Principles of Math 12 - Trigonometry II Practice Exam ANSWERS

Principles of Math 12 - Trigonometry II Practice Exam ANSWERS

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Unformatted text preview: Trigonometry II Practice Exam - ANSWERS ANSWERS 1. C NR 2) 28 19. B 28. D 2. D NR 3) 1 20. D NR 7. 4 NR 1) 1.50 11. C 21. D 29. B 3. A 12. C NR 5. 2413 30. C 4. A 13. A 22. A 31. C 5. D 14. C 23. C 32. A 6. A NR 4) 0.7 24. B 33. B 7. D 15. A 25. B 8. A 16. C NR 6. 0.5 9. C 17. B 26. A 10. B 18. A 27. D Copyright © Barry Mabillard, 2006 Principles of Math 12 Trigonometry II Practice Exam - ANSWERS www.math12.com 1 www.math12.com 1. Analyze each of the possible answers to see which one gives back sin 75˚ Look at sin 45˚ cos 30˚ + cos 45˚ sin 30˚ Comparing this expression with the identity sin ( A + B ) = sin A cos B + cos A sin B , it can be seen that the value of A is 45˚, and B is 30˚ Therefore, sin ( A + B ) = sin ( 450 + 300 ) = sin 750 The answer is C. The graph of tan5x can be obtained by horizontally stretching the original graph by a 1 factor of . This will make the arms of the graph much closer to each other. 5 1 Multiply the original general solution by to obtain the new general solution. 5 1⎛ π π nπ ⎞ ⎜ − + nπ ⎟ = − + 5⎝ 4 20 5 ⎠ The answer is D. 2. cos 2 x + sin x . (You will get the same sin 2 x answer if you use the left side, but since the right side is already in terms of sinx and cosx, it’s easier to use.) NR #1 Substitute 2.1 radians into the right side cos 2 2.1 + sin 2.1 = 1.50 sin 2 2.1 *Make sure you have your calculator in radian mode. **Type in cos 2 2.1 as ( cos ( 2.1) ) 2 The answer is 1.50 3. Substitute the height of 14 m for h(t), then graph both sides and find the point of intersection. The x-value will give the time 14 m is first reached. 14 = −6 cos π 20 t +9 t = 16.3 s The answer is A. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 2 www.math12.com 4. Use the identity cot 2 x = csc2 x − 1 cot 2 x + csc x − 4 ( csc 2 x − 1) + csc x − 4 csc2 x + csc x − 5 The answer is A. The equation cos 2 x + cos x + 1 = 3 can be transformed into the equation ⎛x⎞ ⎛ x⎞ cos 2 ⎜ ⎟ + cos ⎜ ⎟ + 1 = 3 using a horizontal stretch by a factor of 3. ⎝3⎠ ⎝3⎠ (Remember that the reciprocal is what goes inside the brackets.) 5. The new general solution is found by multiplying the original general solution by 3. 3 ( 2nπ ) = 6nπ What this means is that the new solutions will be 6π , 12π , 18π ... The answer is D. 6. Start with the relationship tan A = sin A cos A sin A cos A tan A cos A = sin A sin A cos A = tan A tan A = Now plug in the corresponding expressions sin A = m m2 and tan A = 3 n n sin A tan A m cos A = n2 m n3 m n3 cos A = × 2 nm n2 cos A = m cos A = The answer is A. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 3 www.math12.com Use the identities 1 + tan 2 x = sec 2 x and cos 2 x = 1 − sin 2 x 7. 1 + tan 2 x 1 − sin 2 x sec 2 x cos 2 x 1 cos 2 x cos 2 x 1 1 • 2 cos x cos 2 x 1 cos 4 x 1 cos 2 x sec 2 x The answer is D. 8. Use the identity 1 + tan 2 x = sec 2 x substitute 5 for tan 2 x 7 5 = sec 2 x 7 75 + = sec 2 x 77 12 sec 2 x = 7 1+ The answer is A. 9. Use the identity cos ( 2 A ) = cos 2 A − sin 2 A From the given expression cos 2 ( 4π ) − sin 2 ( 4π ) , we can see the A - value is 4π Now replace 4π for A on the left side: cos ( 2 A ) = cos ⎡ 2 ( 4π ) ⎤ = cos ( 8π ) ⎣ ⎦ The answer is C. