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EXAM 1 PRACTICE SOLUTIONS 1. Given that the point P = (− 52 , √ 21 ) 5 is on the unit circle what are the exact values of all six trigonometric functions? • sin = − 52 • cos = √ 21 5 √ • tan = − 2 2121 • csc = − 52 • sec = √ 5 21 21 √ • cot = − 21 . 2 2. Find the exact value of the following ) = sin(3 · 2π − π2 ) = − sin( π2 ) = −1 a. sin( 11π 2 √ b. sin(−300◦ ) = 3 2 c. cos( 10π ) = − 12 3 d. cos(405◦ ) = √ 2 2 The following illustration is the labeling we used for the following two questions 3. Solve the right triangle if a = 6, b = 4. √ √ Solution c = 52 = 2 13, A = 56.3◦ , B = 33.7◦ . 4. Solve the right triangle if B = 25◦ , c = 13. Solution A = 65◦ , b = 5.49, a = 11.78 5. Find the exact value of each of the remaining trigonometric functions of θ given that cos θ = 3 5 and 3π 2 Solution cos θ = < θ < 2π. x hyp = 3 5 implying that x = 3 and hyp = 5 and so by the Pythagorean theorem y = 4. Since the angle is in the fourth quadrant sin θ < 0. Hence • sin θ = − 54 , tan θ = − 43 , sec θ = 53 , csc θ = − 54 , cot θ = − 34 . 6. Find the exact value of each of the remaining trigonometric functions of θ given that tan θ = − 31 and sin θ > 0. Solution Since tan θ < 0 and sin θ > 0 it must be that θ ∈ ( π2 , π) and if we write tan θ = 1 y x we have that y = 1 2 EXAM 1 PRACTICE SOLUTIONS and x = −3. By the Pythagorean theorem the hypotenuse of the right triangle is √ √ √ √ • sin θ = 1010 , • cos θ = − 3 1010 , • cot θ = −3, • csc θ = 10, sec θ = − 310 √ 10. Hence 7. For the function f (x) = − 43 cos(3x − π4 ) + 1 a. What is the amplitude? 3/4 b. What is the period? 2π 3 c. What is the phase shift? π 12 d. What is the maximum value f (x) can take? 3/4+1 e. What is the minimum value f (x) can take? -3/4+1 8. For the graph below, write two equations for the graph, one using the function sine and the other using the function cosine. −3 sin(3x + π) + 1 or 3 cos(3(x + π2 )) + 1 or −3 cos(3(x − π2 )) + 1