Solutions to Some Additional Practice for the Midterm Exam

Transcription

1 Haberman MTH Solutions to Some Additional Practice for the Midterm Exam. a. Convert into radians. rad. 60 rad. b. Convert radians into degrees. rad. rad. 60 rad Find the arc-length spanned by an angle measuring on a circle of radius 0 feet. In part a of # we found that rad. Thus, s rθ 0 and we see that the arc-length spanned by an angle measuring on a circle of radius 0 feet is feet.. Evaluate the following expressions: a. ( ( (ce is the reference angle for rd and e is negative in the quadrant

2 b. (ce is the reference angle for th and e is negative in the quadrant c. cos(00 cos(60 (ce 60 is the reference angle for 00 d. cos 7 cos cos 6 (ce is coterminal with 7 (ce is the reference angle for nd and coe is negative in the quadrant e. tan cos co ( ( ( s (ce is the reference angle for nd and coe is negative in the quadrant f. ( 00 cos( 600 sec 6 cos 60 (ce 60 is coterminal with 600

3 . If ( θ and θ is in the second quadrant, find the exact value of the following expressions. a. cos( θ We can use the Pythagorean identity to find cos( θ : ( θ ( θ + cos ( θ + cos ( θ ( θ ( θ 6 6 cos cos cos (note that we take the negative square root ce coe is negative in the second quadrant b. ( θ cos ( θ (ce the graph of y cos( θ shifted right units becomes the graph of y ( θ c. θ + ( θ (ce the period of y ( θ is d. tan( θ ( θ cos( θ e. sec( θ cos( θ

4 f. csc( θ ( θ. Find all of the solutions to the following equations on the interval [ 0, : a. cos( θ cos( θ cos( θ In the first quadrant, we know that θ is a solution. In the second and third quadrant coe is negative, so there are no solutions. In the fourth quadrant we know that θ Thus the solution set on the interval is a solution. [ is { } 0,,. b. ( θ + ( θ + ( θ Since ( x when x + k (where k, we need : θ + k + k θ θ + k Since 0 θ <, we can only take k 0 and k in the statement above. Thus, the solution set is {, }.

5 6. Find all of the solutions to the following equations: a. ( x + ( x + ( x ( x Since this e value isn't one we are familiar with, we will need to use the inverse e function (i.e., arce function. ( x k ( x +, or x + k where k b. 7 + cos( t 7 + cos( t cos( t cos( t cos( t t + k, k OR t + k, k t + k, k 6 OR t + k, k 6

6 6 7. Use the e and coe functions to find the coordinates of the point P in Figure. Coordinates of P: ( cos 7, 7, ( (, P 7 6 Figure 8. Sketch a graph of the function gt ( t amplitude of g. +. State the period, midline, and First, let s write the function in standard form : ( t ( t gt + + Thus, this is a function of the form ( gt A w t h + k where A, w, h, and k 0. Since A the function has amplitude units. Ug the fact that w, we can find the period: period. So the period is units. w Since k 0, the midline is y 0. Since h, we need to start our e wave at t, i.e., shift the wave left units. Below is the graph of gt ( ( t +. A graph of ( ( + gt t.

7 7 9. Find the misg sides a and b and the misg angle A for the right-triangle below. (The triangle may not be drawn to scale. We can find A easily, ug the fact that the sum of the angles in a triangle is 80 : We can find b ug e: (0 A A A 70 b b (0.7 and we can use coe to find a: cos(0 a a cos(0.698 (you won't be expected to obtain approximations on the exam ce you won't have a calculator (you won't be expected to obtain approximations on the exam ce you won't have a calculator

8 8 0. a. Find possible algebraic rule for the function y f( t graphed in Figure. Figure : Graph of y f( t. Let s write a rule involving e, so our rule will have the form ( w and we need to determine the values of A, w, h, and k. px A x h + k The midline is the line midway between the function s maximum and minimum output values. The function s maximum output value is and its minimum output value is. Since 8 is the average of these values, the midline is y 8 so k 8. The amplitude is the distance between the function s maximum output value,, and its midline y 8, which is units. Therefore, A. The function completes one period between x 0 and x. Thus, the period of the function is 0. To find w we need to solve : w w w Near the y-axis, the graph of y ( x is increag and passes through its midline, so we need to look for a spot in the graph of y px where it shows this behavior, and one such spot is on the y-axis, at x 0 so we don t need a horizontal shift. Therefore, an algebraic rule for the graphed function is f( t ( t + 8. (There are MANY other answers.

9 9 b. Use your answer to part a to find exact solutions to f( t 0 f( t 0 ( t ( t ( t t + k or t + k, k t + k + k or t, k t + k or t + k, k. Evaluate the following: a. cos y x [ ] cos (ce the range of cos is 0, b. 7 c. cos cos ( 7 ( ( (ce the range of y ( x is, ( ( cos ( 6 [ ] (ce the range of is y cos ( x 0,