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# TRIGONOMETRY MATH 117

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This 66 page Class Notes was uploaded by Sadie Schroeder on Wednesday September 30, 2015. The Class Notes belongs to MATH 117 at Western Kentucky University taught by David Neal in Fall. Since its upload, it has received 7 views. For similar materials see /class/216740/math-117-western-kentucky-university in Applied Mathematics at Western Kentucky University.

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Date Created: 09/30/15

Dr Neal Spring 2009 MATH 117 Solving for Angles We can use the inverse trig functions on a calculator to solve for the angles that give a particular trig function value In other words given a specific value v we can find the angles 6 such that sine v or cose v or tane v We always take the angles to be between 00 and 360 and in each case there will be two solutions However the calculator will only give one angle so we will need to find the appropriate symmetric angles in the correct quadrants to give the solutions Example To the nearest 100th find the angles between 00 and 360 such that a sine 0850444 c cose 091706 e tane 018101 b sinG 70916921 d cose 7078801 f tane 45 Solution a There are two angles where sine 0850444 one in the lst Quadrant 121740 5826quot and one in the 2nd Quadrant 180 7 5826 H The command sin 10850444 gives 6 58260 The corresponding angle in the 2nd Quadrant is 180 7 5826quot 12174quot b There are two angles where sine 66480 70916921 one in the 3rd Quadrant and one in the 4th Quadrant Initially we can ignore the negative to find the reference angle in the lst Quadrant sin 10916921 gives 6 6648 The corresponding angles in the 3rd and 4th Quadrants are 180 6648 246480 and 360 7 6648quot 293520 Thus the solutions are 246480 and 293520 180 5548 350 5548 c There are two angles where cos 09l706 one in the 1st Quadrant and one in the 4th Quadrant The command cos1091706 gives 6 235 The corresponding angle in the 4th Quadrant is 360 7 2350 3365 Thus the solutions are 235 and 3365 d There are two angles where cose 7078801 one in the 2nd Quadrant and one in the 3rd Quadrant Initially we ignore the negative to find the reference angle in the 1st Quadrant cos1078801 gives 6 38quot The corresponding angles in the 2nd and 3rd Quadrants are 180 i 38quot which give 142 and 218 Thus the solutions are 142 and 218 e There are two angles where tane 018101 one in the 1st Quadrant and one in the 3rd Quadrant The command tan 1018101 gives 6 1026 The corresponding angle in the 3rd Quadrant is 180 10260 19026 f There are two angles where tane 715 one in the 2nd Quadrant and one in the 4th Quadrant The command tan 115 gives 6 5631 The corresponding angles in the 2nd Quadrant and 4th Quadrants are 180 7 5631 12369 and 360 7 5631quot 30369 Thus the solutions are 12369 and 30369 Dr Neal Spring 2009 235 360 7235 3365 142 38quot 1 180i3 J 218 1026quot 19026quot 180 7 5631 5531quot 360 7 5631 Dr Neal Spring 2009 MATH 117 Latitude and Longitude A point on Earth is designated by its latitude and longitude For instance Bowling Green KY is at 37 N 8629OW The latitude 4 is measured North or South from the Equator Therefore 90 S S 90 where northern latitudes are positive and southern latitudes are negative The longitude 6 is measured East or West from the Prime Meridian which passes from the North Pole to the South Pole just east of London England We always use 180 S6 S1800 where Eastern longitudes are positive and Western longitudes are negative Rectangular Coordinates The radius of the Earth is approximately p 39632 miles Given the latitude q and the longitude 6 we can convert the point into rectangular coordinates x y z The center of Earth becomes the origin 0 0 0 The xiaXis then moves outward from 000 to the intersection of the Equator with the Prime Meridian which is just south of Ghana Africa in the Gulf of Guinea This point has x y z coordinates 39632 0 0 The North Pole creates the x y z coordinates 0 0 39632 The rectangular coordinates are given by x p cos cos0 y p cost sine z psin For Bowling Green KY 1 370 N and 9 786290 W thus x 39632 cos37 cos 8629 m 2048 y 39632 cos37 sin 8629 m 7315852 2 39632 sin37 m 238511 Dr Neal Spring 2009 Sydney Australia is at 335508 1511 E thus 3355 8 15110 thus x 39632 cos 3355 cos1511 m 7289161 y 39632 cos 3355 sin1511 159626 2 39632 Sln 33550m 7219032 Journey through the Center of the Earth Let39s find the direct linear distance from Bowling Green to Sydney Using the coordinates 2048 7315852 238511 and 7289161 159626 7219032 we find the distance to be Jx1 x22 y1 y22 z1 222 J3096412 4754782 4575432 m 728905 miles Journey Across the Surface of the Earth To find the actual distance across the face of the Earth from Bowling Green to Sydney we need the angle a between the two vectors u p 61 11 39632 78629 37quot and v p 62 12 39632 1511quot 73355 which is given by the formula a cos71cos 1 cos 2 cos61 62 sin 1 sin 2 In this case we obtain a cos 1cos 1 cos 2 cos61 62sin 1 sin 2 cosilcos37 cos 3355 cos 8629 1511 sin37 sin 3355 cosilcos37 cos 3355 cos 23739 sin37 sin 3355 m 1337329359o Once we have a then the distance along the great circle through the two points is given by the arclength formula TE 800 X 39632 miles Distance Across Earth a X 1 Dr Neal Spring 2009 Therefore the distance from Bowling Green to Sydney is TE 800 X 39632 m 9250426 miles Dist 1337329359o X 1 Exercise Find the distance from Berlin 523 N 132 E to Buenos Aires 344 S 5825 W Answer on next page Dr Neal Spring 2009 Answer Angle in Between a cosilcos523 cos 344 cos132quot 5825 sin523 sin 344 cosilcos523 cos 344 cos7145 sin523 sin 344 n 10664807360 Dist 1066480736 X181 gtlt 39632 7376942 miles 00 D