Trigonometry Topics Accuplacer Review July 12 2016
Trigonometry Topics Accuplacer Review – revised July 2016You will not be allowed to use a calculator on the Accuplacer Trigonometry test.For more information, see the JCCC Testing Services website at http://www.jccc.edu/testing/ or call 913-4694439.1. If cos θ 54 and tan θ 0 , find the value for csc θ .a.43b. 53c. 54d.35c.541d.352. Find the value of sin θ if θ is acute and tan θ 54 .a.34414b.3. If 0 θ π and tan θ 3 , find the value of sec θ .2a.13c. 3b. 10d.104. Give all values of θ between 0 and 2π for which tan θ 3 .a.πb.32π3c.4π3d.2π3d. 61 πd.3 91100d. and5π3( ).5. Evaluate sin 1 a.1276πb. 6 π and7116 π11c. 6 π6. If sin θ 3 and 0 θ π , find the value of sin(2θ ) .10a. 0.62b. 0.3c.3 91507. What is the exact value for sec(210 ) ?a.12b. c. 23223π cos(3θ )then what is ( f g )(π ) ?8. If f (θ ) 1 sin θ and g (θ ) 2a. 29. If sin a b.351 32c. 0d. 2πand cos b 54 , and we know a and b are acute angles, find cos(a b) .a. 1b. 0c. 257d.72510. Choose the function below with the smallest amplitude:a.f (θ ) 12 sin(θ )g (θ )b. 13cos(2θ 1)c. f (θ ) 3sin ( 13 θ )d. g (θ ) sin( 2θ 1)
11. If ABC is a right triangle, with ACB as the right angle and θ as BAC , find tan θ .aca.b.abc.bad.cb30 and AC 1 . Find the length of PB .12. In the following picture, ABC a.b. 23c. 1.5d.32d.7413. Find the value for cos θ if θ lies in the 3rd Quadrant and csc θ 43 .a. 34b.34c. 740 , find all solutions for 0 θ 2π .14. Given tan θ 2 tan θ 1 2a. 0b.πc.415. Which of the following has the largest value?a. sin π6b. cos 4π( x)16. Find the maximum value of the function f a.13b.πand43π4π4and5π4d. csc π4c. tan 23π13d.sin(3 x 2) 1 .c. 143d. 2317. If an acute angle θ is part of a right triangle where the side opposite of θ is 5 units long and the hypotenuse is9 units long, what is cot θ ?a.2 145b.95c.53d.5 71418. Find the value for the angle x:a. tan 1()34b. sec 1()c. cos 1532()43d. csc 1()45
19. Which of the following has the highest maximum value on its graph?a. b. yy 2 sin x13 c. y sin(5 x 1)cos(2 x) 2 d. y 5sin x 420. If the angle of elevation of the sun is θ degrees, and a man’s shadow along level ground is 4.5 feet long, whichexpression gives the height of the man in feet?a. 4.5sin θb. 4.5tan θc. 4.5cot θd. 4.5cscθ3π421. A circle has a radius 16.4 cm. Find the length of the arc cut by a central angle ofa. 12.3π cmb. 16.4 cmc. 135 cm()3 14 sin x π2 ?22. What is the range of the function f ( x) a.( , )b.( π2 , 52π )[ 14 , 14 ]c. c. 2d. 16.4π cmd.[ 2.75, 3.25]( ) 123. Simplify csc sin .2 23a.b.24. Simplifya.12 1d. 2232b.a. tan θ26. If sin θ 23cos 12π cos π4 using the following identity: cos a cos b cos(a b) cos(a b) .25. At the angle θ a.3233π41 343 24c.d. 1, which of the following has the largest value?b. sec θc. cos θd. sin θand cos θ 0 , find tan θ .25b.52c. 325d. 927. Which value of θ causes tan θ to be undefined?ππa. 0b. 6πc. 2d. 44 sin ( 2 x 1)c. f ( x) ( x) sin( x 1)d. f 28. Which of the following has a phase shift of exactly zero?f ( x) 3sin(2 x 2)a. b. f ( x)(13cos(2π x) 3)29. Simplify the following: tan sin 1 1 .a.π6 2 π3b.c. 3d.3330. If cot θ y and cos θ w , express csc θ in terms of y and w.a.yww1b. yc. w31d. y
31. Choose the function that matches the following graph:42 5π 4π 3π 2π ππ2π3π4π5π6π 2 4b. f ( x ) 2cos ( x ) 1 a. f ( x ) 2sin(2 x ) 1( )( )c. f ( x ) 2sin 2x 1d. f ( x ) 2cos 2x 132. Which identity below is not true?a. 2 csc(2θ ) sec θ csc θb. cot 2 θ 1csc θ 1d. tan(2θ ) c. cos 4 (θ ) 2sin 2 (θ ) cos 2 (θ ) sin 4 (θ ) 12sin θ cos θcos 2 θ 1233. If two strings are tied to the top of a pole and each is stretched tight so that it connects to level ground in a straightline, each string would create a hypotenuse of a right triangle. One string is longer than the other, so they formdifferent angles with the ground. The shorter string hits the ground 18 feet from the pole’s base and forms an angle of45º with the ground. The longer string forms an angle of θ with the ground. Which expression gives the length of thelonger string in feet?a. 9 2 sin θb. 9sin θc. 18cscθd. 18 tan θ(θ ) 3sin(2θ 1) 4 ?34. What is the period of the function f a. 4π1b. 3c. 335. For all values of x for which the expression is defined, cos 1 x 1c. sec xa. cos xb. ( cos x )d. πd. none of these36. A triangle contains an angle A which measures 34π , and an angle B which measures π6 . The length of the sideopposite angle A measures 12 cm. What is the length of the side opposite angle B ?b. 12 cmc. 6 3 cma. 6 2 cm4d. 15 cm
