Chapte 5: Tigonometic Functions of Angles In the pevious chaptes we have exploed a vaiety of functions which could be combined to fom a vaiety of shapes. In this discussion, one common shape has been missing: the cicle. We aleady know cetain things about the cicle like how to find aea, cicumfeence and the elationship between adius & diamete, but now, in this chapte, we exploe the cicle, and its unique featues that lead us into the ich wold of tigonomety. Section 5. Cicles... 99 Section 5. Angles... 04 Section 5. Points on Cicles using Sine and Cosine... 5 Section 5.4 The Othe Tigonometic Functions... 5 Section 5.5 Right Tiangle Tigonomety... Section 5. Cicles To begin, we need to emembe how to find distances. Stating with the Pythagoean Theoem, which elates the sides of a ight tiangle, we can find the distance between two points. The Pythagoean Theoem states that the sum of the squaes of the legs of a ight tiangle will equal the squae of the hypotenuse of the tiangle. In gaphical fom, given the tiangle shown, a + b = c a b c We can use the Pythagoean Theoem to find the distance between two points on a gaph. Example Find the distance between the points (-, ) and (, 5) By plotting these points on the plane, we can then daw a tiangle between them. We can calculate hoizontal width of the tiangle to be 5 and the vetical height to be. Fom these we can find the distance between the points using the Pythagoean Theoem: dist = 5 dist = 4 + = 4 This chapte is pat of Pecalculus: An Investigation of Functions Lippman & Rasmussen 0. This mateial is licensed unde a Ceative Commons CC-BY-SA license.
00 Chapte 5 Notice that the width of the tiangle was calculated using the diffeence between the x (input) values of the two points, and the height of the tiangle was found using the diffeence between the y (output) values of the two points. Genealizing this pocess gives us the geneal distance fomula. Distance Fomula The distance between two points x, ) and x, ) can be calculated as dist = ( x y x ) + ( y ) ( y ( y Ty it Now. Find the distance between the points (, 6) and (, -5) Cicles If we wanted to find an equation to epesent a cicle with a adius of centeed at a point (h, k), we notice that the distance between any point (x, y) on the cicle and the cente point is always the same:. Noting this, we can use ou distance fomula to wite an equation fo the adius: = ( x h) + ( y k ) (h, k) (x, y) Squaing both sides of the equation gives us the standad equation fo a cicle. The equation of a cicle centeed at the point (h, k) with adius can be witten as ( x h) + ( y k) = Notice a cicle does not pass the vetical line test. It is not possible to wite y as a function of x o vice vesa. Example Wite an equation fo a cicle centeed at the point (-, ) with adius 4 Using the equation fom above, h = -, k =, and the adius = 4. Using these in ou fomula, ( x ( )) + ( y ) = 4 simplified a bit, this gives ( x + ) + ( y ) = 6
Section 5. Cicles 0 Example Wite an equation fo the cicle gaphed hee. This cicle is centeed at the oigin, the point (0, 0). By measuing hoizontally o vetically fom the cente out to the cicle, we can see the adius is. Using this infomation in ou fomula gives: ( x 0) + ( y 0) = simplified a bit, this gives x + y = 9 Ty it Now. Wite an equation fo a cicle centeed at (4, -) with adius 6 Notice that elative to a cicle centeed at the oigin, hoizontal and vetical shifts of the cicle ae evealed in the values of h and k, which is the location of the cente of the cicle. Points on a Cicle As noted ealie, the equation fo a cicle cannot be witten so that y is a function of x o vice vesa. To elate x and y values on the cicle we must solve algebaically fo the x and y values. Example 4 Find the points on a cicle of adius 5 centeed at the oigin with an x value of. We begin by witing an equation fo the cicle centeed at the oigin with a adius of 5. x + y = 5 Substituting in the desied x value of gives an equation we can solve fo y + y = 5 y y = ± = 5 9 = 6 6 = ± 4 Thee ae two points on the cicle with an x value of : (, 4) and (, -4) Example 5 In a town, Main Steet uns east to west, and Meidian Road uns noth to south. A pizza stoe is located on Meidian miles south of the intesection of Main and Meidian. If the stoe advetises that it delives within a mile adius, how much of Main Steet do they delive to?
0 Chapte 5 This type of question is one in which intoducing a coodinate system and dawing a pictue can help us solve the poblem. We could eithe place the oigin at the intesection of the two steets, o place the oigin at the pizza stoe itself. It is often easie to wok with cicles centeed at the oigin, so we ll place the oigin at the pizza stoe, though eithe appoach would wok fine. Placing the oigin at the pizza stoe, the delivey aea with adius miles can be descibed as the egion inside the cicle descibed by x + y = 9. Main Steet, located miles noth of the pizza stoe and unning east to west, can be descibed by the equation y =. To find the potion of Main Steet the stoe will delive to, we fist find the bounday of thei delivey egion by looking fo whee the delivey cicle intesects Main Steet. To find the intesection, we look fo the points on the cicle whee y =. Substituting y = into the cicle equation lets us solve fo the coesponding x values. x x + x = ± = 9 = 9 4 = 5 5 ±.6 This means the pizza stoe will delive.6 miles down Main Steet east of Meidian and.6 miles down Main Steet west of Meidian. We can conclude that the pizza stoe delives to a 4.47 mile segment of Main St. In addition to finding whee a vetical o hoizontal line intesects the cicle, we can also find whee any abitay line intesects a cicle. Example 6 Find whee the line f ( x) = 4x intesects the cicle ( x ) + y = 6. Nomally to find an intesection of two functions f(x) and g(x) we would solve fo the x value that would make the function equal by solving the equation f(x) = g(x). In the case of a cicle, it isn t possible to epesent the equation as a function, but we can utilize the same idea. The output value of the line detemines the y value: y = f ( x) = 4x. We want the y value of the cicle to equal the y value of the line which is the output value of the function. To do this, we can substitute the expession fo y fom the line into the cicle equation.
