Math 115 Spring 2014 Written Homework 10-SOLUTIONS Due Friday, April 25

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1 Math 115 Spring 014 Written Homework 10-SOLUTIONS Due Friday, April 5 1. Use the following graph of y = g(x to answer the questions below (this is NOT the graph of a rational function: (a State the domain and range of g in interval notation. Solution: There is a point on the graph corresponding to every x value except x =. Hence, the domain of g is (, (, 5]. The range is (, 1 (1, ] (b What is g(? What is g(? What is g(4? Solution: We are looking for the y-values corresponding to the points where x =, x = and x = 4. We see that g( =, g( is undefined (as x = is not in the domain of g, and g(4 =. (c For what values of x does (i g(x =?, (ii g(x =?, (iii g(x = 1? Solution: We start by looking at the set of points on the graph of g that intersect the horizontal lines, y =, y = and y = 1, respectively. For y = we see that y = intersects the graph at the points (1,, (, and the line interval from (, to (0,. Hence, the solutions to the equation g(x = are the set {x R x 0, x = 1 or x = }. For y =, there is only one point of intersection with the graph; (,. g(x = when x =. Hence, For y = 1, there is no point of intersection with the graph. Hence, there are no solutions to the equation g(x = 1.

2 (d Determine all of the following limits from the graph of y = g(x: x g(x x g(x x + g(x x 0 g(x x 4 g(x x 4 + g(x x g(x Solution: g(x = 1 x g(x = x x + g(x = x 0 g(x does not exist (since the limit from the right of x = 0 is 1 and the limit from the left of x = 0 is x 4 g(x = 1 g(x = x 4 + x g(x =

3 . Express the lengths a and b in the figure below in terms of θ. Solution: Use the right-triangle definitions of the trig ratios and then solve for a and b: sin θ = a 4 a = 4 sin θ and cos θ = b 4 b = 4 cos θ. A boat approaches a 0-ft lighthouse whose base is at sea level. Let b be the distance between the boat and the base of the lighthouse. Let L be the distance between the boat and the top of the lighthouse. Let θ be the angle of elevation between the boat and the top of the lighthouse. (a Express b as a function of θ. Solution: To model this situation, we can use a right triangle: Using the right-triangle definitions, we know tan θ = 0 0. Thus, b = b tan θ.

4 (b Express L as a function of θ. Solution: Using the same right triangle from above, sin θ = 0 0. Thus, L = L sin θ. 4. (a In a circle of radius r, an arc of length 10 is swept out by an angle of radians. What is the exact radius of the circle? Solution: The radian was defined in such a way that the arc length, s, is equal to the radius times the angle, when the angle is in radians. i.e. Here we have that s = 10 and θ =. Thus, s = rθ. 10 = r r = 10. (b In a circle of radius r =, what is the exact length of the arc swept by an angle of 10? Solution: Before we can apply the arc length formula, we must have an angle in radians. Thus, here we must convert the angle from degree measure into radian measure. 10 ( π 180 = 10π 180 = 1π 18 = 7π. Now we have r = and θ = 7π. s = rθ = ( 7π = 7π. The length of the are subtended by the angle 10 on the circle is 7π. 5. Suppose θ is an angle in standard position (meaning it starts on the positive x-axis whose radian measure is an integer value between 0 and π. (a If the angle falls in the third quadrant, what is the radian measure θ? Solution: The angles in the third quadrant are between the angles π = and = Thus, the only integer value between these is 4. So, θ = 4. π (b If the terminal side of the angle had fallen in any other quadrant, could you still answer the question with a single value? Explain. Solution: If the terminal side had fallen in quadrant I, we would be able to answer this question because those angles fall between 0 and π = Hence, in quadrant I, θ would be equal to 1. In quadrants II and IV, there would be two possible values for θ; and in II and 5 and in IV.

5 . For each angle θ given below, when sketched in standard position, determine the quadrant in which it lies. (Hint: in some cases, it might help to determine a coterminal angle that is between 0 and π. You should justify your answer in some way - either with a brief explanation or a picture. (a θ = 7π Solution: θ = 7π = π + π. π is full revolutions and then π is less than π terminal side of θ will end up in the first quadrant. so the (b θ = 11π 4 Solution: θ = 11π 4 = π π 4 clockwise π which is more than π 4 in the third quadrant.. π is 1 full revolution clockwise an then we continue but less than π. Thus, the terminal side of θ will end up (c θ = 5.5 Solution: θ = 5.5 is between the quadrantal angles π = and π = Thus, the terminal side of θ will end up in the fourth quadrant. 7. (a Suppose θ is an angle in standard position which intersects the unit circle at the point (x, y. If y = 1, what are the possible values of cos θ? If you know that cos θ is positive, in 8 which quadrant does the angle lie? Solution: The equation of the unit circle is x + y = 1. If we plug y = 1 8 we get two possible values of x and thus two possible values of cos θ: into the equation, x + y = 1 x + ( 1 8 = 1 x = 1 x = 4 x = ± 8 The possible values are cos θ = and cos θ =. If we know that cos θ is positive, then 8 8 the angle must lie in quadrant I because both the x and the y coordinate of the corresponding point on the unit circle are positive.

