Eam Review Practice Questions for Midterm - Math 00Q - 0Fall The following is a selection of problems to help prepare ou for the second midterm eam. Please note the following: there ma be mistakes the distribution of problems on this review sheet does not reflect the distribution that will be on the eam. learning math is about more than just memorizing the steps to certain problems. You have to understand the concepts behind the math. In order to test our understanding of the math, ou ma see questions on the eam that are unfamiliar, but that rel on the concepts ou ve learned. Therefore, ou should concentrate on learning the underling theor, not on memorizing steps without knowing wh ou re doing each step. (Some eamples of unfamiliar problems can be found in the sample problems below, to give ou an idea.) Now trig functions. Angles can be measured in degrees or radians, and though we primaril use radians, it s worth knowing how to convert. More importantl, ou should know the unit circle ver well!. Convert 0 to radians. 0 = radians You can memorize a conversion formula, or just remember that 0 degrees equals radians. 0 = 0, so = radians corresponds to 0 degrees.. Convert radians to degrees. = 90. Which of the following angles are coterminal?, 5, 9, 7, 9 and 7 all correspond. To see this add + = 9. On the other hand, if we substract = 7.. What is an angle in the interval [, ] that corresponds to the angle with )? (i.e., is coterminal We can find the answer b the addition of since our starting angle is below our desired range. Notice that + =. 5. What is an angle in the interval [, 0] that corresponds to the angle 5 with 5 )? (i.e., is coterminal Just as above, 5 the desired range. =. In this case we substract in order to move into. Draw the unit circle. Label the angles 0,,,,, 7, 7,,, 5. Label the coordinates of the points on the unit circle that correspond to those angles. 7. Find an angle in [0, ) that is coterminal with the angle 5. Page of 9
Eam Review Since 5 = 5 + 0 = 5 + 0 5 is coterminal with the angle 5 8. Find an angle in [0, ) that is coterminal with the angle 5.. Since 5 = 5 + 0 5 = 5 + is coterminal with the angle 5 5. 9. Find the reference angle for the angle 7. (The reference angle is the angle in [0, ] formed between a given angle and the -ais, so for eample, the reference angle for is.) The reference angle for 7 is. (The angle 7 is in the fourth quadrant, ou want to find how far past 7 ou have to go to get to the -ais that s.) 0. Find the reference angle for the angle 5. The reference angle for 5 is.. Find the reference angle for the angle 7 5. The reference angle for 7 5 is 5. Do ou understand what the trigonometric functions are? both using triangles or the unit circle? How to calculate them,. Define sine, cosine, tangent, cosecant, secant, cotangent. What are their domain and range? Let θ be the angle formed starting with the positive -ais and going counterclockwise if θ is positive and clockwise if θ is negative. Then the corresponding terminal point on the unit circle has coordinates (cos(θ), sin(θ)). The other trig functions are defined as follows: tan(θ) = sin(θ) cos(θ), cot(θ) = cos(θ) sin(θ), sec(θ) = cos(θ), csc(θ) = sin(θ) Function Domain Range sin(θ) (, ) [, ] cos(θ) (, ) [, ] tan(θ) all real numbers ecept n for odd integers n (, ) cot(θ) all real numbers ecept n for all integers n (, ) sec(θ) all real numbers ecept n for odd integers n (, ] [, ) csc(θ) all real numbers ecept n for all integers n (, ] [, ) For angles between 0 and, these trig functions can be equivalentl defined using right triangles. In a right triangle, sin(θ) = opposite hpotenus tan(θ) = opposite adjacent sec(θ) = hpotenuse adjacent. What is sin(0 )?, cos(θ) = adjacent hpotenus, cot(θ) = adjacent opposite, csc(θ) = hpotenuse opposite. sin(0 ) =. What is csc( 5 )? Page of 9