Answers to Trigonometry Topics Accuplacer .29.30.31.32.33.34.35.36.adbbdacbdacbcddaSolutions to Trigonometry Topics Accuplacer Review1we should find sin θ . Using sin 2 θ cos 2 θ 1 , and substituting the known value, we havesin θ2sin 2 θ ( 54 ) 1 . Solving for sin θ we get sin θ 53 . Since tan θ is negative and cos θ is positive, θ is in1. Since csc θ quadrant IV, where sin θ 0 . So sin θ 53 , and thus csc θ 53 .2. We can use tan θa hypotenuse of length of opposite sidelength of adjacent side54to imagine a right triangle with legs of length 5 and 4 units, and henceSo then sin θ52 4 2 41 . length of opposite sidelength of hypotenuse541.3. Using the identity that 1 tan 2 θ sec 2 θ , we get 1 ( 3) sec 2 θ and thus sec θ is either2Since θ is said to be between 0 andπ210 or 10 ., then cos θ must be positive and thus its reciprocal sec θ must also bepositive, hence sec θ 10 .4.tan θ is negative in quadrant II and quadrant IV and tan θˆ 3 when the reference angle, θˆ π3 . Thus, thequadrant II angle with a reference angle of θˆ π3 would beangle of θˆ π3 would beθ 53π. The two solutions areθ 23πθ 23π, and the quadrant IV angle with a referenceθ 53πand.5. The inverse sine is a function, so that for each value in the domain there is one and only one value in the range. The π π domain of the inverse sine function is [ 1,1] , and the range is , . The correct answer is in quadrant IV, but 2 2 must be written as 61 π instead of116 πto be in the range.5
6. The identity needed is sin 2θ 2sin θ cos θ , but we also need the value of cos θ . Since 0 θ π , cos θ must291be positive. Then we can find cos θ using the identity sin θ cos θ 1 . So ( 103 ) cos θ 1 and cos 2 θ 1002which means cos θ sin 2θ. Therefore911022( )2 ( 103 ) 10913 91502.7. Since 210 is in quadrant III, sec(210 ) 0 and uses the reference angle 30 . Thus we have that sec(210 ) 1cos 210 (8. Since )1 cos 30 ( ) 1( )3 2( f g ) (π ) ( f ( g (π ) ) ) ,( f g ) (π ) f ( π2 ) 23.we first findπ cos(3π )g (π ) 21 sin ( π2 ) 1 ( 1) 2 . π ( 1) π22so9. We would like to use the identity cos( a b) cos a cos b sin a sin b , so we need the two missing trig values. Toget them we just apply sin 2 θ cos 2 θ 1 to obtainget thatcos(a b) ( 54 )( 54 ) ( 53 )( 53 ) 1625 259 cos a 72545and sin b 35. Substituting these values, we10. Since the amplitude is the magnitude (absolute value) of the coefficient in front of the sin θ or cos θ , we see that ofthese options,11. Since tan θ13 is the smallest. Note that if no value is present, you assume an amplitude of 1.length of opposite sidelength of adjacent sidewe have that tan θ)12. Using the larger right triangle, we have tan(30a 1 3ab. length of opposite sidelength of adjacent sideThen we can use the smaller right triangle, PCB , to get cos(30 )PB 323 21BCPB3to show that BC 3 .( ) ( 3)and hence 2 PB 2or 1.5.13. First we use the value of csc θ to see that sin θ 34 and then the identity sin 2 θ cos 2 θ 1 to show thatcos θ 1 ( 34 ) 2cos θ 7474. Since the angle is in quadrant III, it has to be the negative option, so.14. Factoring the equation, since it is quadratic in form, we get ( tan θ 1)( tan θ 1) 0 , so tan θ 1 . tan θ ispositive in quadrant I and quadrant III and tan θˆ 1 when the reference angle, θˆ π4 . Thus, the quadrant I angleθ π4 , and the quadrant III angle with a reference angle of θˆ π4. The two solutions are θ π4 and θ 54π .with a reference angle of θˆ π4 would bewould beθ 54π15. Comparing the values, we have sin π6 12, cos 4π 1 , tanπcsc 4 has the largest value.62π3 3 , and csc π4 2 . Since2 1so
