MATH 3 Santa Monica College Trigonometry and Precalculus Practice Exam

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NAME:_______________________________________________________ DATE:_________________ MATH 3: FINAL EXAM, FALL 2021. INSTRUCTIONS: Show all your work. You will be graded on your methods as well as on the accuracy of your final answer. HONOR CODE: I, as an honest student, agree to do this exam on my own, without the assistance of anyone else. I know that there is no actual benefit for me if I do not do the work myself, since taking an exam involves my individual integrity. Initials: _______________________ 5 𝑠𝑖𝑛𝜃 = − 7 𝑎𝑛𝑑 𝑠𝑒𝑐𝜃 > 0 . Find the remaining trigonometric function values. 1. 2. From a window 30 feet above the street, the angle of elevation to the building across the street is 50° and the angle of depression to the bottom of the building is 20°. Find the height of the building across the street. Window Street 3. A sector has an area of 24𝜋 𝑖𝑛2 . The central angle of the circle is 60°. What is the arc length of the circle? 4. The tires of a car have radius 15.0 inches, and are turning at the rate of 970 rpm (revolutions per minute). How fast is the car traveling in miles per hour (mph)? 5280 𝑓𝑡 = 1 𝑚𝑖. 5. 𝜋 6 𝜋 3 Find the amplitude, period, phase shift, domain, range and graph of: 𝑦 = 2𝑠𝑖𝑛 ( 𝑥 − ) . 1 6. (a) Graph: 𝑓(𝑥) = 𝑡𝑎𝑛 (6 𝑥) . (b) 𝐷𝑂𝑀(𝑓) = 𝑅𝐴𝑁(𝑓) = 1 𝜋 7. Graph: 𝑦 = 𝑠𝑒𝑐 (5 𝑥 + 4 ) . 8. Verify the identity: 1 + 𝑡𝑎𝑛𝜃𝑡𝑎𝑛(2𝜃) = 𝑠𝑒𝑐(2𝜃) 9. Use a half-angle formula to find the exact value of 𝑠𝑖𝑛67.5° . 10. Find the exact value of: 𝑐𝑜𝑠 [𝑡𝑎𝑛−1 12 5 3 − 𝑠𝑖𝑛−1 (− 5)] 11. Solve for 𝜃 , 𝑤ℎ𝑒𝑟𝑒 0 ≤ 𝜃 < 2𝜋 . (a) 𝑠𝑖𝑛𝜃 + 1 = 2𝑐𝑜𝑠 2 𝜃 (b) √3𝑐𝑜𝑡(2𝜃) − 1 = 0 12. The Vietnam Veterans Memorial in Washington D.C., is V-shaped with equal sides of length 246.75 feet. The angle between these two sides measures 125°12′. Find the distance between the ends of the two sides. 13. Using the triangle in problem 12, find the area of the triangle determined by the two sides.. 14. For the plane curve 𝑥 = 𝑡 + 2 , 𝑦 = 𝑡 2 , 𝑓𝑜𝑟 − 1 ≤ 𝑡 ≤ 1 . (a) Find a rectangular equation of the curve. (b) Graph the curve. 15. Convert the polar coordinates of (3, 5𝜋 ) 6 to rectangular coordinates. √3 3 , ) 2 2 16. Convert the rectangular coordinates of ( to polar coordinates. 17. Convert the polar equation 𝑟(𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃) = 3 to a rectangular equation. 18. Convert the rectangular equation 𝑥 2 + 𝑦 2 = 14𝑦 to a polar equation. 19. Graph the cardioid 𝑟 = 2 + 2𝑐𝑜𝑠𝜃 . 20. Find the exact distance between the points (−3, −3, −1) 𝑎𝑛𝑑 (2,1, −4) . 21. The vector 𝑣⃗ has initial point (−4, −6,2) and terminal point (−2,3, −5) . Write 𝑣⃗ in the form 𝑎𝑖 + 𝑏𝑗 + 𝑐𝑘 , that is find the position vector. 22. Given the vector 𝑣⃗ = 4𝑖 − 𝑗 + 8𝑘 . (a) Find the magnitude of 𝑣⃗ , 𝑡ℎ𝑎𝑡 𝑖𝑠 ‖𝑣⃗‖ . (b) Find the unit vector for 𝑣⃗ . 23. Find the angle 𝜃 between the vectors 𝑢 ⃗⃗ = 2𝑖 + 𝑗 + 2𝑘 𝑎𝑛𝑑 𝑣⃗ = 3𝑗 + 4𝑘 . Round off your answer to one decimal place. 24. Given the vectors 𝑢 ⃗ = 2𝑖 − 𝑗 + 2𝑘 𝑎𝑛𝑑 𝑤 ⃗⃗ = 𝑖 + 2𝑗 − 3𝑘 , find the cross product, that is find 𝑣×𝑤 ⃗⃗ . 25. Referring to the vector found in problem 24, show that 𝑣 × 𝑤 ⃗⃗ is orthogonal to vector 𝑣 also from problem 24.
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Explanation & Answer:
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Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.Here's the final solution, i'll send the remaining 3 problems next.

