 22 Areas of Triangles

23 23 Areas of Triangles The area of a triangle is  base  height. If  is an acute angle, then the height of the triangle in Figure 16(a) is given by h = b sin . Thus the area is  base  height And we know and, Thus, = sin  (a)

24 24 Areas of Triangles If the angle  is not acute, then we see that the height of the triangle is h = b sin(180° –  ) = b sin 

25 25 Example – Finding the Area of a Triangle Find the area of triangle ABC shown.


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