Section 5. Cicles 0 ( x ) + y = 6 we eplace y with the line fomula: y = 4x ( x ) + (4x) = 6 expand x 4x + 4 + 6x = 6 and simplify 7x 4x + 4 = 6 since this equation is quadatic, we aange it to be = 0 7x 4x = 0 Since this quadatic doesn t appea to be factoable, we can use the quadatic equation to solve fo x: ( 4) ± ( 4) 4(7)( ) 4 ± 8 x = =, o appoximately x = 0.966 o -0.7 (7) 4 Fom these x values we can use eithe equation to find the coesponding y values. Since the line equation is easie to evaluate, we might choose to use it: y = f (0.966) = 4(0.966) =.864 y = f ( 0.7) = 4( 0.7) =.9 The line intesects the cicle at the points (0.966,.864) and (-0.7, -.9) Ty it Now. A small adio tansmitte boadcasts in a 50 mile adius. If you dive along a staight line fom a city 60 miles noth of the tansmitte to a second city 70 miles east of the tansmitte, duing how much of the dive will you pick up a signal fom the tansmitte? Impotant Topics of This Section Distance fomula Equation of a Cicle Finding the x coodinate of a point on the cicle given the y coodinate o vice vesa Finding the intesection of a cicle and a line Ty it Now Answes. 5 5. ( x 4) + ( y + ) = 6. x + ( 60 60 / 70x) = 50 gives x = 4 o x = 45.9 coesponding to points (4,48) and (45.9,.8), with a distance between of 4. miles.
04 Chapte 5 Section 5. Angles Since so many applications of cicles involve otation within a cicle, it is natual to intoduce a measue fo the otation, o angle, between two lines emanating fom the cente of the cicle. The angle measuement you ae most likely familia with is degees, so we ll begin thee. An angle is the measue between two lines, line segments o ays that shae a stating point but have diffeent end points. It is a otational measue not a linea measue. Measuing Angles A degee is a measuement of angle. One full otation aound the cicle is equal to 60 degees, so one degee is /60 of a cicle. An angle measued in degees should always include the unit degees afte the numbe, o include the degee symbol. Fo example, 90 degees = 90 When measuing angles on a cicle, unless othewise diected we measue angles in standad position: measued stating at the positive hoizontal axis and with counteclockwise otation. Example Give the degee measue of the angle shown on the cicle. The vetical and hoizontal lines divide the cicle into quates. Since one full otation is 60 degees= 60, each quate otation is 60/4 = 90 o 90 degees. Example Show an angle of 0 on the cicle. An angle of 0 is / of 90, so by dividing a quate otation into thids, we can sketch a line at 0.
Section 5. Angles 05 Going Geek When epesenting angles using vaiables, it is taditional to use Geek lettes. Hee is a list of commonly encounteed Geek lettes. θ ϕ o φ α β γ theta phi alpha beta gamma Woking with Angles in Degees Notice that since thee ae 60 degees in one otation, an angle geate than 60 degees would indicate moe than full otation. Shown on a cicle, the esulting diection in which this angle points would be the same as anothe angle between 0 and 60 degees. These angles would be called coteminal. Afte completing thei full otation based on the given angle, two angles ae coteminal if they teminate in the same position, so they point in the same diection. Example Find an angle θ that is coteminal with 800, whee 0 θ < 60 Since adding o subtacting a full otation, 60 degees, would esult in an angle pointing in the same diection, we can find coteminal angles by adding o subtacting 60 degees. An angle of 800 degees is coteminal with an angle of 800-60 = 440 degees. It would also be coteminal with an angle of 440-60 = 80 degees. The angle θ = 80 is coteminal with 800. By finding the coteminal angle between 0 and 60 degees, it can be easie to see which diection an angle points in. Ty it Now. Find an angle α that is coteminal with 870, whee 0 α < 60 On a numbe line a positive numbe is measued to the ight and a negative numbe is measued in the opposite diection to the left. Similaly a positive angle is measued counteclockwise and a negative angle is measued in the opposite diection, clockwise.
06 Chapte 5 Example 4 Show the angle 45 on the cicle and find a positive angleα that is coteminal and 0 α < 60 Since 45 degees is half of 90 degees, we can stat at the positive hoizontal axis and measue clockwise half of a 90 degee angle. Since we can find coteminal angles by adding o subtacting a full otation of 60 degees, we can find a positive coteminal angle hee by adding 60 degees: 45 + 60 = 5 5-45 Ty it Now. Find an angle β is coteminal with 00 whee 0 β < 60 It can be helpful to have a familiaity with the commonly encounteed angles in one otation of the cicle. It is common to encounte multiples of 0, 45, 60, and 90 degees. The common values ae shown hee. Memoizing these angles and undestanding thei popeties will be vey useful as we study the popeties associated with angles 80 50 0 0 5 40 Angles in Radians While measuing angles in degees may be familia, doing so often complicates mattes since the units of measue can get in the way of calculations. Fo this eason, anothe measue of angles is commonly used. This measue is based on the distance aound a cicle. 90 70 60 00 45 5 5 0 0 0 Aclength is the length of an ac, s, along a cicle of adius subtended (dawn out) by an angleθ. θ s
Section 5. Angles 07 The length of the ac aound an entie cicle is called the cicumfeence of a cicle. The cicumfeence of a cicle is C = π. The atio of the cicumfeence to the adius, poduces the constant π. Regadless of the adius, this constant atio is always the same, just as how the degee measue of an angle is independent of the adius. To expand this idea, conside two cicles, one with adius and one with adius. Recall the cicumfeence (peimete) of a cicle is C = π, whee is the adius of the cicle. The smalle cicle then has cicumfeence π () = 4π and the lage has cicumfeence π () = 6π. Dawing a 45 degee angle on the two cicles, we might be inteested in the length of the ac of the cicle that the angle indicates. In both cases, the 45 degee angle daws out an ac that is /8 th of the full cicumfeence, so fo the smalle cicle, the aclength = (4 π ) = π, and fo the lage cicle, the length of the ac o aclength 8 = (6 π ) = π. 8 4 Notice what happens if we find the atio of the aclength divided by the adius of the cicle: π Smalle cicle: = π 4 π Lage cicle: 4 = π 4 The atio is the same egadless of the adius of the cicle it only depends on the angle. This popety allows us to define a measue of the angle based on aclength. A adian is a measuement of angle. It descibes the atio of a cicula ac to the adius of the cicle. In othe wods, if s is the length of an ac of a cicle, and is the adius of the cicle, then s adians = Radians also can be descibed as the length of an ac along a cicle of adius, called a unit cicle.