Eam Review csc( 5 ) = sin( 5 ) = = 5. If t = 0, what are sin(t), csc(t), and cot(t)? sin(t) = csc(t) = sin(t) = cot(t) = cos(t) sin(t) = = =. If cot(t) = and t is in the interval [, ], then what is sin(t)? Since cot(t) = cos(t) sin(t) = t = 5 is the angle in [, ] where cot(t) =. sin(t) = sin( 5 ) = 7. If cos(t) = and t is in the interval [, ], then what is tan(t)? t = is the angle in [, ] where cos(t) =. tan(t) = sin(t) cos(t) = = 8. Given that cos(θ) = 7 and θ is in quadrant IV, find sin(θ). Opposite side of θ is ± 7 = ± 5 Since sin(θ) is negative in quadrant IV, so sin(θ) = 5 7 9. Given that tan(θ) = 5 and θ is in quadrant II, find csc(θ). Hpotenuse side of θ is + ( 5) = so csc(θ) = sin(θ) = = 0. Use the triangle below to answer these questions. C β B α (a) Sa α = and B = 8. What is A? (b) Sa β = and A = 0. What is C? (c) Sa A = 5 an B = 0. What is α? A (a) tan( ) = 8 = Therefore, = 8 (b) cos( ) = 0 So = 0 = (c) tan(α) = 0 5 = So α = arctan() Page of 9
Eam Review. Is sin( 9 0 ) closest to -, 0, or? Note that the angle 9 0 is almost, but just barel less than. On the unit circle, then, it s just above the -ais, so the sine is close to 0. What about going backwards? If ou know the values of trig functions, can ou work backwards to find out what the angle might be?. If sin(t) =, what might t be? Give all possible solutions in [, ]. t =,. If sin t =, what might t be? Give all possible solutions in [0, ]. t =. If cos t =, what might t be? Give all possible solutions in [0, ]. t =, 5 5. If tan t =, what might t be? Give all possible solutions in [0, ]. t =, 7. Find all values of t in the interval [0, ] such that sin t = 0. t = 0,, 7. Find all values of t in the interval [0, ] such that csc t =. i.e., sin(t) = so sin(t) = and t =, 5. 8. Find all values of t in the interval [0, ) such that cot t =. i.e., tan(t) =, so tan(t) =. Then t =, 7. Like an other function, we sometimes want to graph trigonometric functions. Can ou? 9. Graph sin(), cos(), and tan(). Page of 9
Eam Review sin() 0.5 0.5 cos() 0.5 0.5 5 tan() 5 0. Graph f() = sin(). Page 5 of 9
Eam Review sin(). Graph g() = cos( ). Can ou find another function that has this same graph? cos( ) 0.5 0.5 This is the same as sin().. Graph h() = sin(). sin( ) Page of 9
Eam Review. Where does csc() have a vertical asmptote? The function csc() is the same as sin(), so it is undefined (and has a vertical asmptote) whenever sin() = 0. That s at multiples of : 0,,, etc.. Graph sec(). sec() 5 5 5. Graph 5 sin( ) +.5. 5 sin( ) +.5. Graph cos(( + )). Page 7 of 9
Eam Review cos(( + )) 0.5 0.5 Save problems 7- for our final eaminations(not incleded toeam)for eam review continue with Pb.7-80) Can ou solve equations that involve trigonometric functions? 7. Find all solutions to sin() + = 0 in [0, ). sin() + = 0 sin() = = 7, 8. Find all solutions to cos =. cos = cos = = 0,,,... = k 9. Find all solutions to sin = sin in [0, ). sin = sin sin = sin = = 0. Find all solutions to cos = 0 in [0, ). Page 8 of 9
Eam Review cos = 0 cos = cos = ± =,,, 5. Find all solutions to cos() = in [0, ). cos() = =, 5, 7, =, 5, 7,. Find all solutions to tan () =. tan () = tan() = ± =,,, 5 + k. Find all solutions to sin + = sin in [0, ). sin + = sin sin = sin = = 5, 7. Find all solutions to cos( ) = 0 in [0, ). cos( ) = 0 cos( ) = 0 =,, 5,... = +, +, 5 +, 7 +, 9 +, + Page 9 of 9