Neal 5pm 2cm MATH 117 Simeosine anhs Each mgammemeruhmm eah he graphed as a mum at x when he vaname x1 alwaysmm ans lnparum anthefumuam y nx ycasx are de nedfax a x and are eyehe Thaus nae shape unhe graphxzpeats Itself avenhe pemds 74n x 72n 727 st EISxSZn 27mg Ansx n 1 whmh are sempxy wrapamunds huh unn male The Sine anh ohe cycle afthesmegaphamursas x rangesfmm n in 27K radxans and me shape a thegraphmnhemalded amund39hevaluesamunrgat x quot 3quot and 27x 3397 Om Cycle at y xinx Twa Cycles Muluplyxrghy amgauve re ux nae gaph aham nae xeaxes and hmxupxymg by a mnstahua mueasesuaexargemea S s a rather 39hahe1 S 1 Winn124 Anylimde m D Neal 5pm 2cm PhaseShi Gwen the gaph ht yfx we can shit the gaph m the nght by e um wALh the mhmah x ee We shu39uathe left e uhnmm Lhafunman y x e 7 L I 3 Similarly we can shu39uhe me graphm the m ax nght by a unam angle wnh the mhmah mute name Graph me cycle eaLh unhe fallawmgfunmms Label we paint an the xe amulan 3th when the mats and the peaks am a y e 25mx 3 1 y 725mb eJ Snlu nn a The me gaphxs sh 39udta 7 whzre me cycle hegms ohe cycle ends J Sanaw dede the cycle Imth a a exlergvh 2n which at e 27 3 3 27 mm equal pieces aflerglh 27quot 2 Ta da sa we must add lenglhs hr g mm the WWW e We m e i me New 2h sum e7 5 D Neal 5pm 2cm h Thenegauve magnphls shl edta whzre an cycle h2gms NW was the WWW 2n ml W W 1quotth 271mm lm W l L m the mum palm mfg WWW Om yd WWW W as x W m n m M mm mm W lwlpmmmmllmmwlwmw glam M 7 I n j 2n 1 n 2n Om Cycle 0 y wxx Twa Canvpleke Cycles D Neal 5m 2m 5 1 n J 7 2n 2 2 2 5 y 2 Winn122 Winn m m1 2 Graph an cycle mh fth fullnw rgfunmms Label rm paint an the x7 was um shww when the mats and the peaks am a y3caserJ h yi4 asx SnluLmn a The mm graph shu ted m g whzre an cycle h2gns mm the cycle 1 u 2n mm equal pieces aflenglh 7quot and add these lengths u mm the umng pmm gig D Neal 5m 2m h Thenegauve mm gnth shu39tedta 7 whzxe an cycle hegms mm the m M a M W W W 1quotth m m m M a L mm the umng pmm uf 7 Dr Neal Spring 2009 MATH 117 Spherical to Rectangular Coordinates A point in threeidimensional space is uniquely determined by its spherical coordinates p 6 4 where p is the length or radius of the sphere 6 is the sideways angle measured from the positive xiaXis and q is the vertical angle measured upward or downward from the x y plane to the point x5 y x y 0 We generally take 790 S q S 900 and 0 S 6 S 360 although sometimes we take 7180 S 6 S 180 Of course p Z 0 Now given the spherical coordinates p 6 1 we wish to find the rectangular coordinates x y z The z coordinate From the relationship sin i we have 2 p sin p The radiusr on the xy plane From the relationship cos L we have r p 0054 p The X and y coordinates As usual x r 0050 and y r sine By using r p 0051 we have x p cos 0056 and y p 0054 sine The rectangular coordinates of a point p 6 1 are given by x p cos 0058 y p cos sine z psin Example 1 Compute the rectangular coordinates of the two points u 12 120 30quot and v 20 270 745quot Dr Neal Spring 2009 For point u p 12 6 120 and q 30quot Z 12 sin30 12 gtlt 6 12 J3 1 A x10005300005120 12X7gtltE3J J 0 3 3 120 y1200530 sin120 12gtlt7gtlt79 S Thus u 3J3 9 6 270 7450 For v p 20 e 270 and 745 2 Z 20 sin 45 20 X 2 10J J5 x 20 cos 45 0052700 20 X 7 X 0 0 2 y 20 cos 45 sin270 20 X g X 1 10 Thusv 0 10J 10J Dr Neal Spring 2009 The Angle Between Two Points in Spherical Coordinates Let u pl 61 11 and v p2 62 12 be two given points We will now see how to find the angle between u and v based upon just the given angles To do so we will use the formula for the rectangular coordinates and then simplify the dot product The rectangular coordinates of u and v are u plcos 1 c0591 p1 c0s 1 sinel p1 simpl and v p2 cos 2 c0562 p2 cos 2 sin 62 p2 sin 2 The dot product is then u v p1 cos 1 cosel gtlt p2 cos 2 cosez p1c0s 1 sinel gtlt p2 cos 2 sin62 p1 sin 1 gtlt p2 sin 2 p1 p2c0s 1 cos 61 cos 2 c0592 cos 1 sin91c0s 2 sinez sin 11 sin 12 p1 p2c0slt1 c0s 2c0s61c0s02 sinOl sin 62 sin 1 sin ltgt2 p1p2c0s 1cos 2 cos61 62 simpl sin 12 Because u and v are on spheres with radii p1 and p2 we have p1 and p2 Thus u v p1p2cos 1cos 2cos91 92sin 1 sin 2 IHIIIVI pl p2 cos pl cos 2 cos61 92 sin 1 sin 2 So the angle in between u and v is a c0s71cos 1cos 2 cos61 62 sin 1 sin 2 Example 2 Find the angle between u 12 120 30quot and v 20 270 745 Solution The lengths 12 and 20 of the two vectors are not relevant to the angle in between All we need are the directional angles For u 91 120 in the x y plane and 1 30 upward For v 62 270 in the xy plane and 2 745 downward Dr Neal Spring 2009 270 y 7450 The angle in between u and v is x cosilcos30 cos 45 cos1200 2700 sin30o sin 450 cosilcos30quotcos 45quotcos 150quotsin30o sin 45 71 J5 J5 J3 1 5 s 7 X 7 X X 0 2 2 2 1 3J2 J7 4 5 cos cos 8 4 8 m1521144o Exercise Find the angle between u 10 100 60quot and v 6 720 730 Answer on next page Dr Neal Spring 2009 Use a cosilcos 1 cos 2 cos61 92 sin 1 simpz with 1 60quot 2 730 01 100 02 720 01 c0571cos60 cos 30 cos100 20 sin60o sin 30 c0571cos60 cos 30 cos120 sin60o sin 300 1 J3 1 J3 4 x gtlt 2 2 2 2 45 J5 46m cos cos 8 4 8 1305quot Dr Neal Spring 2009 MATH 117 Spherical Coordinates Points in threeidimensional space can be labeled in rectangular coordinates x y z and plotted with the positive axes oriented as shown below Z Example 1 Plot the points u 4 6 7 v 5 78 5 and w 74 6 712 Solution The key to plotting points is drawing a box with edges that are parallel to the axes The point 4 6 7 is shown below along with three other corner points on the x y xz and yz planes 7 06 7 4 6 7 40 7 6 4 60 Dr Neal Spring 2009 Below are the points v 5 78 5 and w 74 6 712 and some other vertices Z 0785 5 578 74 46 8 Illl 11 I vvv v I II I y 5 78 0 5 74 6 712 712 0 6 712 Vectors in 