08 Chapte 5 Since adians ae the atio of two lengths, they ae a unitless measue. It is not necessay to wite the label adians afte a adian measue, and if you see an angle that is not labeled with degees o the degee symbol, you should assume that it is a adian measue. Consideing the most basic case, the unit cicle, o a cicle with adius, we know that otation equals 60 degees, 60. We can also tack one otation aound a cicle by finding the cicumfeence, C = π, and fo the unit cicle C = π. These two diffeent ways to otate aound a cicle give us a way to convet fom degees to the length of the ac aound a cicle, o the cicumfeence. otation = 60 = π adians ½ otation = 80 = π adians ¼ otation = 90 = π / adians Example 5 Find the adian measue of a d of a full otation. Fo any cicle, the aclength along a thid otation would be a thid of the cicumfeence, π C = (π ) =. The adian measue would be the aclength divided by the adius: π π adians = = Conveting between adians and degees. π degee = 80 adians π adians o: to convet fom degees to adians, multiply by 80 adian = 80 π degees o: to convet fom adians to degees, multiply by 80 π adians
Section 5. Angles 09 Example 6 Convet 6 π adians to degees Since we ae given a poblem in adians and we want degees, we multiply by When we do this the adians cancel and ou units become degees. To convet to adians, we can use the convesion fom above π π 80 adians = = 0 degees 6 6 π 80 π Example 7 Convet 5 degees to adians In this example we stat with degees and want adians so we use the othe π convesion so that the degee units cancel and we ae left with the unitless measue 80 of adians. π π 5 degees = 5 = 80 Ty it Now 7π. Convet adians to degees 0 Just as we listed all the common angles in degees on a cicle, we should also list the coesponding adian values fo the common measues of a cicle coesponding to degee multiples of 0, 45, 60, and 90 degees. As with the degee measuements, it would be advisable to commit these to memoy. π 5π 6 π 4 π π π π 4 π 6 0, π We can wok with the adian measues of an angle the same way we wok with degees. 7π 6 5π 4 4π π 5π π 6 7π 4
0 Chapte 5 Example 8 Find an angle β that is coteminal with 9, whee 0 β < π 4π When woking in degees, we found coteminal angles by adding o subtacting 60 degees a full otation. Likewise in adians, we can find coteminal angles by adding o subtacting full otations of π adians. 9 π 9 8 π = π π = π 4 4 4 4 The angle 4π is coteminal, but not less than π, so we subtact anothe otation. π 8 π = π π = π 4 4 4 4 The angle π 9π is coteminal with 4 4 Ty it Now 4. Find an angle φ that is coteminal with 7π whee 0 φ < π 6 Aclength and Aea of a Secto Recall that the adian measue of an angle was defined as the atio of the aclength of a s cicula ac to the adius of the cicle, θ =. Fom this elationship, we can find aclength along a cicle fom the angle. Aclength on a cicle The length of an ac, s, along a cicle of adius subtended by angleθ in adians is s = θ θ s Example 9 Mecuy obits the sun at a distance of appoximately 6 million miles. In one Eath day, it completes 0.04 otation aound the sun. If the obit was pefectly cicula, what distance though space would Mecuy tavel in one Eath day?
Section 5. Angles To begin, we will need to convet the decimal otation value to a adian measue. Since one otation = π adians, 0.04 otation = π (0.04) = 0.076 adians. Combining this with the given adius of 6 million miles, we can find the aclength: s = (6)(0.076) =.578 million miles tavelled though space. Ty it Now 5. Find the aclength along a cicle of adius 0 subtended by an angle of 5 degees. In addition to aclength, we can also use angles to find the aea of a secto of a cicle. A secto is a potion of a cicle between two lines fom the cente, like a slice of pizza o pie. Recall that the aea of a cicle with adius can be found using the fomula A = π. If a secto is dawn out by an angle of θ, measued in adians, then the faction of full cicle that angle has dawn out is, since π is one full otation. Thus, the aea of the θπ secto would be this faction of the whole aea: θ θπ Aea of secto = π = = θ π π The aea of a secto of a cicle with adius subtended by an angle θ, measued in adians, is Aea of secto = θ Example 0 An automatic lawn spinkle spays a distance of 0 feet while otating 0 degees. What is the aea of the secto the spinkle coves? Fist we need to convet the angle measue into adians. Since 0 degees is one of ou common angles, you ideally should aleady know the equivalent adian measue, but if not we can convet: π π 0 degees = 0 = adians. 80 6 The aea of the secto is then Aea π = = 6 (0) 04.7 ft
Chapte 5 Ty it Now 6. In cental pivot iigation, a lage iigation pipe on wheels otates aound a cente point, as pictued hee. A fame has a cental pivot system with a adius of 400 metes. If wate estictions only allow he to wate 50 thousand squae metes a day, what angle should she set the system to cove? Linea and Angula Velocity When you ca dives down a oad, it makes sense to descibe its speed in tems of miles pe hou o metes pe second, these ae measues of speed along a line, also called linea velocity. When a cicle otates, we would descibe its angula velocity, o otational speed, in adians pe second, otations pe minute, o degees pe hou. As a point moves along a cicle of adius, its angula velocity, ω, can be found as the angula otation θ pe unit time, t. θ ω = t The linea velocity, v, of the point can be found as the distance tavelled, aclength s, pe unit time, t. s v = t Example A wate wheel completes otation evey 5 seconds. Find the angula velocity in adians pe second. The wheel completes otation = π adians in 5 seconds, so the angula velocity π would be ω =.57 adians pe second 5 Combining the definitions above with the aclength equation, s = θ, we can find a elationship between angula and linea velocities. The angula velocity equation can be solved fo θ, giving θ = ωt. Substituting this into the aclength equation gives s = θ = ωt. http://commons.wikimedia.og/wiki/file:pivot_otech_00.jpg CC-BY-SA
