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Math 25 February 19, 2017 Name: Answer Key D. Arnold Exam #1A Multiple Choice. (50 points) For each of the following questions, determine the correct answer then fill in the corresponding oval on your scantron. (5pts ) 1. Convert 5π/12 radians to degrees. (a) 15◦ (b) 95◦ (d) 4 75◦ (c) 55◦ (e) 45◦ Solution: We know that 180 degrees equals π radians, so: 5π 180 degrees 5π radians = radians · 12 12 π radians = 75 degrees (5pts ) 2. Convert 36 degrees to radians. (a) 4π/9 radians (b) π/8 radians (d) 3π/5 radians (c) π/7 radians (e) 4 π/5 radians Solution: We know that 180 degrees equals π radians, so: 36 degrees = 36 degrees · = (5pts ) π radians 180 degrees π radians 5 3. An arc of a circle subtended by an angle of 80◦ measures 4 inches. What is the radius of the circle? (a) 8/π inches (b) 4 9/π inches (c) 4/π inches (d) 3/π inches (e) 7/π inches Solution: Consider the following image: 4 inches 80◦ r Math 25/Exam #1A – Page 2 of 13 – Name: Answer Key The relation between the central angle θ, radius r, and arc length s is given by the formula: s θ= r However, in order to use this formula, the angle must be measured in radians. Thus: π 80◦ = 80◦ · 180◦ 4π = 9 Next, solve θ = s/r for r: s θ= r rθ = s s r= θ Substitute θ = 4π/9 and s = 4 in. 4 in 4π/9 9 r = in π r= (5pts ) 4. A sector of a circle subtended by an angle of 75 degrees has an area of 5π/6 square inches. What is the radius of the sector? 75◦ r (a) 3/π inches (b) 4π/3 inches (c) 2/π inches (d) 4 inches (e) 4 2 inches Solution: Consider the following image: 75◦ r Math 25/Exam #1A – Page 3 of 13 – Name: Answer Key The area of a sector subtended by an angle θ with radius r is given by the formula 1 A = θr2 2 Let’s solve this equation for r. 2A = θr2 2A r2 = θ r 2A r= θ In order that we may use this formula, our angle θ must be expressed in radian measure. π 180◦ 75◦ = 75◦ · = 5π 12 Now, we can substitute θ = 5π/12 and A = 5π/6 square inches. r 2A r= s θ 2 in 2 5π 6 r= 5π 12 p 2 r = 4 in r = 2 in (5pts ) 5. A young boy ties a tennis ball to the end of a piece of rope that is 36 inches long. He then swings the ball in a circular arc, the ball traveling about his head at 2 revolutions per second. What is the linear speed of the ball in inches per second? (a) 4 144π in/s (b) 36π in/s (c) 36 in/s (d) 72 in/s (e) 72π in/s Solution: Start with a picture. 36 in Math 25/Exam #1A – Page 4 of 13 – Name: Answer Key We will use the formula v = ωr to calculate the velocity of the tennis ball. However, ω must be measured in radians per unit time. rev rad · 2π s s rad = 4π s ω=2 Next: v = ωr rad (36 in) = 4π s in = 144π s (5pts ) 6. Consider the following right triangle. 2 θ 4 Find √ the exact value of csc θ. (b) 2 (a) 2 5/5 √ (d) 4 5 (e) 1/2 (c) Solution: Label the hypotenuse h. h 2 θ 4 Now use the Pythagorean Theorem. h2 = 22 + 42 h2 = 4 + 16 h2 = 20 √ h = 20 √ h=2 5 √ 5/5 Math 25/Exam #1A – Page 5 of 13 – Name: Answer Key Now, the cosecant is the reciprocal of the sine, so let’s first find sin θ. opp hyp 2 = √ 2 5 1 =√ 5 sin θ = Hence: csc θ = (5pts ) √ 5 √ 2 3 7. Given that sec θ = , what is the exact value of cos θ? 3 √ √ √ 3 6 2 (a) (b) (c) 3 √6 √3 3 2 (d) 4 (e) 2 6 Solution: The cosine is the reciprocal of the secant. Hence: 3 cos θ = √ 2 3 √ 3 3 = √ ·√ 2 3 3 √ 3 3 = √6 3 = 2 (5pts ) 8. Given the angle θ = 120◦ , which of the following is its reference angle? (a) 45◦ (b) 5π/6 (c) 30◦ (d) 4 π/3 (e) π/4 Solution: Draw the angle. The reference angle is the angle made with the horizontal axis, but it should be placed and labeled in the first quadrant. Math 25/Exam #1A – Page 6 of 13 – Name: Answer Key y 120◦ 60◦ x Hence, the reference angle is 60◦ . However, that choice is not available, so change 60◦ to radians. πR 60◦ = 60◦ · 180◦ π = radians 3 (5pts ) 9. What is the period of the function f (x) = 3 tan (a) 4 4 π x ? 