3D Every point x y 2 forms a Vector from the origin 0 0 0 to x y 2 Below are the vectors v 5 78 5 and w 74 6 712 545 4 8 Illl ll 4 6 712 Dr Neal Spring 2009 As with vectors in the x y plane a Sedimensional vector u has a length now labeled p But in threeidimensional space a vector has two directional angles labeled 8 and ltgt Collectively p 6 b are the spherical coordinates of the vector u x y 2 Length and Angles The length of vector u x y z is the distance from 00 0 to x y 2 given by puJx2y222 Length or norm The angle 6 is determined solely by the x y coordinates As always 6 is measured from the positive xiaXis and tan 9 y x Then is 6 is given by tan71y x if 6 is in Quadl e tan71yx180 ire is in 11 or 111 tan71yx360 ire is in 1v Sideways Angle The angle 4 is measured from the x y plane up or down to the point x y z By convention we let be positive if z gt 0 and let 4 be negative if z lt 0 Then we always have 790 S d S 90quot P From the illustration to the right we see that Z r Z Z s1n P lyc2y2z2 So the angle 4 is given by I sin71 sinl P lxz y2 22 Vertical Angle Dr Neal Spring 2009 Example 2 Find the spherical coordinates of v 5 78 5 and w 74 6 712 Solution The length of vector v is p l5282 52 J114 For v the angle 6 in the x y plane is in the 4th Quadrant and is determined by the first two coordinates 5 78 So 6tan71 85360o m 302 The vertical angle 4 satisfies sinq so 5 sin 1 j m 2792quot Thus the spherical coordinates of vector v are p J114 0 m 302 and I m 2792 Z v 5 78 5 74 6 4114 2 5 z 712 5 78 14 X w 4413412 The length of vector w is p J42 62 122 14 For w the angle 6 is in the 2nd Quadrant and is determined by the first two coordinates 74 6 So 6tan716 4180o w 12369 The vertical angle 1 satisfies sin 3 so p 39 1 12J N 7590 4 sm 14 Thus the spherical coordinates of vector w are p 14 e 4 12369 and 4 4 459 Dr Neal Spring 2009 Distance and Angle In Between Given two vectors u x1 yl 21 and v x2 y2 22 we can find the vector from u to v the direct distance from u to v and the angle in between u and v The formulas for doing so are the same as for twoidimensional vectors but with an added 2 term Vector fromuto v v u x2 7 x1 y2 iyl 22 7 21 Distance from u to v Iv u kaz x12 y2 yl2 22 202 Dot Product u v x1 x2 yly2 2122 Angle in Between a cosi1 uv Example 3 Let v 5 78 5 and w 74 6 712 Find the vector from v to w the direct distance from v to w and the angle in between v and w Solution The vector from v to w is w v 9 14 717 The distance from v to w is simply the length of w v given by Iw v J92 142 172 I566 m 2379 As found in the previous example J114 and I w 14 The dot product is vw 574 786 5712 7128 So the angle in between vectors v and w is 1 128 489quot 0 cos J114x14J Z v578 5 J114 a 14 X m2379 w 74 6 712 Dr Neal Spring 2009 MATH 117 The Unit Circle 0 1 S arclength 710 10 Xeaxis yeaXis The Equation of the Unit Circle x2 y2 1 Angle 6 is measured counterclockwise from the positive x7 axis There are 360 in the entire circle Angle 6 makes a point x y on the unit circle and an arclength of s units gt cos62sin62 1 The xecoordinate is called the cosine ofe x cose The yecoordinate is called the sine of 6 quotz y sinB The fundamental trigonometric identity is cos2 6 sin2 6 1 The radian measure of angle 6 is the arclength s created by 6 on the unit circle The circumference of a circle is C 21V So the circumference of the unit circle is 21 Whole Circle gt 360 gt 21 radians Halfecircle gt 180 gt T radians 2 2 Quarter Circle gt 900 gt Tn rad Eighth of Circle gt 450 gt n rad Dr Neal Spring 2009 Converting Degrees to Radians One Way Take the whole circumference of 21 and multiply by the proportion of the circle being used 21 X Z 3600 Another J t It 1 o b 11 B 2 2 0 117 J us mu 1 gtlt x WaY p y Z y 180 ecause 3600 Z 180 Example 1 Convert 160 to radians 1600 16 3211 811 nx rad First way 21 x 7 Whole circle 360 36 36 9 16 8 Second way 160quotgtlt 1T n n rad 1800 18 9 Example 2 Graph each angle below and show the orientation Convert each angle to radians For the negative angles find positive angles in both degrees and radians that are corterminal to the original negative angle a 270 b 7135quot c 7220quot Solution Negative angles are measured clockwise from the positive x axis 7135 J 270 11 2711 31 radlans 1800 18 2 7220 a Convert to radians 270 gtlt b 135 gtlt 1200 rad after dividing out a common factor of 45 Dr Neal Spring 2009 b Cont To find a positive coeterminal angle ie a positive angle that ends at the same point on the unit circle simply add 360 or add 21 3 3 8 5 Coeterminal positive angles 135 360 225 and 2n 41T T 22 11 0 Convert to radians 220 X n n n rad 180 18 9 1811 To nd a pos1t1ve coetermmal angle add 360 or add 21 39 11 11 18 7 Coeterminal positive angles 220 360 140 or n 2n Converting Radians to Degrees 180 11 to convert back to degrees Given the radian measure of an angle multiply it by Example 3 Convert each angle to degrees For the negative angles find positive 007 terminal angles in both degrees and radians 1111 a b 7 c 180 11 Solution In each case multiply by 7 180 11 180 a ngtlt 7gtlt18 126 b ngtlt 11gtlt30 330 10 11 6 11 4 180 c gtlt 4 X 60 240 3 11 11 12 Positive coeterminal angle for b 330 360 30 or n Add 360 or add 211 E 6 6 4 6 2 Positive coeterminal angle for 0 240 360 120 or it it Add 360 or add 211 Dr Neal Spring 2009 MATH 117 Areas of Triangles We now will use the rightitriangle trig formulas to find the areas of right triangles equilateral triangles and isosceles triangles We then will use the isosceles triangles to find the area of regular nisided polygons Finally we will use Heron s Formula to find the areas of other scalene triangles RightiTriangle Formulas x2y2z2 z x2y2 x ZZ y2 y 22 x2 Ad39 0 O cosS J sin6 x tan6 p1Z Hyp z Hyp 2 Ad x x