Section 5. Angles Substituting this into the linea velocity equation gives s ωt v = = = ω t t Relationship between linea and angula velocity When the angula velocity is measued in adians pe unit time, linea velocity and angula velocity ae elated by the equation v = ω Example A bicycle has wheels 8 inches in diamete. The tachomete detemines the wheels ae otating at 80 RPM (evolutions pe minute). Find the speed the bicycle is tavelling down the oad. Hee we have an angula velocity and need to find the coesponding linea velocity, since the linea speed of the outside of the ties is the speed at which the bicycle tavels down the oad. We begin by conveting fom otations pe minute to adians pe minute. It can be helpful to utilize the units to make this convesion otations adians adians 80 π = 60π minute otation minute Using the fomula fom above along with the adius of the wheels, we can find the linea velocity v adians inches (4 inches) 60 π = = 5040 minute π minute You may be wondeing whee the adians went in this last equation. Remembe that adians ae a unitless measue, so it is not necessay to include them. Finally, we may wish to convet this linea velocity into a moe familia measuement, like miles pe hou. inches feet mile 60 minutes 5040 π 4.99 minute inches 580 feet hou = miles pe hou (mph) Ty it Now 7. A satellite is otating aound the eath at 7,59 kilometes pe hou at an altitude of 4 km above the eath. If the adius of the eath is 678 kilometes, find the angula velocity of the satellite.
4 Chapte 5 Impotant Topics of This Section Degee measue of angle Radian measue of angle Convesion between degees and adians Common angles in degees and adians Coteminal angles Aclength Aea of a secto Linea and angula velocity Ty it Now Answes. α =50. β = 60. 6 7π 4. 6 5π 5. 7. 55 8 6. 07.4 7. 4.8 adians pe hou
Section 5. Points on Cicles using Sine and Cosine 5 Section 5. Points on Cicles using Sine and Cosine While it is convenient to descibe the location of a point on a cicle using the angle o distance along the cicle, elating this infomation to the x and y coodinates and the cicle equation we exploed in section 5. is an impotant application of tigonomety. A distess signal is sent fom a sailboat duing a stom, but the tansmission is unclea and the escue boat sitting at the maina cannot detemine the sailboat s location. Using high poweed ada, they detemine the distess signal is coming fom a distance of 0 miles at an angle of 5 degees fom the maina. How many miles east/west and noth/south of the escue boat is the standed sailboat? In a geneal sense, to investigate this, we begin by dawing a cicle centeed at the oigin with adius, and making the point on the cicle indicated by some angle θ. This point has coodinates (x, y). If we dop a line vetically down fom this point to the x axis, we would fom a ight tiangle inside of the cicle. No matte which quadant ou adius and angle θ put us in we can daw a tiangle by dopping a pependicula line to the axis, keeping in mind that the value of x & y change sign as the quadant changes. θ (x, y) Additionally, if the adius and angle θ put us on the axis, we simply measue the adius as the x o y with the coesponding value being 0, again ensuing we have appopiate signs on the coodinates based on the quadant. Tiangles obtained with diffeent adii will all be simila tiangles, meaning all the sides scale popotionally. While the lengths of the sides may change, the atios of the side lengths will always emain constant fo any given angle. To be able to efe to these atios moe easily, we will give them names. Since the atios depend on the angle, we will wite them as functions of the angle θ. Fo the point (x, y) on a cicle of adius at an angle of θ, we can define two impotant functions as the atios of the sides of the coesponding tiangle: The sine function: The cosine function: sin(θ ) = cos(θ ) y = x θ x (x, y) y
6 Chapte 5 In this chapte, we will exploe these functions on the cicle and on ight tiangles. In the next chapte we will take a close look at the behavio and chaacteistics of the sine and cosine functions. Example The point (, 4) is on the cicle of adius 5 at some angle θ. Find cos(θ ) and sin(θ ). Knowing the adius of the cicle and coodinates of the point, we can evaluate the cosine and sine functions as the atio of the sides. x y 4 cos( θ ) = = sin( θ ) = = 5 5 Thee ae a few cosine and sine values which we can detemine faily easily because they fall on the x o y axis. Example Find cos( 90 ) and sin( 90 ) On any cicle, a 90 degee angle points staight up, so the coodinates of the point on the cicle would be (0, ). Using ou definitions of cosine and sine, x 0 cos( 90 ) = = = 0 y sin( 90 ) = = = (0, ) 90 Ty it Now. Find cosine and sine of the angle π Notice that the definitions above can also be stated as: On a cicle of adius at an angle of θ, we can find the coodinates of the point (x, y) at that angle using x = cos(θ ) y = sin(θ ) On a unit cicle, a cicle with adius, x = cos(θ ) and y = sin(θ )