16. Since extreme values of sin(3 x 2) are 1 , we get the extreme values of the given function as13and13(1) 1 23( 1) 1 34 . Therefore the function’s maximum value is 23 .17. Using the Pythagorean Theorem, we get the missing 3rd side of the triangle ascot θ length of adjacent side2 145 .length of opposite side we get18. Since cos x we have that sec x 3553 1and therefore x sec92 52 56 2 14 . Then since( 53 ) .19. Since the graphs of cos 𝑥 and sin 𝑥 each have a maximum value of 1, the maximum value of 2 sin x is 2 1.4 .The maximum value of 13 cos(2 x ) 2 is 13 2 2.3 . The maximum value of sin(5 x 1) is 1. The maximum value of y5sin x 4 is 5 4 1 . Therefore1 cos(2 x ) 23has the largest maximum value.20. Consider the right triangle created by the man, his shadow and the distance from the top of his head to the edge of thelength of opposite sideheightheightshadow as the hypotenuse. Then we can say that tan θ length of 4.5 . Therefore his heightadjacent sideshadow lengthis 4.5tan θ feet.21. Since the arc length is the radius multiplied by the angle in radians, we get arc length3π (16.4)43 π (4.1) 12.3πcm.( )( )3 1 14 3 14 sin ( x π2 ) 3 1 14 . So 2 43 3 14 sin ( x π2 ) 3 14 , which gives us the range [ 2.75, 3.25] .22. For all values of θ , we know 1 sin θ 1 , so 1 sin x π2 1 . Then 1 14 14 sin x π2 1 14 , and finally( ) is the angle θ in the interval π θ π23. Working from the inside out, sin 1 32 π . We then find csc ( ) ( ) (sin 1 π3231sin π332)2such that sin θ 32. So 1 3 2 . 3224. Substituting the given values into the identity, we get()(π π cos π πcos 1241242)( )( )cos π3 cos π6 21 2 322 14 341 34 .25. Substituting the value of the angle into each of the choices, we see that tansec( 34π ) 1cos 34π( ) 1 22 2 , cos( 34π ) 22, and sin( 34π ) 22( 34π ) 1 ,. Therefore sin θ has the largest valuesince it is the only one that is positive.26. Using sin 2 θ cos2 θ 1 , and substituting the known value, we havecosθ 27. tan θ 53. Since it is stated that cos θ 0 , it must be 53( 23 )2 cos2 θ 1 . Solving for cosθ we get. Then we can find tan θ sin θ2. 3 cosθ 5532sin θis undefined when the denominator is zero, so since cos π2 0 , tan θ is undefined when θ cosθ7π2.
28. Since the phase shift is determined by a value being added or subtracted in the argument of the trigonometric function, and f ( x ) 13 cos(2π x ) 3 has no value in the argument being added or subtracted, this gives our answer. You couldinput a “ 0” next to the x to see that it has zero phase shift.( 12 ) is the angle θ in the interval π2 θ π2sin31.sin 1 ( 12 ) π6 . We then find tan ( π6 ) cos(( )) 3329. Working from the inside out, sin 1π6π61 sin θ1 wy12. So1232wcosθw y . Solving for sin θ gives y , so we can write sin θ , so we haveysin θsin θθ 30. We have cosθ w and cotθcsc such that sin θ y.w31. From the graph, we can estimate the maximum function value is 1 and the minimum function value is 3 . This tellsus the function has a midline at y 1 and an amplitude of 2. The period of the function appears to be 4π . The onlychoice with a midline at y 1 , an amplitude of 2, and a period of 4π is f ( x ) 2sin ( 2x ) 1 , so this is the correctfunction for the given graph.32. On this problem, you can find the solution by finding equivalent statements to until you see an identity you know or a1contradiction. In this case, we are looking for the contradiction. The contradiction is cot 2 θ 2csc θ 12211sin θsin θtan 2 θ , which is not the same because the right side can be re-written as21 122csc θ 11 sin θ cos θ2sin θas cot θ . (You can verify cot θ and tan θ are not the same by noticing, for example, that cot 22tan 222( π6 ) 3 and( π6 ) 13 , which are different.) You can also show the other three choices are true statements using trigonometricidentities.33. The triangle formed by the shorter string tells us the pole is 18 feet tall since any right triangle containing a 45º anglehas legs of the same length. Then the longer string makes a right triangle with the 18-foot pole opposite the angle θ ,length of opposite sideheight of pole18 and we get that the longer string is 18 18cscθ feetso we can write sin θ length of stringsin θhypotenuselength of stringlong.34. The period of sin θ is 2π . The only transformation that affects the period is the horizontal stretch (or shrink). In thiscase, the coefficient of 2 on the θ would shrink the graph horizontally (and thus shrink the period) so that thefunction’s period is now π .35. The expression cos 1 x refers to the inverse cosine function, which gives the angle θ in the interval 0 θ π suchthat cosθ x . None of the other given expressions mean the same thing, so the answer is “none of these”.36. We can use Law of Sines to find b , the side opposite angle B . We haveget b( )12sin π6 12 12 2sin 34π212 6 2 cm.28sin 34π12 sin π6b. When we solve for b , we
Trigonometry Topics Accuplacer Review – revised July 2016 . You will not be allowed to use a calculator on the Accuplacer
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