zcose and y zsine RightiTriangle Area Given a right triangle we can find the area using l Rt Triangle Area 3 X base gtlt height The base and height are the adjacent and opposite sides of the two acute angles so 1 we also can say Area E X opp gtlt adj or Area Ex y Example 1 Find the areas of the following right triangles 50 391 3911 Dr Neal Spring 2009 550 12 20 iii 1V V Solutions 1 We have the base and height so the area is 3050 750 sq units 1 ii The last side is x J652 562 33 and the area is E3356 924 sq units iii The two lateral sides are given by x 120040o and y 125in40quot So the area is given by 2 1 12 400 400 Area E gtlt120040 gtlt125in40 m 35453 sq units iv To find the height y we use tan55 g and y 8tan55 So the area is given by 1 E X 8 X 8tan55 m 457 sq units v To find the base x we use tan30 g and x So the area is given by x tan30 20 l X 20 X w 34641 sq units 2 tan30 Dr Neal Spring 2009 0 Another general form of rightitriangle area can be given when we have the hypotenuse z and one angle 6 First recall that sin2625inecose so that l l sine cose Esin26 Because xzcosG and yzsin6 we obtain Area Exy 1 22 22 1 22 E xzcosB gtltzs1n6cose s1n0 Es1n26 s1n26 The forms to use are 2 2 Rt Triangle Area Z 0056 sine or Rt Triangle Area ZTsin2 2 ln Example iii above we have Area sin2 gtlt 40 m 35453 sq units Examples iv and V also demonstrate other formulas that can be used If we have an angle 6 in a right triangle with x being adjacent and y being opposite then 2 2 Rt Triangle Area x tane and Rt Triangle Area y 2 2tan 6 1 In most cases though it is easiest to use Area E X base gtlt height and simply find the base and height using rightitriangle trig Equilateral Triangle Area Given an equilateral triangle with three sides of length s and three 600 angles we can 1 h still find the area using Exbasexheight We note that sin60 which gives 3 J5 hsgtltsin60 gtlts 2 Dr Neal Spring 2009 s 60quot I 60quot I 3 The base of the entire triangle is s and the height is s so the area of an 2 equilateral triangle is 12 X s X g s TS J s2 Equilateral Triangle Area Example 2 Find the area of an equilateral triangle with sides of length 20 inches 2 Solution The area is EXTZO m 1732 square inches Isosceles Triangle Area Gwen an 1sosceles tr1angle we can nd the area us1ng E X basex hezght prov1ded we know the base angle 6 and either the base length b or vertical side s Generally we are given only one of b or s Dr Neal Spring 2009 If we have s then 3 s X 0059 and h s X sine which gives lsosceles Triangle Area g X h s2 0050 sine b If we have b then tan 6 h b 2 which gives h Etane The area is then 2 lsosceles Triangle Area g X h thane Example 3 Find the areas of the following isosceles triangles 1 Vertical sides of length 10 inches and a vertical angle of 30quot ii A base of 12 feet and base angles of 400 180 30 Solution 1 If the vertical angle is 30quot then each base angle is 6 T 75 So the height is h 10 gtlt sin75o and half the base is b 2 10 gtlt cos75quot So the area is EX h102 cos75o sin75o 25 inZ 30quot h 10 h 400 75 12 6 b2 i ii ii With b 12 and 6 40quot then h 6 tan40o or h 6tan 40 So the area is b 122 E X h 6 X 6tan 400 Ttan40o 302 m Dr Neal Spring 2009 Regular nisided Polygons A regular n isided polygon makes n congruent isosceles triangles where S S the base of a triangle equals one side s of the polygon If there are n sides then the vertical angle of each interior triangle is V In order to find the area of the V V polygon we first must find the area V of each interior isosceles triangle But now we will do so in terms of V2 the vertical angle V We note that s 2 tan zj L2 so that h V h 2 h tan 3 s 2 s s2 3600 The area of one interior isosceles triangle is then X h V Using V 2 4tan 3 V 1800 and 2 we obtain the area of a regular n isided polygon 7132 Regular nigon area W 4tan n Example 4 Find the area of a regular octagon n 8 with sides of length 10 inches Solution Each vertical angle is V 360 8 45 Bisecting an interior triangle we have 5 l tan225 so h The overall area is then 8 gtlt base gtlt ht h tan225 2 2 8 X 10 8 1 X 10 X or o which gives about 4828427 square inches 2 tan225 4m 180 8 Dr Neal Spring 2009 Scalene Triangles and Heron s Formula Suppose a triangle has sides of length a b and c Then Heron 5 Formula gives the area as Area ss as bs c 1 where SEabc Example 5 Find the area of the following triangle 14 12 20 Solution Let s 12 14 20 23 Then the area is J2323 2023 1423 12 J23 X 3 X 9 X 11 V683 m 8265 sq units Dr Neal Spring 2009 MATH 117 Angular Velocity vs Linear Velocity Given an object with a fixed speed that is moving in a circle with a fixed radius we can define the angular velocity u of the object That is we can determine how fast the radian measure of the angle is changing as the object moves on its circular path linear speed angular speed vmr We always use radians as the unit of measure when working with angular velocity For instance our angular velocities should be in units such as rad sec or rad hr However sometimes we will give the result in other layman s terms such as revolutions per minute or degrees per second so as to give a better understanding of how fast the angle is moving But in order to do calculations with angular velocity we always must use radians rad X s c Note though that radians will be an invisible unit For example the units e will yield ft sec Relationship Between Angular Velocity a and Linear Velocity v Given a fixed speed v and radius r then v 211 vcor 0 0 r tlmeoflrev Example 1 A cylinder with a 25 ft radius is rotating at 120 rpm a Give the angular velocity in rad sec and in degrees per second b Find the linear velocity of a point on its rim in mph Solution a To convert rpm revolutions per minute to radians per second we first