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 4 www.math12.com 10. When the domain is changed to 0 < x < 2π , the solutions on the extreme ends are no longer included, so there are fewer solutions. The answer is B. NR #2 Graph the equation in your calculator. The interval 0 ≤ x ≤ 14π is too crowded to see clearly, so use the interval 0 < x < 2π instead. If there are 4 solutions in the interval 0 ≤ x ≤ 2π , there are 28 in the interval 0 ≤ x ≤ 14π since that interval is 7 times bigger. (We can do this since there are no points of intersection occurring at the extreme ends.) The answer is 28 NR #3 The expression ( sin x + cos x ) must be foiled to get 2 ( sin x + cos x )( sin x + cos x ) = sin 2 x + 2sin x cos x + cos 2 x . The error occurs in Step I. All other steps would be correct if the error did not occur. The answer is 1 Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 5 www.math12.com 11. m sin x cot x =8 4 csc x tan x m sin x cot x = 32 csc x tan x cos x 1 sin x = 32 sin x sin x cos x 32 m cos x = cos x 32 m= cos 2 x m = 32sec 2 x m sin x The answer is C. 12. Solve on your graphing calculator Don’t include the solution at 2π since it‘s not included in the domain. The answer is C. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 6 www.math12.com 13. cos x is undefined when the denominator becomes zero 1 − 2sin x 1 − 2sin x = 0 2sin x = 1 1 2 π 5π x= , 66 sin x = The general solutions for the asymptotes are π 6 ± 2nπ , 5π ± 2nπ 6 The answer is A. 14. π⎞ ⎛ sec ⎜ x − ⎟ 2⎠ ⎝ 1 = π⎞ ⎛ cos ⎜ x − ⎟ 2⎠ ⎝ 1 = π π cos x cos + sin x sin 2 2 1 = cos x ( 0 ) + sin x (1) 1 sin x = csc x = The answer is C. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 7 www.math12.com NR #4) First simplify the expression using the identity 1 + cot 2 x = csc2 x 1 = 0.43 1 + cot 2 x 1 = 0.43 csc 2 x 1 1 sin 2 x sin 2 x = 0.43 Then solve by graphing over the interval 0 ≤ x < π 2 (Make sure you’re in radian mode) The answer is 0.7 15. sin x 1 + tan x sec x sin x 1 + sin x 1 cos x cos x cos x sin x • + cos x sin x cos x + cos x 2 cos x The answer is A Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 8 www.math12.com 16. First draw out the two triangles, and use Pythagoras to determine the unknown side. Now state what you know: 7 3 sin A = sin B = 8 5 15 4 cos A = cosB = 8 5 Plug all of these into cos ( A − B ) = cos A cos B + sin A sin B cos ( A − B ) = cos A cos B + sin A sin B = 15 4 73 •+• 85 85 4 15 21 + 40 40 4 15 + 21 = 40 = The answer is C. 17. First use the identity 1 + cot 2 x = csc 2 x −5csc 2 x + 12 cot 2 x − 9 = 0 −5 (1 + cot 2 x ) + 12 cot 2 x − 9 = 0 −5 − 5cot 2 x + 12 cot 2 x − 9 = 0 7 cot 2 x − 14 = 0 7 cot 2 x = 14 cot 2 x = 2 The answer is B. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 9 www.math12.com 18. Expand cos ( x − y ) − cos ( x + y ) using the sum & difference identities. cos x cos y + sin x sin y − ( cos x cos y − sin x sin y ) = cos x cos y + sin x sin y − cos x cos y + sin x sin y = 2sin x sin y The answer is A. 19. When cos 2 x − sin x intersects the line y = 1 , we can set the two equations equal to 2 each other, then simplify. 1 2 2 2 cos x − 2sin x = 1 cos 2 x − sin x = 2 cos 2 x − 2sin x − 1 = 0 The answer is B. ⎛θ ⎞ ⎛ 2θ ⎞ ⎛ θ ⎞ ⎛ 2θ ⎞ 20. The expression sin ⎜ ⎟ cos ⎜ ⎟ − cos ⎜ ⎟ sin ⎜ ⎟ is generated using the identity ⎝5⎠ ⎝7⎠ ⎝5⎠ ⎝ 7 ⎠ sin ( A − B ) = sin A cos B − cos A sin B ⎛θ ⎞ ⎛ 2θ ⎞ ⎛ θ ⎞ ⎛ 2θ ⎞ sin ( A − B ) = sin ⎜ ⎟ cos ⎜ ⎟ − cos ⎜ ⎟ sin ⎜ ⎟ ⎝5⎠ ⎝7⎠ ⎝5⎠ ⎝ 7 ⎠ θ 2θ A= B= 5 7 Now replace A & B on the left side: ⎛ θ 2θ ⎞ sin ( A − B ) = sin ⎜ − ⎟ ⎝5 7 ⎠ ⎛ 7θ 10θ ⎞ = sin ⎜ − ⎟ ⎝ 35 35 ⎠ ⎛ 3θ ⎞ = sin ⎜ − ⎟ ⎝ 35 ⎠ The answer is D. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 10 www.math12.com 21. Simplify the identity so it can be typed into the TI-83 csc 2 x =5 sec 2 x 1 sin 2 x = 5 1 cos 2 x 1 • cos 2 x = 5 sin 2 x cos 2 x =5 sin 2 x Graph in your TI-83 to get the following: The first point of intersection occurs at 0.21 radians, and there is a period of π 2 = 1.57 The answer is D. NR #5. Simplify each of the following II 1 1 1 1 • cos x = sec x cos x = • 9 9 cos x 9 2 2 2 2 cot x − csc x = cot x − (1 + cot x ) = cot 2 x − 1 − cot 2 x = −1 III 2cos 2 x + 2sin 2 x = 2 ( cos 2 x + sin 2 x ) = 2 (1) = 2 IV 2 cot x - I 2 cos x 2 cos x 2 cos x = =0 sin x sin x sin x The answer is 2413 Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 11 www.math12.com 22. Use the identity cos 2 x = cos 2 x − sin 2 x cos 2 x = cos 2 x − sin 2 x = (1 − sin 2 x ) − sin 2 x = 1 − 2sin 2 x = 1 − 2m 2 The answer is A. 23. 1 1+ 1 + csc x sin x = sin x sin x sin x 1 + = sin x sin x sin x sin x + 1 = sin x sin x sin x + 1 1 = • sin x sin x sin x + 1 = sin 2 x The answer is C. 24. cos ( x + y ) cos y = cos x cos y − sin x sin y cos y cos 450 cos y − sin 450 sin y cos y 2 2 cos y − sin y 2 2 cos y 2 2 cos y sin y 2 2 − cos y cos y 2 2 tan y − 2 2 2 (1 − tan y ) 2 The answer is B. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 12 www.math12.com ⎛ π⎞ 25. First determine the angle as a degree to make things easier: sec ⎜ − ⎟ = sec ( −150 ) ⎝ 12 ⎠ 0 0 0 Then find two angles which give -15˚: sec ( −15 ) = sec ( 30 − 45 ) sec ( 300 − 450 ) = 1 cos ( 300 − 450 ) 1 cos 30 cos 45 + sin 300 sin 450 1 = 2 3 21 • +• 2 222 1 = 6 2 + 4 4 1 = 6+ 2 4 4 = 6+ 2 = 0 0 Now rationalize the denominator 4 6− 2 = • 6+ 2 6− 2 = = = 4 ( 4 4 ( 6− 2 6+ 2 ( ( 6− )( 2) 6−2 6− 2 ) 6− 2 ) ) 4 = 6− 2 The answer is B. NR 6. Compare cos 2 kx − sin 2 kx with the identity cos ( 2 A ) = cos 2 A − sin 2 A In this identity, the right-side variables will be half the left-side variable. Example: cos ( 8 x ) = cos 2 4 x − sin 2 4 x So, following that logic, cos (1x ) = cos 2 1 1 x − sin 2 x 2 2 The answer is 0.5 Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 13 www.math12.com 26. cos 2 x − sin 2 x − 1 + 2sin x (1 − sin x ) − sin 2 2 x − 1 + 2sin x 2sin x − 2sin 2 x 2sin x (1 − sin x ) The answer is A. 27. Draw in the triangles and use Pythagoras to solve for the unknown side Now state what you know: −6 −2 sin x = sin y = 5 85 cos x = 7 85 sec ( x + y ) = cos y = 21 5 1 cos ( x + y ) 1 1 = cos ( x + y ) cos x cos y − sin x sin y 1 7 21 −6 −2 • − • 5 85 85 5 1 = 7 21 12 − 5 85 5 85 = = 1 7 21 − 12 5 85 = 5 85 7 21 − 12 The answer is D Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 14 www.math12.com 28. csc x − sin x 1 − sin x sin x ⎛ sin 2 x ⎞ 1 −⎜ ⎟ sin x ⎝ sin x ⎠ 1 − sin 2 x sin x cos 2 x sin x ⎛ cos x ⎞ cos x ⎜ ⎟ ⎝ sin x ⎠ cos x cot x Get a common denominator Separate out to finish the question The answer is D. NR 7. Graph the equation The answer is 4. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 15 www.math12.com 29. sin x + tan x cos x + 1 sin x sin x + cos x cos x + 1 sin x cos x sin x + cos x cos x cos x + 1 sin x cos x + sin x cos x cos x + 1 sin x cos x + sin x 1 • cos x cos x + 1 sin x ( cos x + 1) 1 • cos x cos x + 1 sin x = cos x = tan x The answer is B. 30. csc 4 x − 1 ( csc 2 x − 1)( csc 2 x + 1) cot 2 x ( csc2 x + 1) The answer is C. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 16 www.math12.com 31. Graph sin 4 x = − 7π 180 24 11π π 82.50 • = 1800 24 52.50 • π 1 in degree mode 2 0 = The general solutions are 7π nπ 11π nπ , ± ± 24 2 24 2 The answer is C. 32. 