4 (b) 2 (c) 8 (d) 2π (e) 4π Solution: The period of y = tan ωx is T = π/ω. Hence: π ω π = π 4 4π = π =4 T = (5pts ) 10. What is the amplitude of the function f (x) = −2 cos (a) 2/π π 2 x−1 ? (c) −2 (b) 4 (d) 4 2 (e) π/2 Solution: The amplitude of A cos ωx is |A|. Hence: Amplitude = | − 2| =2 Math 25/Exam #1A – Page 7 of 13 – Name: Answer Key Essay Questions. (50 points) Do all of your work on graph paper. When finished, arrange your solutions in order, then staple this page atop your work. (10pts ) 1. A statue sits atop a vertical cliff. You are 500 feet from the base of the cliff. Your angle of elevation to the bottom of the statue is 52◦ . Your angle of elevation to the top of the statue is 58◦ . How tall is the statue? Note: First state an exact answer to the problem (no decimals), then use Mathematica to round your answer to the nearest tenth of a foot. Solution: First, an image. x y 58◦ 52◦ 500 This image gives us two equations. tan 52◦ = y 500 and tan 58◦ = x+y 500 Solve the first equation for y. y = 500 tan 52◦ Substitute this result into the second equation and solve for x. x+y 500 x + 500 tan 52◦ tan 58◦ = 500 tan 58◦ = Multiply both sides by 500. 500 tan 58◦ = x + 500 tan 52◦ x = 500 tan 58◦ − 500 tan 52◦ Rounding to the nearest tenth of a foot, x = 160.2 feet, the height of the statue. Math 25/Exam #1A (10pts ) – Page 8 of 13 – Name: Answer Key 2. You are given that tan θ = −3 and π/2 < θ < π. Set up an xy-coordinate system, draw and label the angle θ, then choose and label an appropriate point on the terminal side of the angle θ. Use this information to compute the five remaining trigonometric functions of θ. Make sure each answer is in simple radical form. Note: No right triangles may be drawn for this problem. Solution: Start with an image. (−1, 3) √ y 10 θ x Hence: √ −1 x 10 cos θ = = √ = − r 10 10 √ y 3 3 10 sin θ = = √ = r 10 10 y 3 = = −3 x −1 x −1 1 cot θ = = =− y 3 3 √ √ r 10 sec θ = = = − 10 x −1 √ √ r 10 10 csc θ = = = y 3 3 tan θ = (5pts ea. ) 3. Sketch exactly one period of each of the following functions on graph paper. Clearly scale and label each axis. Each of these problems must be sketched by hand without the aid of a technology. 1 (a) y = − tan x 2 Math 25/Exam #1A – Page 9 of 13 – Solution: The period of y = − tan 1 x 2 Name: Answer Key is: π 1/2 = 2π T = We will start by drawing the graph of y = tan 1 x 2 . y π Now, to draw the graph of y = − tan the x-axis. 1 x 2 2π x , we need only flip the last graph across y π π (b) y = 2 sec x 2 Solution: To start, the period of y = 2 sec T = 2π π x 2 x is: 2π π 2 =4 We’ll start by drawing the graph of y = sec π x 2 . Because the secant is the inverse Math 25/Exam #1A – Page 10 of 13 – Name: Answer Key of the cosine, we lightly draw the cosine curve, then the secant will have vertical asymptotes where the cosine is equal to zero. y 1 2 4 x −1 Now, to draw the graph of y = 2 sec π x 2 , we need only double all the y-values. y 2 2 4 x −2 (c) y = −2 sin(2x − π) Solution: Start by factoring out a 2. y = −2 sin(2x − π) π y = −2 sin 2 x − 2 We’ll begin by drawing the graph of y = −2 sin 2x, which has an amplitude of 2, a period of T = 2π/2 = π, and is reflected across the x-axis. Math 25/Exam #1A – Page 11 of 13 – Name: Answer Key y 2 π/2 x π −2 Next, replace x in y = −2 sin 2x with x − π2 to obtain y = −2 sin 2 x − will shift our previous graph π/2 units to the right. π 2 , which y 2 π/2 π x 3π/2 −2 1 π (d) y = 3 cos x+ 2 4 Solution: Start by factoring out a 1/2. 1 π y = 3 cos x+ 2 4 1 π y = 3 cos x+ 2 2 We’ll begin by drawing the graph of y = 3 cos 21 x, which has an amplitude of 3 and a period of T = 2π/(1/2) = 4π. y 3 π 2π 3π 4π x −3 Next, replace x in y = 3 cos 12 x with x + π2 to obtain y = 3 cos 12 x + shift our previous graph π/2 units to the left. π 2 , which will Math 25/Exam #1A – Page 12 of 13 – Name: Answer Key y 3 x −π/2 π/2 3π/2 5π/2 7π/2 −3 (10pts ) 4. Pictured below is one period of a trigonometric function. Determine its equation. y 2 π x π/2 −3 Solution: This picture appears to be a cosine function with period T = π, amplitude 2, that has been reflected across the x-axis, then shifted downward. So, let’s begin by graphing y = −2 cos 2x, which has period T = 2π/2 = π. Math 25/Exam #1A – Page 13 of 13 – Name: Answer Key y 2 π x π/2 −2 Next, subtract 1 from the equation y = −2 cos 2x, which gives us the equation y = −2 cos 2x − 1 and shifts the previous graph down 1 unit. y 2 π x π/2 −3 That is the given picture. Hence, we have the correct equation, y = −2 cos 2x − 1.