note that there are 21 radians per revolution We then have Dr Neal Spring 2009 rev rev rad 1 min rad 120 120 gtlt 21t gtlt 11 mm mm rev 60 sec sec d d 180 d Them we have 411 41l gtlt E 720 per sec sec sec 11 rad b We use v o r but we must be in the correct units The angular velocity must use radians Thus the linear velocity is then d t 411 gtlt25ft101tL sec sec Converting to mph we obtain 1 t l 39l 10nL10an3600 x m2142 mph sec sec hr 5280 ft Example 2 A tire witha 9 inch radius is rotating at 30 mph Find the angular velocity of a point on its rim Also express the result in revolutions per minute Solution We simply use on vr but we must make sure that v and r have matching length units Here we shall use miles in order to put 0 in rad hr 03 V X l30LleSgtlt 1 X121 gtlt 5280i 211200id r hr 9m ft mile hr 1 Note that the units actually come out as however radians are a suppressed unit with regards to angular velocity So we write rad hr How many revolutions per minute are there with this spinning tire Because one revolution is Zn radians we have 1 rev 1 hr 211200 M 211200 M X gtlt m 560225 rpm hr hr 211 rad 60 mm Planetary Equatorial Velocities We also can find the angular velocity and the linear velocity at the equator of a planet given that we know the radius of the planet r and the time that it takes for the planet to make one 360 rotation on its axis Note One 360 rotation on the axis is not the same as the actual length of a day due to the planet s orbital movement relative to the Sun Dr Neal Spring 2009 Planetary Data 8 Radius of Earth 6378140 km 6378140 m m 39632 miles Mass of Earth m 5974383 X 1024 kg Average Gravity on Earth m 980665 ms2 m 32174 ftsZ 1 Earth Day m 24 hours lOOO 328084ft 1 mile 1609344 km Conversions km 1 m 5280 ft 1 mile Example 3 The radius of Earth is approximately 39632 miles It takes 23h 56m 4ls for the Earth to rotate once on its axis a Find the angular velocity and linear velocity at the equator b Find the linear velocity at 500 N latitude 211 Solution a We now shall use a t1me of 1 rev The angular velocity of Earth s spin is then given by 0 n m 0262516 M g 41 h 60 3600 360 We also could say on 56 4 1 m 150410 per hour 3 39 hr 60 3600 Note that it takes just under 24 hours to rotate 360 so every hour the Earth must rotate a little more than 1 24 th of a circle or just over 15 The linear velocity of Earth spin at the equator is about v 0 r z 0262516 7quot X 39632 milesm 10404 mph Iquot Dr Neal Spring 2009 b At 500 N latitude the angular velocity is the same but the radius is smaller Recall r 39632 cos50 at 500 N So the linear velocity at 500 N is now only 0262516 7quot X 39632 cos50 milesm 66876 mph Iquot Example 4 Jupiter rotates in approximately 9h 50m lts radius is 11194173 times that of Earth s Find the linear velocity at the equator of Jupiter Solution We again use v o r to obtain 211 m X 11194173 gtlt 39632 miles m 2834765 mph vnr Exercises 1 A cylinder with a 2 ft radius is spinning at 450 rpm a Find its angular velocity in degrees per sec b Find the linear speed on the rim in mph 2 If a cylinder with a 6 in radius is spinning at 24 mph find the angular velocity in rpm of a point on its rim 3 a What is the radius of the circle at 28 1539 N latitude b Find the linear velocity of the Earth s rotation at 28 1539 N c Find the distance between points at the following coordinates 28 1539 N 76 08 E and 28 1539 N 53 4439E 4 The radius of Mars is 0532036 times that of Earth One rotation on its axis takes about 24 hr 37 min 22662 sec Find the linear speed of its equatorial spin Dr Neal Spring 2009 Solutions 2 d d l a First 0 450 E X La 9001 1 Converting to deg per sec we have mm re mm d l 39 180 d u 900n1 X E X E 2700 persec mm 60 sec 11 rad b V m 900n x 2 ft gtlt eo x 1 M m6426mph mm hr 5280 ft 2 Because 6 in 05 feet we have m V X l 24 quotmes XL X 5280 253440 id r hr 05 ft mzle hr Then 253440 M X i E X ill l z 67227 rpm hr 211 rad 60 mm 3 a The radius is r m 39632 cos2825 m 3491146 miles 211 23 2 4391 60 3600 b The linear velocity is v u r gtlt 3491146 miles 91648 mph 0 We first find the angle between 76 0839E and 53 4439E which is given by 75 6839 7 53 4439 22 2439 224 The distance between the points is then s0r 224 gtlt gtlt 3491146m13648766 miles 211 rad 24 37 22662 60 3600 4 v u r gtlt 0532036 gtlt 39632 miles w 53805 mph Dr Neal Spring 2009 MATH 117 The Law of Sines Besides SideeAngleeSide SAS and SideeSideeSide SSS there are two other congruence forms that completely determine a triangle These are AngleeSideeAngle ASA and SideeAngleeAngle SAA In each case the third angle can be determined because all the angles must sum to 180 So with ASA and SAA we have all three angles but we must find the other two sides The Law of Cosines will not apply because it requires two sides to find the third side And we may not have a right triangle so we usually can t use rightetriangle trig In this case we use the Law of Sines We again label the sides of the triangle as a b c and the angles as A B and C As usual side a is opposite angle A side b is opposite angle B and side c is opposite angle C Then a b c sinA sinB sinC Law of Sines Example 1 Find the remaining sides in the following triangles 85 55 100quot 300 a v 391 3910 Solution i First let A 850 and C 55quot Then B 180 7 850 7 550 40quot The given side is b 10 To find sides a and c we have Ia 10 gt a10is1n85 m1539498 s1n85o s1n40o s1n40o i gt pm A 127437 s1n55o s1n40o s1n40o ii Let W 30quot and V 180 7 100 7 30quot 50quot Then W 40 V wmm203 andvmm3lll4 gt sin30o sinlOOo sin50o sinlOOo sinlOOo Dr Neal Spring 2009 When we have Side7Angle7Side then we can use the Law of Cosines to find the third side But it may be easier to use the Law of Sines to find the other two angles But now we