Section 5. Points on Cicles using Sine and Cosine 7 Utilizing the basic equation fo a cicle centeed at the oigin, x + y =, combined with the elationships above, we can establish a new identity. x + y = substituting the elations above, θ )) + ( sin( θ )) simplifying, ( cos( = (cos( θ )) + (sin( θ )) = dividing by (cos( θ )) + (sin( θ )) = o using shothand notation cos ( θ ) + sin ( θ ) = Hee cos ( θ ) is a commonly used shothand notation fo (cos(θ )). Be awae that many calculatos and computes do not undestand the shothand notation. In 5. we elated the Pythagoean Theoem a + b = c to the basic equation of a cicle x + y = and now we have used that equation to identify the Pythagoean Identity. The Pythagoean Identity. Fo any angle, cos ( θ ) + sin ( θ ) = One use of this identity is that it allows us to find a cosine value if we know the sine value o vice vesa. Howeve, since the equation will give two possible solutions, we will need to utilize additional knowledge of the angle to help us find the desied solution. Example If sin( θ ) = and θ is in the second quadant, find cos(θ ). 7 Substituting the known value fo sine into the Pythagoean identity, cos ( θ ) + = 7 9 cos ( θ ) + = 49 40 cos ( θ ) = 49 40 40 cos( θ ) = ± = ± 49 7 Since the angle is in the second quadant, we know the x value of the point would be negative, so the cosine value should also be negative. Using this additional infomation, we can conclude that cos( θ ) = 40 7
8 Chapte 5 Values fo Sine and Cosine At this point, you may have noticed that we haven t found any cosine o sine values using angles not on an axis. To do this, we will need to utilize ou knowledge of tiangles. Fist, conside a point on a cicle at an angle of 45 degees, o 4 π. At this angle, the x and y coodinates of the coesponding point on the cicle will be equal because 45 degees divides the fist quadant in half and the x and y values will be the same, so the sine and cosine values will also be equal. Utilizing the Pythagoean Identity, π cos 4 + π sin = 4 π π cos + cos = 4 4 π cos = 4 cos π = 4 since the sine and cosine ae equal, we can substitute sine with cosine add like tems divide since the x value is positive, we ll keep the positive oot π cos = 4 often this value is witten with a ationalized denominato Remembe, to ationalize the denominato we multiply by a tem equivalent to to get id of the adical in the denominato. π cos = = = 4 4 π Since the sine and cosine ae equal, sin = as well. 4 The (x, y) coodinates fo a cicle of adius and angle of 45 degees =,
Section 5. Points on Cicles using Sine and Cosine 9 Example 4 Find the coodinates of the point on a cicle of adius 6 at an angle of 4 π. π π Using ou new knowledge that sin = and cos =, along with ou 4 4 elationships that stated x = cos(θ ) and y = sin(θ ), we can find the coodinates of the point desied: π x = 6 cos = 6 = 4 π y = 6 sin = 6 = 4 Ty it Now. Find the coodinates of the point on a cicle of adius at an angle of 90 Next, we will find the cosine and sine at an angle of 0 degees, o 6 π. To find this, we will fist daw the tiangle on a cicle at an angle of 0 degees, and anothe at an angle of -0 degees. If these two ight tiangles ae combined into one lage tiangle, notice that all thee angles of this lage tiangle ae 60 degees. 0 (x, y) 60 60 60 y y Since all the angles ae equal, the sides will all be equal as well. The vetical line has length y, and since the sides ae all equal we can conclude that y =, o y =. Using this, we can find the sine value: π y sin = = = = 6
0 Chapte 5 Using the Pythagoean Identity, we can find the cosine value: π π cos + sin = 6 6 π cos + 6 cos π = 6 4 = since the y value is positive, we ll keep the positive oot π cos = = 6 4 The (x, y) coodinates fo a cicle of adius and angle of 0 degees =, By taking the tiangle on the unit cicle at 0 degees and eflecting it ove the line y = x, we can find the cosine and sine fo 60 degees, o π, without any additional wok. By this symmety, we can see the coodinates of the point on the unit cicle at 60 degees will be,, giving π cos = and π sin = We have now found the cosine and sine values fo all of the commonly encounteed angles in the fist quadant of the unit cicle. Angle 0 0 0 60 Cosine Sine 0 π π π π, o 0, o 45, o 60, o 90 6 4 0
Section 5. Points on Cicles using Sine and Cosine Fo any given angle in the fist quadant, thee will be anothe angle with the same sine value, and anothe angle with the same cosine value. Since the sine value is the y coodinate on the unit cicle, the othe angle with the same sine will shae the same y value, but have the opposite x value. Likewise, the angle with the same cosine will shae the same x value, but have the opposite y value. As shown hee, angle α has the same sine value as angle θ; the cosine values would be opposites. The angle β has the same cosine value as the angle; the sine values would be opposites. sin( θ ) = sin( α) and cos( θ ) = cos( α) sin( θ ) = sin( β ) and cos( θ ) = cos( β ) (x, y) (x, y) α θ θ β It is impotant to notice the elationship between the angles. If, fom the angle, you measued the shotest angle to the hoizontal axis, all would have the same measue in absolute value. We say that all these angles have a efeence angle of θ. An angle s efeence angle is the size of the smallest angle to the hoizontal axis. (x, y) A efeence angle is always an angle between 0 and 90 degees, o 0 and π adians. θ θ Angles shae the same cosine and sine values as thei efeence angles, except fo signs (positive/negatives) which can be detemined by the quadant of the angle. θ θ
Chapte 5 Example 5 Find the efeence angle of 50 degees. Use it to find cos( 50 ) and sin( 50 ) 50 degees is located in the second quadant. It is 0 degees shot of the hoizontal axis at 80 degees, so the efeence angle is 0 degees. This tells us that 50 degees has the same sine and cosine values as 0 degees, except fo sign. We know that sin( 0 ) = and cos( 0 ) =. Since 50 degees is in the second quadant, the x coodinate of the point on the cicle would be negative, so the cosine value will be negative. The y coodinate is positive, so the sine value will be positive. sin( 50 ) = and cos( 50 ) = The (x, y) coodinates fo a cicle of adius and angle 50 ae, Using symmety and efeence angles, we can fill cosine and sine values at the est of the special angles on the unit cicle. Take time to lean the (x, y) coodinates of all of the majo angles in the fist quadant! π π 90,, ( 0, ) π 0,,, 60,,, π π 5,,, 45,,, 4 4 5π 50,,, 6 π 0,,, 6 π (, ) 80,, 0 0, 0,, 0 ( ) π ( ) 60,,, 0 7π 0,,, 6 5π 5,,, 4 4π 40,,, π 70,, 0 (, ) 5π 00,,, π 0,,, 6 π 7 5,,, 4
