Trigonometry I -- Answers -- Trigonometry I Diploma Practice Exam - ANSWERS 1

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1 Trigonometry I -- Answers -- Trigonometry I Diploma Practice Exam - ANSWERS

2 Formulas These are the formulas for Trig I you will be given on your diploma. a rθ sinθ cosθ tan θ cotθ cosθ sinθ sec θ cscθ cosθ sinθ Answer Sheet. A NR ). 9. B 9. D. C NR ).8. D. C NR ).8. C. C. D. C. B. D. A 4. B. C. D 5. A 4. A 4. D 6. A NR 4) NR 5) C 5. D 5. B 8. C 6. C 6. B 9. B 7. B 7. C. D 8. B 8. D Copyright Trigonometry Barry I Diploma Mabillard, Practice Exam 6 - ANSWERS

3 . Draw the following graphs in your graphing calculator: f g ( θ ) y sinθ ( θ) y sin θ The transformations applied to the original graph are a horizontal stretch by a factor of, and a shift down by two units. As can be seen on the graph, the new range for ( ) The answer is A. g θ is y.. First state all the information you know: The length of arc is 895 km. The angle is 55. However, we never use degrees in the arc length formula, so this π must be converted to radians rad 8 Now use the arc length formula a rθ to solve for the radius. a rθ a 895 r km θ.9599 To find the height of the satellite above the surface of the planet, take the value you just found (height of satellite from the centre of the planet) and subtract the radius of the planet km 5 km km Trigonometry I Diploma Practice Exam - ANSWERS

4 NR #) When a question gives a point, such as,, these values can be put in for x and y. f θ. In the context of this question, x and y are called θ and ( ) π f( θ) acos θ 4 4 π a cos 4 Insert the given point. 4 π π π a cos 4 -, or thinking in degrees, a cos Bring the -4 over to the other side 4 a Since cos 4 4 a a.8 The value of a, to the nearest hundredth, is.8. Look at the graph on the right. Notice that the midline has been shifted three units down, and the amplitude is k. That means: Minimum - - k Maximum - + k The range of the graph is k f θ + k ( ) Trigonometry I Diploma Practice Exam - ANSWERS 4

5 π 4. Draw the graph of y cos θ + This graph is identical to the graph of sinθ, reflected in the x-axis. The answer is B. 5. At the y-intercept, θ. π f ( θ) cos kθ + b π cos k() + b cos b ( ) b since cos b The answer is A. 6. By inspection, the midline is at -7, and the amplitude is 4. To determine the correct answer, it is necessary to check each of the possible answers to see which one actually matches the graph provided. The only equation which is a correct match is f θ 4sin θ 7 ( ) ( ) The answer is A. Trigonometry I Diploma Practice Exam - ANSWERS 5

6 7. a and d will be the same. The b-value will be the same. (see below for proof) 6 6 Degree Mode: b P 6 Radian Mode: b P π Thus, the only value which will change is the c-value, which changes from to At point A, the y-value of both graphs is equal to m. Therefore, f ( k) + g( k) m+ m m 9. First determine the quadrant the angle is found in. cscθ < Based on the diagrams to the right, the angle is in Quadrant IV. Now draw in the triangle and use Pythagoras to solve for the hypotenuse. From the triangle, 4 sinθ 5 The answer is A. Trigonometry I Diploma Practice Exam - ANSWERS 6

7 . First use the information from Based on the unit circle, A < θ < cos A, 9 to solve for the angle A. Then use B 6 + A to determine the value of B. B 6 + A Finally, evaluate secb sec B cos B cos9 undefined NR #) The length of one complete period is units. The graph starts at π, ends at 7π, so the length is 6π b P 6π The answer is. NR #)First determine the quadrant the angle can be found in. Then use Pythagoras to complete the triangle. 4 sinθ 5 cosθ 5 sin θ cos θ The answer is.8 Trigonometry I Diploma Practice Exam - ANSWERS 7

8 . The correct answer is C, since the new graph has a phase shift of c units to the right. This will move all the θ - intercepts by the same amount. (Remember: In a trig graph, the x-intercepts are called θ - intercepts). The only graph which has no effect on the y-intercept is y cosθ, since the horizontal stretch will not transform points on the y-axis. The answer is B.. amplitude: max min b-value: The period is, so b P π c-value: For cosine, it s d-value: max + min π The equation is y 4cos θ First convert to degrees to make the calculation simpler: The reference angle is 6-4 6π 8 9 π Now re-convert to radians: The answer is A. 4 π 8 9 NR #4) Use the arc length formula to determine the length of the curved portion of the sidewalk. a rθ π a.69 (Don t forget to convert to radians! 97.69) 8 a 5 m Add up all the lengths: m + 5 m + m m The answer is. Trigonometry I Diploma Practice Exam - ANSWERS 8

9 5. π < θ < tells you that the angle is in quadrant IV Use Pythagoras to determine the remaining side of the triangle a + b c 4 + b 5 b 5 6 b 9 b 4 Thus, cotθ 6. The amplitude of each graph being quadrupled means all the y-values are being multiplied by 4., 8, 5π, 8 5π, 4 8,, 4 8, 8, 5 π, 8 7. The values can be represented in a triangle as follows. Solve for the hypotenuse using Pythagoras. a + b c k + k ( 6 ) (8 ) 6k + 64k c k c k c c 6k sinθ k 5 8k 4 cosθ k 5 6k tanθ 8k 4 The answer is B. Trigonometry I Diploma Practice Exam - ANSWERS 9