use the reciprocal form sinA sinB sinC a b 0 Example 2 Find the remaining side and angles in the triangle below B 40 Solution To find side c we use the Law of Cosines c2 402 502 24050cos30 c 402 502 2 X 40 X 50 X cos30 m 25217 Now let c 25217 C 30quot a 50 and b 40 Then sinA sin30o sinB 50 25217 40 50 39 300 40 39 300 Then sinA i and sinB i But both of these equations have two 25217 25 217 solutions a first quadrant angle and a second quadrant angle However the two shortest sides in a triangle must have opposite angles that are acute less than 90quot Thus angle B must be less than 900 So B sin 1 m 5247760 Finally A m 180 7 30quot 7 524776 9752240 W 2 82478 which 25217 is not the correct value of A We then need A 180 7 82478 to give an obtuse second quadrant angle Note Using the inverse sine to solve for A we have sin71 Dr Neal Spring 2009 SideASideAAngle SSA A congruence form that may not give a unique triangle is sideisideiangle If the given angle is small enough then there could be two triangles made But if the given angle is obtuse then there will be only one possible triangle Example 3 Consider a triangle with A 20quot b l7 and a 10 Give two possibilities for the dimensions of this triangle Solution From the drawing we see that we have SSA So we cannot use the Law of sinA sinB Cosines and the only relationship we can establish is 811120 8173 Thus sinB which has two solutions which gives One Solution o 17 B sin 1 m 35550 10 10 20 B Another Solution 17 B 180 7 3555 14445 10 20 B 17 10 10 20 B B SideiSideAAngle May Give Two Solutions Dr Neal Spring 2009 MATH 117 Right Triangle Word Problems Here we shall consider various word problems that provide some physical applications of the rightitriangle trig formulas RightiTriangle Formulas x2y2z2 z lx2y2 x ZZ y2 y zz x2 x zcosG and y zsine Angles of Elevation and Depression We often measure an angle from the ground upward an angle of elevation or downward from an imaginary horizontal line an angle of depression But by alternate interior angles an angle of depression will be congruent to an angle of elevation l D sky A A ground ground An Angle of Elevation A An Angle of Depression D Then D 5 Instructions With each problem draw a diagram and completely label all necessary pieces Establish a rightrtriangle trigonometric relationship Give the exact algebraic solution in terms of the given information then give a numerical approximation Example 1 i A 15 ft ladder leans against a wall at an angle of elevation of 60quot a How high up the wall does the ladder rest b How far from the wall is the base of the ladder ii A 50 ft pole has a support wire that runs from its top to the ground with an angle of depression of 75quot a How far from the base of the pole does the wire connect to the ground b How much wire is used Dr Neal Spring 2009 Solutions 1 Let the height be h and the base be b Then a sin60 gt h15sin60o m 1299 ft h 1539 b b cos60 E gt b 1500s60o 75 ft b 750 Let the base be b and the wire be w Then a tan 75 2 gt i m 134 ft b tan75o 3911 50 W 50 50 b sin75 gt W m 5176 ft w s1n75o 750 b Example 2 A at 12 foot plank rests with one end on the ground and the other end upon a 4 foot ledge a How far from the base of the ledge is the far end of the plank b What is the grade ie angle of elevation Solution a 42 b2 122 gt bJ122 42 m 113 ft b sin1i gt A swig m 19470 12 12 Dr Neal Spring 2009 Example 3 Jamie is 539 8quot tall Find the length of her shadow if the angle of elevation of the sun is 3020 Solution Let s be the length of her shadow Then 0 5 8 12 tan302 g S 5398quot 5812 gt s m 97363 ft tan302 Example 4 Some wire connects from a pole to point on the ground at an angle of depression of 80quot On the ground the wire is 45 ft from the pole a How much wire is used b How high up the pole is the wire connected Solution Let the height be h and the wire be w 800 a cos80 w 45 m 259 ft h W w cos80 o h 45 b tan80 A 800 also gt h 45tan80 m 2552 ft Example 5 A 5 ft post is supposed to be vertical but it is 4 inches out of alignment at the top a What is the angle of lean b How high does it reach Solution a Sine 412 i 4quot 5 15 1 gt e sin 15 9 38220 539 h b h2 4 122 52 9 gt h 52 1 32 z 499 ft Dr Neal Spring 2009 Example 6 At a certain distance the angle of elevation to the top of a building is 60quot From 40 feet further back the angle of elevation is 450 Find the height of the building Solution h h tan60 and tan45 y 40 y b gt h ytan60 40 ytan45 600 Thus ytan60 ytan45 40tan45 40tan45 gt y tan60 tan45 40tan45 gtlt tan60 tan60 tan45 Then h ytan60 m 9464 ft Example 7 A building is 60 ft high From a distance at point A on the ground the angle of elevation to the top of the building is 40quot From a little nearer at point B the angle of elevation is 70quot Find the distance from point A to point B Solution 60 60 tan70 and tan40 x d x L and d i tan70 tan40 6039 The distance from point A to point B is 60 60 d x 7 m 49667 ft X B A tan40 tan70 Dr Neal Spring 2009 MATH 117 The Law of Cosines When we have a generic scalene triangle with no right angle then the Pythagorean Theorem no longer applies Moreover we no longer have a hypotenuse so the right triangle trig formulas do not apply either In this case we generally label the sides of the triangle as a b c and the angles as A B and C By convention side a is opposite angle A side b is opposite angle B and side c is opposite angle C The Law of Cosines allows us to find one side in terms of the other two sides and the angle in between those sides SideiAngleiside 02a2b2 2abcosC b2a cZ 2ac cosB a2b202 2bc cosA given aband 4C given acand AB given bcand 4A 2 Note If C 90quot then we have a right triangle Then cosC 0 and so c2 a2 b2 To prove the Law of Cosines for side c first drop a