Trigonometry Topics Accuplacer Revie revised July 0 You ill not be alloed to use a calculator on the Accuplacer Trigonometry test For more information, see the JCCC Testing Services ebsite at http://jcccedu/testing/ or call 9-9- 9 If cos and tanθ < 0, find the value for cscθ c b Find the value of if θ is acute and tan b c If 0 θ and tan, find the value of secθ b 0 c 0 Give all values of θ beteen 0 and for hich tan Evaluate sin b b and c c and If = and 0, find the value of sin( θ ) 0 θ 0 b 0 c 9 0 9 00 What is the exact value for sec(0 )? b c cos( θ) 8 If f( θ) = and g( θ) = then hat is ( f g)( )? b c 0 9 If sin a = and cosb =, and e kno a and b are acute angles, find cos( a+ b) b 0 c 0 Choose the function belo ith the smallest amplitude: f ( θ) = sin( θ) b g( θ) cos(θ ) = + c f ( θ) sin ( θ) = g( θ) = sin( θ )
If ABC is a right triangle, ith ACB as the right angle and θ as BAC, find tanθ a c b a b c b a c b In the folloing picture, ABC = 0 and AC = Find the length of PB b c Find the value for cosθ if θ lies in the rd Quadrant and csc b c Given tan θ tanθ + = 0, find all solutions for 0 θ < 0 b c and and Which of the folloing has the largest value? sin b cos c tan csc Find the maximum value of the function f( x) = sin(x) b c If an acute angle θ is part of a right triangle here the side opposite of θ is units long and the hypotenuse is 9 units long, hat is cotθ? b 8 Find the value for the angle x: 9 c tan b sec c cos csc
9 Which of the folloing has the highest maximum value on its graph? y = sin x b y = cos( x) + c y = sin( x ) y = sin x 0 If the angle of elevation of the sun is θ degrees, and a man s shado along level ground is feet long, hich expression gives the height of the man in feet? b tanθ c cotθ cscθ A circle has a radius cm Find the length of the arc cut by a central angle of cm b cm c cm cm What is the range of the function f( x) sin ( x ) (, ) b [, ] Simplify csc sin ( ) b = +? c (, ) c [, ] cos cos Simplify b using the folloing identity: + cos( a+ b) + cos( ab) cos acosb = c + At the angle, hich of the folloing has the largest value? tanθ b secθ c cosθ If = and cosθ < 0, find tanθ b c 9 Which value of θ causes tanθ to be undefined? 0 b c 8 Which of the folloing has a phase shift of exactly zero? f( x) = sin(x ) b f( x) cos( x) = c f( x) sin ( x ) = + f( x) = sin( x ) 9 Simplify the folloing: tan( sin ) b c 0 If cot y and cos, express cscθ in terms of y and y b y c y
Choose the function that matches the folloing graph: f( x) = sin( x) + b f( x) =cos( x) c f( x ) = sin x x f( x ) = cos + Which identity belo is not true? csc( θ) = secθ cscθ b c cos ( θ) sin ( θ)cos ( θ) sin ( θ) cot θ csc cosθ + + = tan( θ ) = cos θ If to strings are tied to the top of a pole and each is stretched tight so that it connects to level ground in a straight line, each string ould create a hypotenuse of a right triangle One string is longer than the other, so they form different angles ith the groun The shorter string hits the ground 8 feet from the pole s base and forms an angle of º ith the groun The longer string forms an angle of θ ith the groun Which expression gives the length of the longer string in feet? 9 b 9 c 8cscθ 8tanθ What is the period of the function f ( θ) = sin(θ + ) +? b c For all values of x for hich the expression is defined, cos x = cos x b ( cos x) c sec x none of these A triangle contains an angle A hich measures, and an angle B hich measures The length of the side opposite angle A measures cm What is the length of the side opposite angle B? cm b cm c cm cm
Ansers to Trigonometry Topics Accuplacer Revie b c d d d c d 8 a 9 d 0 b b c c d d d a 8 b 9 b 0 b a d b b d a c 8 b 9 d 0 a c b c d d a Solutions to Trigonometry Topics Accuplacer Revie Since cscθ = e should find Using sin θ + cos, and substituting the knon value, e have sin θ + = Solving for e get = ± Since tanθ is negative and cosθ is positive, θ is in quadrant IV, here < 0 So =, and thus csc We can use a hypotenuse of Using the identity that tan = to imagine a right triangle ith legs of length and units, and hence length of opposite side length of adjacent side + = So then tan sec = = length of opposite side length of hypotenuse + θ, e get + = sec θ and thus secθ is either 0 or 0 Since θ is said to be beteen 0 and, then cosθ must be positive and thus its reciprocal secθ must also be positive, hence sec 0 tanθ is negative in quadrant II and quadrant IV and tan ˆ hen the reference angle, ˆ ˆ ould be quadrant II angle ith a reference angle of angle of ˆ ould be The to solutions are and Thus, the, and the quadrant IV angle ith a reference 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 domain of the inverse sine function is [,], and the range is, The correct anser is in quadrant IV, but must be ritten as instead of to be in the range
