Pre-Calc Exam Notes 32

# Pre-Calc Exam Notes 32 - 32 Chapter 1 • Right Triangle...

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Unformatted text preview: 32 Chapter 1 • Right Triangle Trigonometry §1.5 1.5 Rotations and Reﬂections of Angles Now that we know how to deal with angles of any measure, we will take a look at how certain geometric operations can help simplify the use of trigonometric functions of any angle, and how some basic relations between those functions can be made. The two operations on which we will concentrate in this section are rotation and reﬂection. To rotate an angle means to rotate its terminal side around the origin when the angle is in standard position. For example, suppose we rotate an angle θ around the origin by 90◦ in the counterclockwise direction. In Figure 1.5.1 we see an angle θ in QI which is rotated by 90◦ , resulting in the angle θ + 90◦ in QII. Notice that the complement of θ in the right triangle in QI is the same as the supplement of the angle θ + 90◦ in QII, since the sum of θ , its complement, and 90◦ equals 180◦ . This forces the other angle of the right triangle in QII to be θ . y (− y , x ) θ x r ( x , y) r θ + 90◦ y 90◦ θ y Figure 1.5.1 x x Rotation of an angle θ by 90◦ Thus, the right triangle in QI is similar to the right triangle in QII, since the triangles have the same angles. The rotation of θ by 90◦ does not change the length r of its terminal side, so the hypotenuses of the similar right triangles are equal, and hence by similarity the remaining corresponding sides are also equal. Using Figure 1.5.1 to match up those corresponding sides shows that the point (− y, x) is on the terminal side of θ + 90◦ when ( x, y) is on the terminal side of θ . Hence, by deﬁnition, sin (θ + 90◦ ) = x −y x = cos θ , cos (θ + 90◦ ) = = − sin θ , tan (θ + 90◦ ) = = − cot θ . r r −y Though we showed this for θ in QI, it is easy (see Exercise 4) to use similar arguments for the other quadrants. In general, the following relations hold for all angles θ : sin (θ + 90◦ ) = cos θ (1.4) cos (θ + 90◦ ) = − sin θ (1.5) tan (θ + 90◦ ) = − cot θ (1.6) ...
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