1 π 3π sec x = is undefined at , ,... since the cos x 22 denominator becomes zero at these values. 1 π 3π is undefined at , ,... , since these cos 2 x 44 angles will result in the denominator becoming zero. sec 2 x = 33. The reason no answer can be found is because the range of y = cos x exists only between -1 and 1, and there are no intersection points in the equations cos x = 2 and cos x = − 2 . The answer is B. The answer is A. Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 17 www.math12.com Written Response #1 • The equation f ( x ) = 8 − 3sin 2 x should be typed into the graphing calculator using y1 = 8 − 3 ( sin( x) ) 2 Make sure calculator is in radian mode. Window Settings: Domain [-2π , 2π , π Range [0 , 8 , 1] 2 *There are multiple answers for the range, as long as the bottom and top of the graph can be seen clearly. • Draw the graph 6.2 = 8 − 3 ( sin( x) ) and find the intersection points. 2 The period is π = 3.14 radians. The general solutions are 0.89 + nπ , n ∈ I 2.26 + nπ , n ∈ I • • State what you know from the graph: max − min 8 − 5 3 = = = 1.5 a= 2 2 2 2π 2π = =2 b= π P c=0 d= min + max 5 + 8 13 = = = 6.5 2 2 2 The equation is g ( x ) = 1.5cos 2 x + 6.5 Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 7 + sin 2 x = 8 − 3sin 2 x 4 sin 2 x = 1 1 sin 2 x = 4 sin 2 x = 1 4 1 2 Use the unit circle to determine the correct angles. sin x = ± 1 2 π 5π x= , 66 sin x = 18 1 2 7π 11π , x= 66 sin x = − www.math12.com Written Response #2 • cos x 1 − sin x cos = π 1 + sin 6 1 − sin π 6 3 2 1 1− 2 3 2 1 2 3 •2 2 • 1 + sin x cos x cos π π 6 6 3 3 3 • 3 3 3 33 3 3 The graphs are not identical because the graph of y1 = 3 3 3 1 1+ 2 3 2 3 2 3 2 32 • 2 3 3 cos x 3π exists at , but 1 − sin x 2 1 + sin x 3π is undefined at . This point discontinuity is not cos x 2 visible when you use the calculator. the graph of y2 = If you go 2nd Trace Value x= 3π 2 • cos x 1 − sin x cos x 1 + sin x = • Use conjugate 1 − sin x 1 + sin x cos x (1 + sin x ) = (1 − sin x )(1 + sin x ) , notice that you get zero for y1 , and no solution for y2 • cos x 1 + sin x + 1 − sin x cos x cos x ( cos x ) (1 + sin x )(1 − sin x ) = + (1 − sin x )( cos x ) cos x (1 − sin x ) cos 2 x 1 − sin 2 x + cos x (1 − sin x ) cos x (1 − sin x ) = cos 2 x cos 2 x + cos x (1 − sin x ) cos x (1 − sin x ) = 2 cos 2 x cos x (1 − sin x ) = cos x (1 + sin x ) 1 − sin 2 x cos x (1 + sin x ) = cos 2 x 1 + sin x = cos x = 2 cos x 1 − sin x ) ( = Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 19 www.math12.com Written Response #3 • • 1 + cos 2 x sin 2 x 1 + (2 cos 2 x − 1) = 2sin x cos x 2 cos 2 x = 2sin x cos x cos x = sin x = cot x (sin x + cos x) 2 = sin 2 x + 2sin x cos x + cos 2 x = sin 2 x + cos 2 x + 2sin x cos x = 1 + 2sin x cos x • • sin 2 x = sin( x + x) sin( x + x) = sin x cos x + cos x sin x = 2sin x cos x 2 sin x cos x − cos x = 0 cos x(2sin x − 1) = 0 π π 5π 3π x= , , , 626 2 3 sin x = 2sin x • 3sin x − 2sin x = 0 sin x = 0 x = 0, π • ⎛ csc x csc x ⎞ ⎛ 16 ⎞ + 15 ⎜ ⎟ = 15 ⎜ ⎟ 3⎠ ⎝5 ⎝ 15 ⎠ 3csc x + 5csc x = 16 8csc x = 16 csc x = 2 1 2 π 5π x= , 66 sin x = Principles of Math 12 Trigonometry II Practice Exam - ANSWERS 20 www.math12.com ...
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This note was uploaded on 11/05/2011 for the course MATH 24325456 taught by Professor Jack during the Spring '09 term at Adventista de las Antillas.

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