The identity needed is sin cosθ, but e also need the value of cosθ Since be positive Then e can find cosθ using the identity hich means sin θ cos θ 9 cos 0 Therefore sin = + = So 9 9 0 0 0 0 θ, cosθ must 0 + cos θ= and 9 cos 00 Since 0 is in quadrant III, sec(0 ) < 0 = = ( ) = cos 0 cos 0 and uses the reference angle 0 Thus e have that sec(0 ) = 8 Since ( f g) = ( f ( g )), e first find ( f g) f ( ) = = sin = ( ) = ( ) cos g( ) = = = so 9 We ould like to use the identity cos( a+ b) = cos acosb sin asin b, so e need the to missing trig values To get them e just apply sin θ + cos to obtain cos a = and sin b = Substituting these values, e get that cos( a b) + = = = 9 0 Since the amplitude is the magnitude (absolute value) of the coefficient in front of the or cosθ, e see that of these options, is the smallest Note that if no value is present, you assume an amplitude of Since tan e have that tanθ length of opposite side length of adjacent side Using the larger right triangle, e have tan(0 ) Then e can use the smaller right triangle, PB = or First e use the value of cscθ to see that a = b = = = to sho that a length of opposite side length of adjacent side BC PCB, to get cos(0 ) PB = and then the identity BC = = = and hence sin θ cos θ PB = so + = to sho that cos ± = ± Since the angle is in quadrant III, it has to be the negative option, so cos tanθ is Factoring the equation, since it is quadratic in form, e get ( tanθ )( tanθ ) = 0, so tan positive in quadrant I and quadrant III and tan ˆ hen the reference angle, ˆ ith a reference angle of ˆ ould be, and the quadrant III angle ith a reference angle of ˆ ould be The to solutions are and Thus, the quadrant I angle Comparing the values, e have csc has the largest value sin =, cos =, tan =, and csc = Since >
Since extreme values of sin(x ) are ±, e get the extreme values of the given function as () = and ( ) = Therefore the function s maximum value is Using the Pythagorean Theorem, e get the missing rd side of the triangle as length of adjacent side length of opposite side cot e get 9 = = Then since 8 Since cos x = e have that sec x = and therefore x sec = 9 Since the graphs of cos x and sin x each have a maximum value of, the maximum value of sin x is The maximum value of cos( x ) + is + = The maximum value of sin(x ) is The maximum value of sin x is = Therefore y = cos( x) + has the largest maximum value 0 Consider the right triangle created by the man, his shado and the distance from the top of his head to the edge of the length of opposite side height height shado as the hypotenuse Then e can say that tan = = Therefore his height length of adjacent side shado length is tanθ feet Since the arc length is the radius multiplied by the angle in radians, e get arc length = () = () = cm, so sin( x ) Then ( x ) + + ( x ) + So sin( x ) For all values of θ, e kno sin Working from the inside out, ( ) sin csc sin, and finally, +, hich gives us the range [ ] sin is the angle θ in the interval = We then find sin = = = Substituting the given values into the identity, e get cos + cos ( ) cos + + cos + + = = = + = θ Substituting the value of the angle into each of the choices, e see that ( ) sec( ) = cos = =, cos( ) =, and ( ) since it is the only one that is positive Using sin θ cos θ such that = tan =, + =, and substituting the knon value, e have cos ± Since it is stated that cosθ < 0, it must be So sin = Therefore has the largest value + cos Solving for cosθ e get Then e can find tanθ = = = cosθ tanθ = is undefined hen the denominator is zero, so since cosθ = cos 0, tanθ is undefined hen
8 Since the phase shift is determined by a value being added or subtracted in the argument of the trigonometric function, and f( x) = cos( x) has no value in the argument being added or subtracted, this gives our anser You could input a +0 next to the x to see that it has zero phase shift 9 Working from the inside out, sin ( ) sin = We then find tan 0 We have cos and y cscθ = = = y is the angle θ in the interval sin cos = = = = θ such that = So cosθ cotθ = = y, so e can rite y sin Solving for gives = y sin θ, so e have From the graph, e can estimate the maximum function value is and the minimum function value is This tells us the function has a midline at y = and an amplitude of The period of the function appears to be The only choice ith a midline at y =, an amplitude of, and a period of is f( x ) = sin x ( ), so this is the correct function for the given graph On this problem, you can find the solution by finding equivalent statements to until you see an identity you kno or a contradiction In this case, e are looking for the contradiction The contradiction is cot csc θ sin θ sin θ because the right side can be re-ritten as = = = = tan θ, hich is not the same csc θ sin θ cos θ as cot θ (You can verify cot θ and sin θ tan θ are not the same by noticing, for example, that ( ) cot = and tan =, hich are different) You can also sho the other three choices are true statements using trigonometric identities The triangle formed by the shorter string tells us the pole is 8 feet tall since any right triangle containing a º angle has legs of the same length Then the longer string makes a right triangle ith the 8-foot pole opposite the angle θ, length of opposite side height of pole so e can rite 8 8 = = = and e get that the longer string is = 8cscθ feet length of hypotenuse length of string string long The period of is The only transformation that affects the period is the horizontal stretch (or shrink) In this case, the coefficient of on the θ ould shrink the graph horizontally (and thus shrink the period) so that the function s period is no The expression cos x refers to the inverse cosine function, hich gives the angle θ in the interval 0 θ such that cos x None of the other given expressions mean the same thing, so the anser is none of these We can use La of Sines to find b, the side opposite angle B We have get sin b = = = = cm sin sin sin = When e solve for b, e b 8