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Math 111 - Trigonometry
Practice Exam 2
Warning: This study guide is not all inclusive- there may be material on the test which
is covered in the book but not here. This is simply meant to be a supplemental study aid to the homework and class notes
- Graph two periods of y = −2 + 4 cos(π(x − 2)). Find the amplitude, period, and average value. Label the axes in a way to reflect the important characteristics of the graph (i.e. when the function reaches it maximum, minimum, and average value).
-5 -4 -3 -2 -1 0 1 2 3 4 5
1
2
- Below is a data table for a function involving a trigonometric function. x -1 0 1 2 3 4 y 3 7 11 7 3 7
Determine a possible formula for the trigonometric function. Solution: Here are two possibilities. There are others: y = 7 + 4 sin(π 2 x) y = 7 + 4 cos(π 2 (x − 1))
- Consider a function with the below graph:
-5 -4 -3 -2 -1 0 1 2 3 4 5
1
2
3
(a) Find a formula for this function involving sine. Solution: y = 1 + 2 sin
(π 2 (x^ −^ 1)
(b) Find a formula for this function involving cosine. Solution: y = 1 − 2 cos
(π 2 x
- Find the following exact values (a) The period of the function y = tan(4(x − 3)) Solution: π/ 4 (b) The horizontal shift of the function f (x) = 1 − 2 sin(3(x − 4)) Solution: 4 units to the right (c) The average value of the function g(x) = 2 − 2 cos(π(x + π)) Solution: 2 (d) tan(cos−^1 (− 135 )) Solution: -12/ (e) cos(arctan(√3) + sin−^1 (^13 )) (Hint: Use the Cosine of a Sum identity)
Solution: Impossible (domain of arcsin is [− 1 , 1]).
- Use the below graph of y = sin x to approximately find arcsin 1, and mark this point on the appropriate axis.
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1.
-0.
1
Find the domain and range of the following functions. (a) y = sin−^1 (z + 1) Solution: Domain = [− 2 , 0], Range = [−π/ 2 , π/2] (b) y = 3 arccos(w 2 ) Solution: Domain = [− 2 , 2], Range = [0, 3 π] (c) y = tan−^1 (z) + 1 Solution: Domain = (−∞, ∞), Range = (−π/2 + 1, π/2 + 1)
Find an expression for sin(tan−^1 (wz )) that does not involve trigonometric functions. Solution: (^) w √w (^2) + z 2
Find all solutions to cos x = 12 (i.e. find all x values that make this statement true). Which one is cos−1 1 2?
Solution: x =
{ (^) π 53 π^ + 2πn 3 + 2πn The only solution that is in the interval [0, π] (the range of arccos x) is x = π/3, so cos−^1 (1/2) = π/3.
- Consider the following two statements: (a) x = sin−^1 y always means y = sin x (b) y = sin x always means x = sin−^1 y Is statement (a) true or false? Is statement (b) true or false? Explain.
Solution: Recall the definition of inverse sine is that x = sin−^1 y means y = sin x AND x ∈ [−π/ 2 , π/2]. So statement a) is true; the definition tells us that if x = sin−^1 y, then it must be true that y = sin x. However statement b) is false. As an example to see when this is not true, consider x = π. Then y = sin π = 0, but sin−^1 y = sin−^1 0 = 0, and 0 6 = π = x. In fact any number x that is outside the restricted domain [−π/ 2 , π/2] of y = sin x that we used to define inverse sine will show that statement b) is false, exactly because these x cannot satisfy the second part of the definition of inverse sine.
- Is it true that if x = tan 5, then 5 = arctan x?
Solution: The range of arctan x is (−π/ 2 , π/2), and 5 is not in this interval, so the statement cannot possible be true. In general, arctan(tan x) = x only if x ∈ (−π/ 2 , π/2).
- For what values of x will sin(arcsin x) = x? (a) all x (b) [− 1 , 1] (c) [−π 2 , π 2 ] (d) [0, π] (e) none of the above Solution: This statement is true for all x in the domain of arcsin x, which is [− 1 , 1].