Section 5. Points on Cicles using Sine and Cosine Example 6 Find the coodinates of the point on a cicle of adius at an angle of 7π. 6 Note that this angle is in the thid quadant whee both x and y ae negative. Keeping this in mind can help you check you signs of the sine and cosine function. 7π x = cos = = 6 6 7π y = sin = = 6 6 The coodinates of the point ae ( 6, 6) Ty it Now. Find the coodinates of the point on a cicle of adius 5 at an angle of 5 π Example 7 We now have the tools to etun to the sailboat question posed at the beginning of this section. A distess signal is sent fom a sailboat duing a stom, but the tansmission is unclea and the escue boat sitting at the maina cannot detemine the sailboat s location. Using high poweed ada, they detemine the distess signal is coming fom a distance of 0 miles at an angle of 5 degees fom the maina. How many miles east/west and noth/south of the escue boat is the standed sailboat? We can now answe the question by finding the coodinates of the point on a cicle with a adius of 0 miles at an angle of 5 degees. x = 0cos( 5 ) = 0 4. 4 miles y = 0sin ( 5 ) = 0 4. 4 miles The sailboat is located 4.4 miles west and 4.4 miles south of the maina.
4 Chapte 5 The special values of sine and cosine in quadant one ae vey useful to know, since knowing them allows you to quickly evaluate the sine and cosine of vey common angles without needing to look at a efeence o use you calculato. Howeve, scenaios do come up whee we need to know the sine and cosine of othe angles. To find the cosine and sine of any othe angle, we tun to a compute o calculato. Be awae: most calculatos can be set into degee o adian mode, which tells the calculato which units the input value is in. When you evaluate cos(0) on you calculato, it will evaluate it as the cosine of 0 degees if the calculato is in degee mode, o the cosine of 0 adians if the calculato is in adian mode. Most compute softwae with cosine and sine functions only opeates in adian mode. Example 8 Evaluate the cosine of 0 degees using a calculato o compute. On a calculato that can be put in degee mode, you can evaluate this diectly to be appoximately 0.9969. On a compute o calculato without degee mode, you would fist need to convet the π angle to adians, o equivalently evaluate the expession cos 0 80 Impotant Topics of This Section The sine function The cosine function Pythagoean Identity Unit Cicle values Refeence angles Using technology to find points on a cicle Ty it Now Answes. cos( π ) = sin( π ) = 0. π x = cos = * 0 = 0 π y = sin = * =. 5 5,
Section 5.4 The Othe Tigonometic Functions 5 Section 5.4 The Othe Tigonometic Functions In the pevious section, we defined the sine and cosine functions as atios of the sides of a tiangle in the cicle. Since the tiangle has diffeent vaiables thee ae 6 possible combinations of atios. While the sine and cosine ae the pominent two atios that can be fomed, thee ae fou othes, and togethe they define the 6 tigonometic functions. Fo the point (x, y) on a cicle of adius at an angle of θ, we can define fou additional impotant functions as the atios of the sides of the coesponding tiangle: The tangent function: The secant function: The cosecant function: The cotangent function: tan(θ ) y = x sec(θ ) = x csc(θ ) cot(θ ) = y x = y θ x (x, y) y Geometically, notice that the definition of tangent coesponds with the slope of the line fom the oigin out to the point (x, y). This elationship can be vey helpful in thinking about tangent values. You may also notice that the atios defining the secant, cosecant, and cotangent ae the ecipocals of the atios defining the cosine, sine, and tangent functions, espectively. Additionally, notice that using ou esults fom the last section, y sin( θ ) sin( θ ) tan( θ ) = = = x cos( θ ) cos( θ ) Applying this concept to the othe tig functions we can state the othe ecipocal identities. Identities The othe fou tigonometic functions can be elated back to the sine and cosine function using these basic identities sin( θ ) cos( θ ) tan( θ ) = sec( θ ) = csc( θ ) = cot( θ ) = = cos( θ ) cos( θ ) sin( θ ) tan( θ ) sin( θ )
6 Chapte 5 These elationships ae called identities. These identities ae statements that ae tue fo all values of the input on which they ae defined. Identities ae always something that can be deived fom the definitions and elationships we aleady know. These identities follow fom the definitions of the functions. The Pythagoean Identity we leaned ealie was deived fom the Pythagoean Theoem and the definitions of sine and cosine. We will discuss the ole of identities moe afte an example. Example 5π Evaluate tan( 45 ) and sec 6 Since we know the sine and cosine values fo these angles, it makes sense to elate the tangent and secant values back to the sine and cosine values. sin(45 ) tan( 45 ) = = = cos(45 ) Notice this esult is consistent with ou intepetation of the tangent value as the slope of the line fom the oigin at the given angle a line at 45 degees would indeed have a slope of. 5π sec = = 6 5π cos 6 =, which could also be witten as Ty it Now 7π. Evaluate csc 6 Just as we often need to simplify algebaic expessions, it is often also necessay o helpful to simplify tigonometic expessions. To do so, we utilize the definitions and identities we have established.
