Math 95 Final Review PDF

Math 95 Final Review Part One: Exact answers only (no decimals). Rationalize all answers. 5 1. Given 𝑠𝑒𝑐 θ = 2 and 𝑡𝑎𝑛 θ < 0, find 𝑠𝑖𝑛 θ and 𝑐𝑜𝑡 θ. 2. Find 𝑐𝑜𝑠 θ and 𝑡𝑎𝑛 θ if θ is in standard position and (2, − 6) is on the terminal side. 3. Sketch...

Math 95 Final Review Part One: Exact answers only (no decimals). Rationalize all answers. 5 1. Given 𝑠𝑒𝑐 θ = 2 and 𝑡𝑎𝑛 θ < 0, find 𝑠𝑖𝑛 θ and 𝑐𝑜𝑡 θ. 2. Find 𝑐𝑜𝑠 θ and 𝑡𝑎𝑛 θ if θ is in standard position and (2, − 6) is on the terminal side. 3. Sketch a picture with the coordinates of the unit circle on the terminal side clearly labeled assuming the given angle is in standard position. Use the picture to find the exact value: a. 𝑠𝑖𝑛(4π/3) b. 𝑡𝑎𝑛(5π) c. 𝑠𝑒𝑐(3π/4) 4. Graph one complete period of the equation 𝑦 =− 2𝑐𝑜𝑠 4𝑥 − ( π 2 ) + 4. Label all key values on the axes. π 5π 5. Find an equation of the sine function whose period goes from 4 to 4 with maximum value 8 and minimum value 2. 6. Evaluate: a. 𝑐𝑜𝑠 (𝑠𝑖𝑛 ) −1 5π 4 𝑡𝑎𝑛(𝑐𝑜𝑠 (− )) −1 5 b. 7 π 7. Suppose 𝑐𝑠𝑐 θ = 3 and 2 < θ < π. Find the value of 𝑠𝑖𝑛 2θ. 7 3π 𝑡 8. Suppose 𝑠𝑖𝑛 𝑡 =− 8 and 2 < 𝑡 < 2π. Find the value of 𝑐𝑜𝑠 2. 4 12 9. If α and β are second-quadrant angles such that 𝑠𝑖𝑛 α = 5 and 𝑡𝑎𝑛 β =− 5 , find 𝑠𝑖𝑛 (α + β). 10. Find all solutions of 𝑥: 2 a. 1 − 𝑠𝑖𝑛 𝑥 = 𝑐𝑜𝑠 𝑥 1 b. 𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 2𝑥 + 2 = 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 2𝑥 c. 𝑐𝑜𝑠 𝑥 = 𝑠𝑖𝑛 2𝑥 Part Two: Round all answers to one decimal place. 𝑜 𝑜 11. Find the area of ∆𝐺𝐻𝐽 with 𝐺= 80 , 𝐻= 40 , and 𝑗 = 7. 4. 𝑜 12. Solve ∆𝐷𝐸𝐹 with 𝐸= 115 , 𝑑 = 4. 6, and 𝑓 = 7. 3. 𝑜 13. Solve ∆𝐴𝐵𝐶 with 𝐴= 24. 2 , 𝑎 = 6. 3, and 𝑏 = 12. 4.  Answers 21 2 21 1. 𝑠𝑖𝑛 θ =− 5 ; 𝑐𝑜𝑡 θ =− 21 10 2. 𝑐𝑜𝑠 θ = 10 ; 𝑡𝑎𝑛 θ =− 3 3. Draw individual graphs for each angle. a. Coordinates − ( 1 2 ,− 2 3 ); 𝑠𝑖𝑛 4π 3 =− 2 3 0 b. Coordinates (-1,0); 𝑡𝑎𝑛(5π) = −1 = 0 c. Coordinates − ( 2 2 , 2 2 ); 𝑠𝑒𝑐 3π 4 =− 2 4. This is a cosine graph that is reflected (starting at the bottom). Your graph should have the following key values labeled: ( π 8 ,2 ; )( π 4 ,4 ; )( 3π 8 ,6 ; )( π 2 ,4 ;)( 5π 8 ,2 ) 5. The graph has a maximum of 8 and minimum of 2, so there is a vertical shift of 5 with amplitude 3. Also, the period is 5π 4 − π 4 = π. So, 𝑦 = 3 𝑠𝑖𝑛⎡2 𝑥 − ⎣ ( π 4 )⎤⎦ + 5. 6. Evaluate: a. 𝑐𝑜𝑠 (𝑠𝑖𝑛 ) = −1 5π 4 3π 4 𝑡𝑎𝑛(𝑐𝑜𝑠 (− )) =− −1 5 2 6 b. 7 5 4 2 7. 𝑠𝑖𝑛 2θ =− 9 𝑡 8+ 15 8. 𝑐𝑜𝑠 2 =− 4 56 9. 𝑠𝑖𝑛 (α + β) =− 65 10. Solutions of 𝑥: π a. 𝑥 = 0 + 2π𝑘, 𝑥 = π + 2π𝑘, 𝑥 = 2 + 2π𝑘 2π 2π𝑘 4π 2π𝑘 b. 𝑥 = 9 + 3 , 𝑥 = 9 + 3 π 3π π 5π c. 𝑥 = 2 + 2π𝑘, 𝑥 = 2 + 2π𝑘, 𝑥 = 6 + 2π𝑘, 𝑥 = 6 + 2π𝑘 11. Area = 20.0 𝑜 𝑜 12. 𝑒 = 10. 1, 𝐷 = 24. 4 , 𝐹 = 40. 6. Note: answers may vary slightly due to rounding and which angle you solve for first. 𝑜 𝑜 13. Triangle 1: 𝐵 = 53. 8 , 𝐶 = 102. 0 , 𝑐 = 15. 0 𝑜 𝑜 Triangle 2: 𝐵 = 126. 2 , 𝐶 = 29. 6 , 𝑐 = 7. 6