Eam Review 5. Find all solutions to sec = cos in [0, ). cos = cos = cos = cos cos = ± =,, 5, 7. Find all solutions to sin sin = 0 in [0, ). sin sin = 0 ( sin + )(sin ) = 0 sin =, =, 7, 7. Find all solutions to sin cos + cos = 0 in [0, ). sin cos + cos = 0 cos (sin + ) = 0 cos = 0 or sin = =, 8. Find all solutions to sin() = cos(). sin() = cos() sin cos = tan = =, 7,, 5 + k,... = 8, 7 8, 8, 5 8 + k Sometimes ou need to use a trigonometric identit to help solve an equation. You ll need to use trig identities in the following problems. Page 0 of 9
Eam Review 9. Find all solutions to cos + sin = in [0, ). cos + sin = ( sin ) + sin = 0 sin sin + = 0 ( sin ) (sin ) = 0 sin = or sin = =,, 5 50. Find all solutions to sin() = cos in [0, ). sin() = cos sin cos = cos sin cos cos = 0 ( cos sin ) = 0 cos = 0 or sin = =,,, 5. Find all solutions to sin(t) tan(t) = 0 in [0, ). sin(t) tan(t) = 0 sin(t) cos(t) sin(t) = 0 ( sin(t) ) cos(t) = 0 sin(t) = 0 or cos(t) = t = 0,,,,... or t =, 7, 9, 5,... t = 0,,,, 8, 7 8, 9 8, 5 8 5. Find all solutions to cot + csc = 0 in [0, ). Page of 9
Eam Review cot + csc = 0 cot + ( + cot ) = 0 cot = 0 cot = ± =,,, 5 5. Find all solutions to cos() = cos in [0, ]. cos() = cos cos = cos cos cos = 0 ( cos + ) (cos ) = 0 cos = or cos = = 0,, Just like with other functions we ve studied, trigonometric functions have inverses. Well, partial inverses (since trig functions aren t one-to-one). Do ou know what the inverse trig functions are? Do ou know their domain and range? Can ou compute them for some values? Can ou graph them? 5. What is arccos? θ = 55. What is arcsin? θ = 5. What is arctan 0? Let arctan 0=θ. Note: θ must be in ( /, /). tan(θ) = 0. Therefore, θ = 0. 57. What is sin (sin( ))? Let sin (sin( ))=θ. Note: θ must be in [ /, /]. since is coterminal with sin (sin( ))=sin (sin( Answer: θ = ))= Page of 9
Eam Review 58. What is tan(arctan( ))? Note: tan(arctan()) = if is in (, ). Since - is in (, ) Answer: tan(arctan( ))=- 59. What is arccos(cos( 9 ))? Note: arccos(cos(θ)) = θ if θ is in the range of the arccosine function, i.e. [0, ]. Since cosine function is an even function cos( 9 ) = cos( 9 ) arccos(cos( 9 ))=arccos(cos( 9 ))= 9 Answer: 9 0. What is tan(sin ( ))? Let sin ( )=θ. Then sin θ=. Using SOH,CAH,TOA O=, H=, A= and tan θ= = So tan(sin ( )=tan θ= Answer:. What is tan(arccos( 5 ))? Let arccos( 5 )=θ. Then cos θ= 5. Using SOH,CAH,TOA A=5, H=, O= and tan θ= 5 So tan(arccos( 5 Answer: 5 )=tan θ= 5. Epress a solution to the equation tan( ) = 5 using inverse trigonometric functions. arctan(5) = = arctan(5) +. Epress a solution to the equation sin() + = using inverse trigonometric functions. sin() = = arcsin( ) = arcsin( ). Graph the function cos (). Page of 9
Eam Review = arccos() 5. Graph the function arctan(). = tan (). Graph the functions csc () and csc(). = csc () Page of 9
Eam Review 0.5 = csc() 0.5 Note that csc() = sin() = sin(). Can ou appl our knowledge to real-world applications? 7. How long a ladder do ou need if ou want to reach a window that s 0 feet off the ground? You re on uneven terrain, so leaning a ladder an steeper than 0 degrees is unsafe. The ladder would be the hpotenuse of a triangle that has height 0 feet and is leaned at an angle of 0 degrees. Then sin(0 ) = 0, where is the length of the ladder. Therefore, 0, so =. = 0 8. An equilateral triangle is circumscribed around a circle of radius 5 (thus, the center of the triangle and the center of the circle are at the same point). What is the area of the triangle? Equilateral triangles have angles