perpendicular to side b from angle B as shown below b b We now can apply the Pythagorean Theorem to the resulting two triangles First d2e2 a2sothat e2 612 512 Alsofb d Wethenhave c2e2f2a2 d2b d2 a2 d2b2 2bdd2 a2b2 2bd But then cosC 1 thus d acosC So the result becomes c2 a2 b2 2 ab cos C a Dr Neal Spring 2009 Example 1 Find the remaining side of the triangle 100 Solution Because we have SideeAngleeSide we can use the Law of Cosines to find the third side which we will call 6 Then c2 152 122 2 X 15 x12 gtlt cos100 so 6 152 122 2 X 15 x12 gtlt cos100 e 207729 SideeSideeSide If we have the three sides a triangle then we can find any of the angles using the Law of Cosines To solve for angle C we have 2 2 2 1 02a2b2 2abcosC gt 2ab cosCaZb2 c2 gt cosC a 2 22 Czcosel m 2ab 7 b2 2 2 7 2 2 b2 Similarly A cos 1 and 3 cos 1 L 2bc 2ac Recall that 005716 is always between 00 and 180 or between 0 an 11 in radians so we will always have one unique solution for any angle Example 2 1 Find angles A and B in the triangle below ii Find the height of the triangle iii Compute the area with E X base gtlt height and with Heron s Formula 20 10 Dr Neal Spring 2009 Solution 1 We know that 202 102 152 2 X 10 X 15 X cosA So then 2 2 2 cosA m 2 and A cos 1 7sje1044775 2 X10 X15 300 300 Likewise 102 152 202 2 X 15 X 20 X cosB Thus 152202 102 525 2 X 15 x20 600 52 cosB and B cos71 n 289550 600 11 Once we have angle A then we have its supplement 6 1800 A But angle A and angle 6 1800 A will have the same sine value one is in the 1st Quadrant the other is in the 2nd Quadrant In the drawing below we see that sine E so h 10sine 10in180 A1051nA e10sin1044775 6 A B iii The area of the original triangle is then ix 15 X h m X 15 X 10 sin1044775 n 7261844173 sq units 10 15 20 Using Heron s Formula we let s T 225 Then the area is J225225 20225 15225 10 J225 x25 gtlt75gtlt 125 e 7261843774 The slight difference comes from rounding off angle A cosil 5J to 1044775 The answers would be identical if we use i X 15 X 105m cosilT73J Dr Neal Spring 2009 Example 3 1 Find the third side of the triangle below 11 Find angle B in the triangle 111 Find the area 40quot 22 Solution 1 Call the third side c Then c a2 12 2ab cosC I162 222 2 X 16 gtlt22 gtlt cos40 e 14167 3911 We now have 222 e162 141672 2X 16 X 14167 gtlt cosB which gives 2 2 2 coshw and 3200571 2X16gtlt14167 162 141672 222 9345190 2 X 16 gtlt14167 111 We can find the height by sin40 116 which gives h 16sin40 The area is then 1 E X 22 gtlt16sin40 w 11313 square units 16 h 400 22 1622 14167 Using Heron s Formula we have s m T 260835 and Area e J260835260835 16260835 22260835 14167 e 11313 no noticeable difference Dr Neal Spring 2009 Example 4 Find the direct distance from Point A to Point C in the following cases i From Point A move 20 feet in a direction 250 South of West to Point B Then from Point B move 25 feet in a direction 30quot East of South to Point C ii From Point A move 10 feet in a direction 400 East of North to Point B Then from Point B move 12 feet in a direction 200 South of East to Point C Solution 1 We have two sides 2039 and 2539 but we must find the angle in between and the apply the Law of Cosines to find the distance b from Point A to Point B From the second drawing below we see that this angle is 85quot Then the desired distance is b I202 252 2 X 20 X 25 X cos85 m 30624 ft ii Similarly we find the angle in between to be 180 7 500 7 200 110 Then the distance is b 102 122 2 X 10 X 12 X cos110 m 180578 ft D Neal 5m 2m MATH 117 Tangent and Comng Gmphs The gaphs n km and ycax are penadmhke the m and mm mp1 hwwevex an tyne has lenglh n n nanm 1 1 1 4 I I 3 Once leaf m Omc leaf mlx On We um Wemmpmmnwmnz and mm me We a an nwemmphmwmn and n h an we Lhwugh we mm mm ph ex XE m be the pmntufane cycle Thehancplvmngpmntsfm earhgl a are AswALh an and came gaphs we can sh 39t and campus at expand the gaphs Mun the 32112an funmans yambx0 and yacmbxc Lhelerglh ufane cycle an005 Phshmkm CounsemRoots lisnmkPex TangenAAsynywtzs lslAsympkPex CompuAsympwkzs Phsmnkm D Neal 5pm 2cm range anh Ta label ve paints and xeaxas stan at x the phase sham a Mama gm apemam aha gm apemam Colagem anh aeaaaavepaasmae am 5 aahepaaawga apema Banglzs ON a damned gaph nae man ruheaahs State an mm a me mhmahs State a asymmaus unhe ruhmahs a y4hh2x h y2m3xr a Salaam a pa he phase shm same 2er Z 3 gg Wampum phase shen ah nae mddh than subtract and add lAJLh ht nae penad twa umes eaeh m ablam x E The penad 1s Naw dede me penad mm A piece a lengvh Fm Subtract 7 a E 1 2 1J L 2M0 m 2 26 26 4 Rants PhShmkPex 42J Ma D Neal 5pm 2cm The penad 1s 2 E Naw dame the pznad mm A piece a length Put the phase sh 39t m the mmdle eheh rhp graph h Phase shm Salve 3x Subtract ea Subtract E 2quot E 3 a me Phshmma ex 3 Ma e FhaseShm smveAXaTn gt x ThepememZ 4 mm pieces aflergvh Fax a mtmmgraph putthephase sh 39t anhe m Naw dede me pznad Wm Phshmma ex 3 WEJ m a me Mammyquot k la w D Neal 5m 2m anlw at Sam and Cami The hancgraphs uf ysecx and ycscx can he ahtamed w rm ypahug ycasx and y ux de mrg asymgmus the mats um nupmmurg Lhegraphs 3 5n Fm x yzsecx casx 7797 y f 2 2 y nx ycscx Dr Neal Spring 2009 MATH 117 Right Triangle Trig Previously we have seen the right triangle formulas x rcose and y rsine where the hypotenuse r comes from the radius of a circle and x is adjacent to 6 and y is opposite of 6 But these formulas hold in any right triangle with hypotenuse 2 even outside the context of a circle radius hYP r y rsine Z yzsin6 opp opp I I xrcose xZCOSG adj adj Solving for 0056 and sine we have x Adjacent smO 2 H ypotenuse 2 H ypotenuse y Opposite cosO 7 The Other Trig Functions We define four other trigonometric functions the tangent the cotangent the secant and the cosecant as follows O t 