Section 5.4 The Othe Tigonometic Functions 7 Example sec( θ ) Simplify tan θ ( ) We can simplify this by ewiting both functions in tems of sine and cosine sec( θ ) cos( θ ) = tan( θ ) sin( θ ) cos( θ ) To divide the factions we could invet and multiply cos( θ ) = cos( θ ) sin( θ ) cancelling the cosines, = = csc( θ ) sin θ simplifying and using the identity ( ) By showing that identity: that sec tan sec( θ ) can be simplified to csc ( θ ) tan( θ ) ( θ ) = csc( θ ). ( θ ), we have, in fact, established a new Occasionally a question may ask you to pove the identity o establish the identity. This is the same idea as when an algeba book asks a question like show that ( x ) = x x +. The pupose of this type of question is to show the algebaic manipulations that demonstate that the left and ight side of the equations ae in fact equal. You can think of a pove the identity poblem as a simplification poblem whee you know the answe you know what the end goal of the simplification should be. To pove an identity, in most cases you will stat with one side of the identity and manipulate it using algeba and tigonometic identities until you have simplified it to the othe side of the equation. Do not teat the identity like an equation to solve it isn t! The poof is establishing if the two expessions ae equal and so you cannot wok acoss the equal sign using algeba techniques that equie equality. Example Pove the identity + cot( α ) = sin( α) + cos( α) csc( α) Since the left side seems a bit moe complicated, we will stat thee and simplify the expession until we obtain the ight side. We can use the ight side as a guide fo what might be good steps to make. In this case, the left side involves a faction while the ight side doesn t, which suggests we should look to see if the faction can be educed.
8 Chapte 5 Additionally, since the ight side involves sine and cosine and the left does not, it suggests that ewiting the cotangent and cosecant using sine and cosine might be a good idea. + cot( α) Rewiting the cotangent and cosecant csc( α) cos( α) + sin( α ) = To divide the factions, we invet and multiply sin( α) cos( α) sin( α) = + sin( α ) Distibuting, sin( α) cos( α) sin( α) = + sin( α) Simplifying the factions, = sin( α) + cos( α) Establishing the identity. Notice that in the second step, we could have combined the and cos( α ) sin( α) befoe inveting and multiplying. It is vey common when poving o simplifying identities fo thee to be moe than one way to obtain the same esult. We can also utilize identities we have aleady leaned while simplifying o poving identities. Example 4 Establish the identity cos + sin ( θ ) ( θ ) = sin ( θ ) Since the left side of the identity is moe complicated, it makes sense to stat thee. To simplify this, we will have to eliminate the faction. To do this we need to eliminate the denominato. Additionally, we notice that the ight side only involves sine. Both of these suggest that we need to convet the cosine into something involving sine. Recall the Pythagoean Identity told us cos ( θ ) + sin ( θ ) =. By moving one of the tig functions to the othe side, we can establish: sin ( θ ) = cos ( θ ) and cos ( θ ) = sin ( θ ) Utilizing this, we now can establish the identity. We stat on one side and manipulate:
cos ( θ ) + sin( θ ) sin ( θ ) + sin( θ ) ( sin( θ ))( + sin( θ )) + sin( θ ) sin( θ ) = = Cancelling the like factos = Establishing the identity Section 5.4 The Othe Tigonometic Functions 9 Utilizing the Pythagoean Identity Factoing the numeato We can also build new identities by manipulating aleady established identities. Fo example, if we divide both sides of the Pythagoean Identity by cosine squaed, cos ( θ ) + sin ( θ ) = Splitting the faction on the left, cos ( θ ) cos ( θ ) cos ( θ ) sin ( θ ) + = Simplifying and using the definitions o tan and sec cos ( θ ) cos ( θ ) cos ( θ ) + tan ( θ ) = sec ( θ ) Ty it Now!. Use a simila appoach to establish that cot ( θ ) + = csc ( θ ) Identities Altenate foms of the Pythagoean Identity + tan ( θ ) = sec ( θ ) cot ( θ ) + = csc ( θ ) Example 5 If tan( θ ) = and θ is in the d quadant, find cos(θ ). 7 Thee ae two appoaches to this poblem, both of which wok equally well. Appoach y Since tan(θ ) = and the angle is in the thid quadant, we can imagine a tiangle in a x cicle of some adius so that the point on the cicle is (-7, -). Using the Pythagoean Theoem, we can find the adius of the cicle: ( 7) + ( ) =, so = 5.