of size. If the triangle is circumscribed around the circle, that means the distance from the center of the triangle to one of its edges is the same as the radius of the circle, so it s 5. Then ou can make a right triangle, with a line connecting the center of the circle to the edge of the triangle and another line connecting the center of the circle to a verte of the triangle. This cuts one of the angles of the equilateral triangle in half, so the right triangle has an angle of. The opposite side to that angle has length 5, so sin( ) = 5, or in other words, = 0 is the length of the line connecting the center of the circle to a verte of the triangle. Therefore, the height of the equilateral triangle is 5 (draw a picture to help see wh that s the case). Then, let s find the length of one side of the equilateral triangle using cos( ) = z 0, where z is half the length of one side of the equilateral triangle. That means that z = 5, so the length of the base of the equilateral triangle is 0. Therefore, the area of the triangle is 5 0. 9. A rope hanging straight down from a pole is feet longer than the pole (that is, the rope hangs straight down and there s feet of etra rope ling on the ground). When the rope is picked up and stretched taut, it hits the ground 8 feet awa from the pole. How tall is the pole? Draw a picture of the situation. Call the length of the pole. Then we have a triangle with legs and 8, with hpotenuse +. Using the pthagorean theorem we have Now we can solve this equation: + 8 = ( + ) + = + 8 + Page 5 of 9
Eam Review = 8 + 8 = + = 70. The number of hours of dalight in a da can be modeled as a sinusoidal curve. Assume that the longest da of the ear is June, which has 5. hours of dalight, and the shortest da of the ear is December, which has 9. hours of dalight (this is close to the real numbers in CT). Let s start the clock on Januar, and sa Jan corresponds to t = 0. Find a sinusoid to model the number of hours of dalight as a function of the da of the ear. Use this to estimate the number of hours of dalight there will be tomorrow. The numbers in this problem are mess, so it s not a great eam problem...but it s kind of fun. If sunlight ranges from 5. to 9., then there is a difference of 5. - 9. =. between our ma and min, so the amplitude is. =.05. The midline, then, is 9. +.05 =.5. The period should be 5, of course (es, in real life it would not be eactl 5, more like 5...but it s close enough for us). There is a shift, which might be easiest to see b counting backwards the lowest point on the sine curve should be Dec, 0 das before Jan (which is t = 0). Thus, the high point is halfwa between t = 0 and t = 55, a ear later, so the high point is approimatel 7. The curve should pass the midline eactl halfwa between the low point and the high point; halfwa between 0 and 7 is 8. Thus, the curve should be shifted to the right b 8. A curve we could use is.05 sin( 5 (t 8)) +.5. As this is being written, it s late October, almost Halloween, which is the 0th da of the ear (so would correspond to t = 0 in our model). The predicted number of hours of sunlight on October is then.05 sin( 5 (0 8)) +.5 = 0.. (The actual number of hours of sunlight is 0...not bad!) 7. A 7 inch (diameter) biccle wheel turns at 0 revolutions per minute. There is a nail in the wheel. Find a function to model the height of the nail above the ground at an point in time, assuming we start the clock when the nail hits the pavement. Minimum value is 0, maimum value is 7, so midline is.5 and amplitude is.5. 