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EXAM 1 PRACTICE SOLUTIONS 1. Given that the point P = ( 2 , 5 p 21 ) 5 is on the unit circle what are the exact values of all six Cos = -2/5, Sin = sqrt(21)/5, Tan = -sqrt(21)/2, Cot = -2/sqrt(21) p p p 21 • tan = 2 2121 • csc = 52 • sec = 5 2121 • cot = . Sec = -5/2, Csc = 5/sqrt(21) 2 trigonometric functions? 2 5 • sin = p 21 5 • cos = 2. Find the exact value of the following a. sin( 11⇡ ) = sin(3 · 2⇡ 2 d. cos(405 ) = = sin( ⇡2 ) = 1 3 2 b. sin( 300 ) = c. cos( 10⇡ )= 3 ⇡ ) 2 p 1 2 p 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. p p 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 ✓ given that cos ✓ = 3 5 and 3⇡ 2 < ✓ < 2⇡. Solution cos ✓ = 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 ✓ < 0. Hence 4 , tan ✓ 5 • sin ✓ = = 4 , sec ✓ 3 = 53 , csc ✓ = 5 , cot ✓ 4 = 3 . 4 6. Find the exact value of each of the remaining trigonometric functions of ✓ given that tan ✓ = 1 3 and sin ✓ > 0. Solution Since tan ✓ < 0 and sin ✓ > 0 it must be that ✓ 2 ( ⇡2 , ⇡) and if we write tan ✓ = 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 p p p p 10 • sin ✓ = 1010 , • cos ✓ = 3 1010 , • cot ✓ = 3, • csc ✓ = 10, sec ✓ = 3 7. For the function f (x) = 3 4 cos(3x ⇡ ) 4 p 10. Hence +1 a. What is the amplitude? 3/4 b. What is the period? 2⇡ 3 c. What is the phase shift? ⇡ 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 + ⇡) + 1 or 3 cos(3(x + ⇡2 )) + 1 or 3 cos(3(x ⇡ )) 2 +1