0 Chapte 5 Now we can find the cosine value: x 7 cos( θ ) = = 5 Appoach Using the + tan ( θ ) = sec ( θ ) fom of the Pythagoean Identity with the known tangent value, + tan ( θ ) = sec ( θ ) + 7 = sec ( θ ) 5 = sec ( ) 49 θ 5 sec( θ ) = ± = ± 49 5 7 Since the angle is in the thid quadant, the cosine value will be negative so the secant value will also be negative. Keeping the negative esult, and using definition of secant, sec( θ ) = = cos( θ ) cos( θ ) = 5 7 5 7 7 7 5 = 5 5 Inveting both sides Ty it Now 7 π. If sec( φ ) = and < φ < π, find tan( φ ) and sin( φ ) Impotant Topics of This Section 6 Tigonometic Functions: Sine Cosine Tangent Cosecant Secant Cotangent Tig identities
Section 5.4 The Othe Tigonometic Functions Ty it Now Answes. -. cos ( θ ) + sin sin θ ( θ ) = cos sin ( θ ) sin + ( θ ) sin ( θ ) ( θ ) = sin ( θ ) cot ( θ ) + = csc ( θ ). sin( φ ) = 40 7 tan( φ ) = 40
Chapte 5 Section 5.5 Right Tiangle Tigonomety In section 5. we wee intoduced to the sine and cosine function as atios of the sides of a tiangle dawn inside a cicle, and spent the est of that section discussing the ole of those functions in finding points on the cicle. In this section, we etun to the tiangle, and exploe the applications of the tigonometic functions on ight tiangles sepaate fom cicles. Recall that we defined sine and cosine as y sin(θ ) = x cos(θ ) = Sepaating the tiangle fom the cicle, we can make equivalent but moe geneal definitions of the sine, cosine, and tangent on a ight tiangle. On the ight tiangle, we will label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle. θ x (x, y) y Given a ight tiangle with an angle of θ sin( θ ) = cos( θ ) = tan( θ ) = opposite hypotenuse adjacent hypotenuse opposite adjacent hypotenuse θ adjacent opposite A common mnemonic fo emembeing these elationships is SohCahToa, fomed fom the fist lettes of Sine is opposite ove hypotenuse, Cosine is adjacent ove hypotenuse, Tangent is opposite ove adjacent. Example Given the tiangle shown, find the value fo cos(α ) The side adjacent to the angle is 5, and the hypotenuse of the tiangle is 7, so adjacent 5 cos( α ) = = hypotenuse 7 α 7 5 8
Section 5.5 Right Tiangle Tigonomety When woking with geneal ight tiangles, the same ules apply egadless of the oientation of the tiangle. In fact, we can evaluate the sine and cosine of eithe the othe two angles in the tiangle. Adjacent to α Opposite β α Adjacent to β Opposite α β Hypotenuse Example Using the tiangle shown, evaluate cos(α ), sin(α ), cos(β ), and sin(β ) 4 adjacent to α cos( α ) = = hypotenuse 5 oppositeα 4 sin( α ) = = hypotenuse 5 adjacent to β 4 cos( β ) = = hypotenuse 5 opposite β sin( β ) = = hypotenuse 5 α 5 β Ty it Now. A ight tiangle is dawn with angle α opposite a side with length, angle β opposite a side with length 56, and hypotenuse 65. Find the sine and cosine of α and β. You may have noticed that in the above example that cos( α ) = sin( β ) and cos( β ) = sin( α). This makes sense since the side opposite of α is the same side as is adjacent to β. Since the thee angles in a tiangle need to add to π, o 80 degees, then π π π the othe two angles must add to, o 90 degees, so β = α, and α = β. π Since cos( α ) = sin( β ), then cos( α) = sin α.
4 Chapte 5 Identities The cofunction identities fo sine and cosine π π cos( θ ) = sin θ sin( θ ) = cos θ In the pevious examples we evaluated the sine and cosine on tiangles whee we knew all thee sides of the tiangle. Right tiangle tigonomety becomes poweful when we stat looking at tiangles in which we know an angle but don t know all the sides. Example Find the unknown sides of the tiangle pictued hee. opposite Since sin( θ ) =, hypotenuse 7 sin( 0 ) = b Fom this, we can solve fo the side b. a 0 b b sin( 0 ) = 7 7 b = sin(0 ) 7 To obtain a value, we can evaluate the sine and simplify 7 b = = 4 To find the value fo side a, we could use the cosine, o simply apply the Pythagoean Theoem: a + 7 = b a + 7 = 4 a = 47 Notice that if we know at least one of the non-ight angles of a ight tiangle and one side, we can find the est of the sides and angles. Ty it Now. A ight tiangle has one angle of π and a hypotenuse of 0. Find the unknown sides and angles of the tiangle.
Section 5.5 Right Tiangle Tigonomety 5 Example 4 To find the height of a tee, a peson walks to a point 0 feet fom the base of the tee, and measues the angle to the top of the tee to be 57 degees. Find the height of the tee. We can intoduce a vaiable, h, to epesent the height of the tee. The two sides of the tiangle that ae most impotant to us ae the side opposite the angle, the height of the tee we ae looking fo, and the adjacent side, the side we ae told is 0 feet long. 57 The tigonometic function which elates the side opposite of the angle and the side adjacent to the angle is the tangent. 0 feet opposite h tan(57 ) = = Solving fo h, adjacent 0 h = 0 tan(57 ) Using technology we can appoximate a value h = 0 tan(57 ) 46. feet The tee is appoximately 46. feet tall. Example 5 A peson standing on the oof of a 00 foot building is looking towads a skyscape a few blocks away, wondeing how tall it is. She measues the angle of declination to the base of the skyscape to be 0 degees and the angle of inclination to the top of the skyscape to be 4 degees. To appoach this poblem, it would be good to stat with a pictue. Although we ae inteested in the height, h, of the skyscape, it can be helpful to also label othe unknown quantities in the pictue in this case the hoizontal distance x between the buildings and a, the height of the skyscape above the peson. 4 x 0 To stat solving this poblem, notice we have two ight tiangles. In the top tiangle, we know the angle is 4 00 ft 00 ft degees, but we don t know any of the sides of the tiangle, so we don t yet know enough to wok with this tiangle. a h
6 Chapte 5 In the lowe ight tiangle, we know the angle of 0 degees, and we know the vetical height measuement of 00 ft. Since we know these two pieces of infomation, we can solve fo the unknown distance x. opposite 00 tan( 0 ) = = Solving fo x adjacent x x tan( 0 ) = 00 00 x = tan(0 ) Now that we have found the distance x, we know enough infomation to solve the top ight tiangle. opposite a a tan(4 ) = = = adjacent x 00 tan(0 ) a tan(0 ) tan(4 ) = 00 00 tan(4 ) = a tan(0 ) 00 tan(4 ) = a tan(0 ) Appoximating a value, 00 tan(4 ) a = 47.4 feet tan(0 ) Adding the height of the fist building we detemine that the skyscape is about 47.4 feet tall. Impotant Topics of This Section SOA CAH TOA Cofunction identities Applications with ight tiangles Ty it Now Answes 56 56. Sin ( α) = Cos ( α) = Sin ( β ) = 65 65 65 π adjacent A. Cos = = so, adjacent = 0 hypoteuse 0 Sin π Opposite O = hypoteuse = 0 Missing angle = 0 degees O cos( β ) = 65 π 0 Cos = 0 = π 0 sin = 0 = so, Opposite = 0 π 6