EXAMPLE 1

1. √ √ √ √ 3 3 3 3 a. cos–1 2 2 2 2 When0θπor0°180°,the angle whose cosine is ≤ ≤ ≤ θ ≤ a. π 30° cos–1 = θ = 6 cos–1 = θ = EXAMPLE 1 Evaluate inverse trigonometric functions Evaluate the expression in both radians and degrees. SOLUTION

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

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

4. Solve the equationsinθ = – where180° < θ < 270°. ≤ ≤ interval –90° θ 90°, the angle whose Use a calculator to determine that in the 5 5 5 8 8 8 – sine is – is sin–1– 38.7°. This angle is in Quadrant IV, as shown. EXAMPLE 2 Solve a trigonometric equation SOLUTION STEP 1

5. 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: 180° + 38.7° = 218.7° θ 5 –  = sin 218.7° – 0.625 8 EXAMPLE 2 Solve a trigonometric equation STEP 2 Use a calculator to check the answer. CHECK :

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

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

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

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