10 8. 6π is equivalent to 6 π on the unit circle since they are co-terminal angles. Evaluate the value of tan using the unit circle. 6 tan 6 sin 6 cos 6 Now rationalize the denominator. The answer is B. 9. Draw the graph of wt () cos t sin( t ) + 8using technology in radian mode. Using the minimum/maximum features of the calculator, the lowest point of a cycle occurs at 6.7 cm, and the highest point at 9. cm. The answer is B. Trigonometry I Diploma Practice Exam - ANSWERS

11 . Draw in the asymptotes as shown in the diagram to the right. The first asymptote occurs at π, and every other asymptote is a multiple of π units away.. To answer this question, determine which of the angles does not reduce to 5. A B π C π D rad Only 6π will give an angle that isn t or 5. a-value 5 6m The period is 4 s. π b-value: Period 4 c-value: None d-value: m π The equation is ht () 6cos t+ 9. Use a negative since the cosine graph starts at the bottom of the ride instead of the top. Trigonometry I Diploma Practice Exam - ANSWERS

12 5. Position B, which is 5 away from the starting position, is.75 6 revolutions Multiply this by the period for one complete revolution to find the time. 4 s.75 5 s. 4. Graph the following: π y 6cos t+ 9 y Determine the time a rider spends over m for one cycle, then multiply by to get the complete time for the entire ride. The total time is.7 seconds. 5. When the Ferris Wheel rotates counter-clockwise, there will be no change to the shape of the graph, since the height with respect to time will be identical. Look for the answer which will result in exactly the same graph. The answer which gives the same graph is y f( t 4), since moving the cosine shape exactly one full period to the right will have everything re-align. The answer is B. 6. When the ride speeds up, the only thing to be affected is the period. Since the period is related to the b-value, the b-value will change. The answer is B. Trigonometry I Diploma Practice Exam - ANSWERS

13 7. The graph is f ( θ ) sin4θ is shown to the right. There are θ - intercepts between θ < π (Note that you don t include the one at π) So, there are asymptotes. 8. a-value 6 6 cm The period is seconds. b-value: π Period c-value: None d-value: cm The equation is ht ( ) cosπt+. 9. First isolate sinθ sinθ sinθ π Solving from the unit circle, θ, Each solution repeats for every co-terminal angle, so the general solutions are π θ ± nπ, ± nπ Trigonometry I Diploma Practice Exam - ANSWERS

14 . From the equation, determine the min/max positions, and the length of a cycle. y.sin ( t- 65) + 6. b Period Minimum: Maximum: Period A window setting that allows us to see all these points is x: [, 6, ], y: [-5,, 5] π θ sin θ π sin θ. First factor the out of the brackets: g ( ) [ ] This tells you there is a horizontal stretch by, then a phase shift π units right.. The graph of f ( θ ) cot 4 θ is shown to the right. The first asymptote occurs along the y-axis, and every other asymptote is a multiple of 45 away. nπ The domain is x R, x ± 4 The answer is A. Trigonometry I Diploma Practice Exam - ANSWERS 4

15 Written Response # Equation of Sunrise a-value The period is 65 days. b-value: Period 65 c-value: days left, since the maximum point is on Dec. d-value: T( x).85cos ( x+ ) Equation of Sunset a-value The period is 65 days. b-value: Period 65 c-value: 7 days right, since the maximum point is on June. d-value: T( x).875cos ( x 7) Use a negative since the cosine graph is upside-down The following transformations are required to transform f ( x ) cos x into the graph representing the sunset time in Yellowknife: Vertical stretch by a factor of.875 Reflection in the x-axis. Horizontal stretch by a factor of 65 (Remember to use the reciprocal for a horizontal stretch) Translation of 7 units right, and units up. Graph y.85cos ( x+ ) y 4 Determine the intersection points, which occur on days and 5. Yellowknife experiences a sunrise earlier than 4 AM for 5 days each year. Feb. 5 is the 46 th day of the year Graph y.85cos ( x+ ) nd Trace Value: x 46. Sunrise 8.54 Graph y.875cos ( x 7) nd Trace Value: x 46. Sunset 6.69 Length of daylight hours. (Can also be written as 8 hours, 9 minutes.) Trigonometry I Diploma Practice Exam - ANSWERS 5

16 Written Response # If it takes second for the tire to go around twice, that means the time for one spin is.5 seconds. The period is.5 s. Parameter Value a 4 b.57 c d 4 ht ( ) 4sin.57t+ 4 Contact between the wheel and ground occurs when the height, ht ( ), is zero. Solve the equation: 4sin.57t + 4 by graphing to obtain the t-intercepts. The first contact occurs at.5 seconds, and every future contact occurs one period later. The general formula is: Time of contact.5 +.5n The fifth contact will be.5 +.5(4).5 s a-value: 9 instead of 4 Period: Still.5 seconds b-value: No change, since no change in period c-value: No change d-value: Midline is at 9 instead of 4. Trigonometry I Diploma Practice Exam - ANSWERS 6

17 Written Response # a - value b - value Phase Shift Vertical Displacement Period 4π Domain x ± nπ Range y, y x-intercepts None y-intercepts None Asymptotes (general equation) x ± nπ The equation of the graph is g ( θ ) sin θ The asymptotes in f ( θ ) occur at the θ - intercepts of g ( θ ) π π f csc π csc 6 5π csc 5π sin Now rationalize the denominator: Trigonometry I Diploma Practice Exam - ANSWERS 7