Trigonometry Exam Solutions
Problem 1
Solution Steps
1) First, since the sine identity is negative, and also the cosine identity is negative, then the
given angle must be located in the 3rd quadrant.
2) Then, the tangent identity should be positive, and thus applying the Pythagorean theorem
of trigonometry should give.
cos(𝜃) = −√1 − 𝑠𝑖𝑛2 (𝜃) = −

2√6
7

3) Using the reciprocal identity of trigonometry, we get the csc and sec as shown.
sec(𝜃) =

1
7√6
=−
cos(𝜃)
12

csc(𝜃) =

1
sin((𝜃))

=−

7
5

4) Finally, using the Pythagorean rule again we get the tangent and cotangent identities as
follows.
tan(𝜃) = √𝑠𝑒𝑐 2 𝜃 − 1 =
cot(𝜃) =

5√6
12

1
2√6
=
tan(𝜃)
5

Problem 2
Solution Steps
1) According to the next figure, we need to find the height of the building across the street.
2) Starting with the lower triangle, the lower height of the building h1 should be equal to the
distance between the window and the ground which is 30 ft.
ℎ1 = 30 𝑓𝑡
3) Using the right triangle trigonometry, we can get the distance (d) between the building
and the window to be.
tan(20°) =

𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 30
=
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝑑

𝑑 = 30 ∙ tan(20°) = 10.9191 𝑓𝑡

4) Then for the higher right triangle, to get the upper height of the building we apply right
triangle trigonometry to get.
𝑡𝑎𝑛(50°) =

ℎ2
ℎ2
=
𝑑
30 ∙ tan(20°)

ℎ2 = 30 ∙ tan(20°) ∙ 𝑡𝑎𝑛(50°)
5) Finally, the height of the building should be.
𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔 = ℎ1 + ℎ2 ≈ 43.012885 𝑓𝑡

Problem 3
Solution Steps
1) First, since the sector area, central angle and radius are related by the following formula.
𝐴=

1 2
𝑟 𝜃
2

2) Then, by substituting with the given values we get.
24𝜋 =

1 2 60°𝜋 1 2 𝜋
∙𝑟 ∙
= ∙𝑟 ∙
2
180° 2
3

3) Solving for r should give.
𝑟 = 12 𝑖𝑛

Problem 4
Solution Steps
1) First, since the number of revolutions per minute given represents the angular velocity
divided by 2 pi.
2) Then, the car tire covers 970 complete cycles in 1-minute represented by the
circumference of the circle.
3) The distance covered by 1-complete cycle should be.
𝐶 = 2𝜋𝑟 = 30𝜋 𝑖𝑛
4) Multiplying this to the given rpm gives the total distance covered by the tire in 1-minute
to be.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑖𝑛𝑐ℎ𝑒𝑠 𝑝𝑒𝑟 𝑚𝑖𝑛𝑢𝑡𝑒 = 970 × 30𝜋 = 29100𝜋 𝑖𝑛/𝑚𝑖𝑛
5) Converting from inches to miles per hour to get.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑚𝑖𝑙𝑒𝑠 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟 = 86.572 𝑚𝑝ℎ

Problem 5
Solution Steps
1) For the given trigonometric function below.
𝜋
𝜋
𝑦 = 2 sin ( 𝑥 − )
6
3
2) The amplitude should be the constant term multiplied to the function as follows.
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 = 2
3) While the period can be evaluated by solving the following equation to get.
2𝜋 𝜋
=
𝑇
6
𝑃𝑒𝑟𝑖𝑜𝑑 = 𝑇 = 12
4) The phase shift is the amount of phase added to the function at (x=0) to give.
𝑃ℎ𝑎𝑠𝑒 𝑠ℎ𝑖𝑓𝑡 =

𝜋
3

5) Since the given function is a periodic function, then its domain should be.
𝐷𝑜𝑚𝑎𝑖𝑛 = (−∞ , ∞)
6) The range of the function will be the set of values between Max. value and Min. value.
𝑅𝑎𝑛𝑔𝑒 = [−2,2]

Problem 6
Solution Steps
1) For the given function, the domain will depend on its period, so by solving the following
equation the period is.
𝜋 1
=
𝑇 6

𝑃𝑒𝑟𝑖𝑜𝑑 = 𝑇 = 6𝜋
2) Then, the domain of this function should be.
𝐷𝑜𝑚𝑎𝑖𝑛 = (−3𝜋 , 3𝜋) 𝑜𝑟 (6𝜋𝑛 + 3𝜋 , 6𝜋𝑛 + 6𝜋)
3) While the range of this function should be all the set of real numbers.
𝑅𝑎𝑛𝑔𝑒 = (−∞ , ∞)

Problem 7
Solution Steps
1) The given function represents the reciprocal of cosine function which is sec.
2) Thus, with the cosine function ranging from {-1} to {1} passing through zero, then the
secant function will have values extending to infinity and negative infinity.
3) To get the period of this function, we solve the following equation.
2𝜋 1
=
𝑇
5
𝑃𝑒𝑟𝑖𝑜𝑑 = ...

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