6 (b Suppose sin α = 4 5 and π < α < π. Find cos α, tan α, cot α, sec α, and csc α. Solution: Since π < α < π, we know that the angle lies in quadrant III. One method would be to draw a right triangle in quadrant III with the opposite side length 4 and the hypotenuse 5. This is a -4-5 triangle so the adjacent side is. Then we can use the right triangle trig definitions to determine the other trig values - remembering that the angle is in the third quadrant. Another method would be to use the equation for the unit circle: x + y = 1 or the equivalent identity cos (x + sin (x = 1. By plugging in sin α = y = 4, we can then solve for 5 cos α = y = 5. Either way, we get the following solutions: cos α = 5, tan α = 4, cot α = 4, sec α = 5, csc α = 5 4 (c Suppose cos β = and β lies in quadrant II. Find sin β, tan β, cot β, sec β, and csc β. Solution: One method would be to draw a right triangle in quadrant II with the adjacent side length and the hypotenuse. This is a triangle so the opposite side is 1. Then we can use the right triangle trig definitions to determine the other trig values - remembering that the angle is in the second quadrant. Another method would be to recognize the point on the Unit Circle here. The point is in ( the second quadrant with an x-coordinate of. Thus, the point is, 1. Either way, we get the following solutions: sin β = 1, tan β = 1 = 1, cot β =, sec β =, csc β = 8. Find( the exact value of each of the following. You must justify your answer in some way. 9π (a cos Solution: 9π = 8π + π = 4π + π. Thus the angle is coterminal to π. Thus, cos ( 9π ( π = cos = 0

7 (b sin ( 4π Solution: The terminal side of 4π ends up in quadrant so it s sine value will be negative. The reference angle here is π. Using either a special right triangle or the unit circle, we find that ( 4π ( π sin = sin = ( 11π (c tan Solution: The terminal side of 11π ends up in quadrant 4 so it s tangent value will be negative. The reference angle is π. Using either a special right triangle or the unit circle, we find that (d ( csc 5π 4 ( 11π tan ( π = tan = sin π cos π = 1 = 1 Solution: The angle 5π 4 will fall in quadrant with a reference angle of θ r = π 4. Using either a special right triangle or the unit circle, we find that ( csc 5π ( π = csc = sin π = 1 1 = 4 (e ( 48π sec Solution: The angle 48π = 1π is a quadrantal angle that is coterminal to π. Thus, ( 48π 1 sec = sec(1π = cos 1π = 1 cos π = 1 1 = 1 ( (f cos 11π Solution: The angle 11π = 9π π = π π will fall in quadrant 1 with a reference angle of θ r = π. Using either a special right triangle or the unit circle, we find that ( cos 11π ( π = cos = 1

8 9. As we did with the sine function in lecture, state the properties of the cosine function. (a State the domain and range of f(x := cos x. Solution: The domain is x R. The range is f(x [ 1, 1]. (b Determine all values of x where f(x is zero (i.e. the x-intercepts. Solution: The zeros occur when x = π + kπ where k is any integer. (c Determine the y-intercept for the graph y = cos x. Solution: The y-intercept occurs at (0, Use your knowledge of transformations to sketch the graph of at least two periods/cycles of each function. ( (a f(x := sin x + π 4 Solution: We begin with the graph of y = sin x: Next we reflect the graph over the x axis to get the graph of y = sin x: Finally, we shift the previous graph to the left π ( 4 to get the graph of y = sin x + π : 4

9 (b g(x := cos(x + 4 Solution: We begin with a graph of y = cos x: Then we shift the graph up 4 to get the graph of y = cos(x + 4:

10 (c ( h(x := cos x π + 4 Solution: We begin with a graph of y = cos x. Then we shift the graph to the right π 4 to get the graph of y = cos(x π 4 Now applying the vertical stretch cause by A = yields the graph y = cos ( x π 4. Lastly, we apply the vertical shift up. This results in the final graph y = cos ( x π 4 +.

11 Remember - there is more than correct answer for each of these graphing questions. ANY complete cycles is correct. Some of you may have used a different method on part (c and ended up with the following graph:

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