0 revolutions per minute means one period is.5 seconds. For variet, let s graph this one as a cosine function. If we use a negative cosine function, no shift is needed. Thus, an equation for the height of the nail is.5 cos(.5 t) +.5 inches at t seconds. 7. A weight hangs at the end of a spring, causing it to bounce up and down. At its lowest point, the weight sits 0 cm above the ground, and at its highest point, it reaches 00 cm above the ground. Ever 0 seconds, the weight bounces up and down times. The height of the weight above the ground can be modeled b a sine curve. Supposing the weight starts from its lowest point at time t = 0 seconds, find a sinusoid to model the height of the weight at time t. This sine curve goes between 0 and 00, so the midline is 80 and the amplitude is 0. It goes through periods in 0 seconds, so one period is 0 =.5 seconds. To model this as a sine curve, we need to appl a shift the graph passes the midline at.5 =.5 seconds. Thus, a function to model this is 0 sin(.5 (t.5)) + 80. 7. A plane is fling at an altitude of,000 feet. Looking down, the pilot sees a ship at an angle of depression of 0 degrees. He sees a submarine at an angle of depression of 0 degrees in the same direction. How far apart are the ship and submarine? Draw a picture of the situation. We get two right triangles, Triangle A with the angle 0 with opposite side,000 and Triangle B with the angle 0 with opposite side Page of 9
Eam Review,000. To find the distance between the ship and submarine we need find the difference between the adjacent side lengths of these two triangles. For Triangle A, call the adjacent side a. Then we have tan(0 ) = 000 a = 000 a a = 000 000 = a For Triangle B, call the adjacent side b. Then we have tan(0 ) = 000 b 000 = b = b 000 000 Now the distance between the ship and submarine is a b which is: = b 000 000 = 0000( ) feet Here are a few additional problems that didn t make it into an of the above sections. 7. Calculate sin( ) without using a calculator. Use the difference identit for sine: ( sin ) ( ) = sin = = ( ) cos ( ) ( ) cos sin 75. Simplif the epression using identities as needed: ( cos + cos ). It helps to first factor this equation: cos + cos = ( cos ) Then we see that we can appl the pthagorean identit sin = cos to obtain: ( cos ) = (sin ) = sin. 7. Calculate cos( 7 8 ) without using a calculator. Page 7 of 9
Eam Review Use the half angle formula for cosine: cos ( 7 8 ) = + cos( 7 8 ) = + cos( 7 ) = + = + Now, taking the square root of both sides we get ( ) 7 + cos = ± 8 but since the angle 7 8 be negative. Thus, is in the second quadrant, we know that the cosine of this angle should cos ( ) 7 + =. 8 77. Simplif tan +. Use the pthagorean identit tan + = sec and sec = cos and we see that 78. True or false? cos + sin tan = sec. tan + = sec = cos. True. We need to start with cos + sin tan and transform it into sec. We use a series of trig identities: cos + sin tan = cos + sin sin cos = cos + sin cos = cos + sin cos = (cos + sin ) cos = () cos = cos = sec (pthagorean identit) 79. True or false? cot csc = cos (csc ) Page 8 of 9
Eam Review True. We would like to start with cot csc of trig identities: and transform it into cos (csc ). We ll use a series cot csc ( ) cot = cot csc ( ) cot = (csc ) (pthagorean identit) csc ( cos ) = (csc ) sin sin switching to sine and cosine = cos (csc ) 80. Show work to prove that sec θ cos θ = sec θ. We need to start with identities: sec θ cos θ and transform it into sec θ. We use a series of trig sec θ cos θ = sec θ cos θ cos θ = cos θ cos θ cos θ = cos θ = cos θ = sec θ Page 9 of 9