1. EXAMPLE 1 Evaluate inverse trigonometric functions Evaluate the expression in both radians and degrees. a.cos –1 3 2 √ SOLUTION a. When 0 θ π or 0° 180°, the angle whose cosine is ≤≤≤θ≤ 3 2 √ cos –1 3 2 √ θ = π 6 = 3 2 √ θ = = 30°

2. EXAMPLE 1 Evaluate inverse trigonometric functions Evaluate the expression in both radians and degrees. b.sin –1 2 SOLUTION sin –1 b. There is no angle whose sine is 2. So, is undefined. 2

3. EXAMPLE 1 Evaluate inverse trigonometric functions Evaluate the expression in both radians and degrees. 3 ( – ) c.tan –1 √ SOLUTION c. When – < θ <, or – 90° < θ < 90°, the angle whose tangent is – is: π 2 π 2 √3 ( – ) tan –1 3√θ = π 3 – = ( – ) tan –1 3√θ = –60° =

4. EXAMPLE 2 Solve a trigonometric equation Solve the equation sin θ = – where 180° < θ < 270°. 5 8 SOLUTION STEP 1 sine is – is sin –1 – 38.7°. This 5 8 5 8 – Use a calculator to determine that in the interval –90° θ 90°, the angle whose ≤≤ angle is in Quadrant IV, as shown.

5. EXAMPLE 2 Solve a trigonometric equation STEP 2 Find the angle in Quadrant III (where 180° < θ < 270° ) that has the same sine value as the angle in Step 1. The angle is: θ 180° + 38.7° = 218.7° CHECK : Use a calculator to check the answer. 5 8 sin 218.7°– 0.625 = – 

6. GUIDED PRACTICE for Examples 1 and 2 Evaluate the expression in both radians and degrees. 1.sin –1 2 2 √ ANSWER π 4, 45° 2.cos –1 1 2 ANSWER π 3, 60° 3.tan –1 (–1) ANSWER π 4, –45° –

7. GUIDED PRACTICE for Examples 1 and 2 Evaluate the expression in both radians and degrees. 4.sin –1 (– ) 1 2 π 6, –30° – ANSWER

8. GUIDED PRACTICE for Examples 1 and 2 Solve the equation for 270° < θ < 360°5. cos θ = 0.4; ANSWERabout 293.6° 180° < θ < 270°6. tan θ = 2.1; ANSWERabout 244.5° 270° < θ < 360°7. sin θ = –0.23; ANSWERabout 346.7°

9. GUIDED PRACTICE for Examples 1 and 2 Solve the equation for 180° < θ < 270°8. tan θ = 4.7; ANSWERabout 258.0° 90° < θ < 180°9. sin θ = 0.62; ANSWERabout 141.7° 180° < θ < 270°10. cos θ = –0.39; ANSWERabout 247.0°