1 Ad I Tangent tanB X Cotangent cote f x Adjacent tane y Opposite 1 H t 1 H t Secant 56306 E M Cosecant csc 6 i M cose x Adjacent s1n6 y Opposite Finding the Trig Function Values and Angles Given Two Sides Given any two sides of a right triangle we can use the Pythagorean Theorem to find the third side Then we can compute the six trig function values of the acute angles in the triangle by using the above ratios We also can use a calculator to approximate the measure of the angles Dr Neal Spring 2009 Example 1 In the triangle below a Find the remaining side b Find the six trig function values of the acute angles c Find the approximate degree measures of the acute angles 56 Solution By the Pythagorean Theorem 62 332 562 so 6 l332 562 J4225 65 With respect to angle a With respect to angle 5 adj 55 Opp 33 hyp 55 adj 33 opp 56 hyp 65 ad 56 ad 33 cos cosB hyp 65 hyp 65 smot Opp g sin Opp E hyp 65 hyp 65 tana 0pp tan 0pp adj 56 adj 33 1 56 1 33 cota cotf tana 33 tan3 56 1 65 1 65 seca SB 3 coso 56 cos 33 1 65 1 65 csc a cscf s1na 33 s1n 5 56 To find the measures of a and 3 put your calculator in degree mode and evaluate COS 15665 and COS 13365 The COS 1 inverse cosine button is 2ND COS Then 56 33 a cos 1 j m 3051 and 3 cos j m 5949 65 65 Dr Neal Spring 2009 Example 2 In the triangle below a Find the remaining side b Find the six trig function values of the acute angles c Find the approximate degree measures of the acute angles d 6 85 157 A Solution Because 852 d2 1572 132 6 we have dJ1572 852 J17424 132 85 157 A With respect to angle A With respect to angle 6 adj 85 opp 132 hyp 157 adj 132 opp 85 hyp 157 cosh z 008661 611232 hyp 157 hyp 157 sinx smez z hyp 157 hyp 157 opp 132 opp 85 tan t adj 85 adj 132 1 85 1 132 001 cot tank 132 tanG 85 1 157 1 157 sec sec 2 cos 85 cose 132 1 157 1 157 csc7 csce smk 132 sme 85 The measures of 7 and 6 are 7 cosilj m 57220 and 6 cosilj m 3278 Dr Neal Spring 2009 Finding the Remaining Sides Given a Side and an Angle Given a side and one acute angle in a right triangle then the other angle is simply the complement of the first ie they add up to 90quot The other two sides can be found by using the appropriate right triangle ratio Example 3 Find the remaining pieces of the following right triangles using only the given labeled information on each a 55 40 I 30quot 391 a ii iii a Solution 1 First on 50quot which is the complement of the given 400 angle But a is not needed to find the other sides We need to apply the appropriate formula below cose sin67 tan67 H p Part 1 7 With respect to 6 40quot we have adj a opp b hyp 12 A So cos40 H 12 gt a 120040o m 919 via calculator O b and sin40 pr gt b 12in40o m 771 via calculator Part ii 7 With respect to B 55quot we have adj 8 opp w hyp 2 So 005550 E 8 2 m 1394757 via calculator z cos55o O and tan 55 gt w 8tan55 m 11425 via calculator J Dr Neal Spring 2009 Part iii 7 With respect to 8 30quot we have adj a opp 20 hyp c Sosin30 2 ci 240 Hy c s1n30o l 2 O 20 20 20 andtan30 gt a 20J Adj a tan30o i J Right Triangle Trig on the xy Plane Any set of x y coordinates determines an angle 6 by considering the segment from 0 0 to x y The trig function values can be quickly found by using x cose s1n6 2 tanB z z x where we always take 2 Ix2 y2 Example 4 Find the six trig function values of the positive angle 6 determined by the following points Then find the approximate measure of the angle 1 79 40 ii 780 739 iii 88 7105 Solution In each case we first need to find the hypotenuse 2 Then we apply the right triangle trig formulas to find the six trig function values Finally we39ll use tan 1 yx to find the angle 6 In each case though we will have to add either 180 or 360 to adjust for the correct quadrant of 6 Part 391 First x 79 y 40 and 2 J92 402 J168 41 Then ln degree mode tan 1 4079 m 77732 which is in the 4th Quadrant But 79 40 is in the 2nd Quadrant so add 180 to adjust 6 m 77732quot 1800102680 Dr Neal Spring 2009 Part 3911 First x 780 y 739 and 2 J802 392 J792 89 So cos6 Q sine 2 tane 2 cot secS cscG 89 89 80 39 80 39 In degree mode tan 1 739780 m 2599 which is in the 1st Quadrant But the given point 780 739 is in the 3rd Quadrant so add 180 to adjust Then 6 m 2599 180 m 205990 Part 111 We have x 88 y 7105 and 2 J882 1052 J18769 137 Thus 105 105 88 137 137 sine tan cote sece csc cose 137 88 105 88 105 g 137 In degree mode tan 1 710588 m 750 which is in the 4th Quadrant as is 88 7105 But to get a positive angle add 360 So we actually have 0 m 310 Finding Other Trig Function Values of 6 Given one trig function value of an angle 6 and its quadrant we can make an x y 2 system to find the other trig function values Always keep the hypotenuse 2 positive and use the appropriate 7 signs for x and y depending on the quadrant Example 5 Find the remaining trig function values of the described angle 0 i 56300 and 0 is in III ii csce 3 and 0 is in II iii cote 132 and 9 is in IV 7 5 Solution 1 For 56306 g then 0056 I And y is also negative because 0 is in 7 z Quadrant III Then y l 22 x2 l49 52 J2 2Jg The other four trig function values are Dr Neal Spring 2009 sine 2 Z JE csc L z 7 y 2 6 3119222 cotezfzi x 5 y 2 6 9 4 ii For 0500 2 then sine 3 2 Butx is negative because 6 is in Quadrant 11 So 2 x 122 y2 81 1 J5 The other four trig function values are cos6 secB 9 z 9 J5 9 4 tan6z i cote x J5 y 4 5 12 iii cote tane2 5 12 5 x 6 is in IV s0 x is pos and y is neg 712 13 z J52 122 13 Quad IV 5 13 COSB sece z 13 x 5 sin6222 2 csee z 13 y 12 Dr Neal Spring 2009 MATH 117 Arithmetic Formulas Given angles A and B there several important arithmetic formulas for the sine and cosine of the angles A i B as well as angles such as A 900 133 2A and B 2 In each case we can obtain an exact formula for the result in terms of the sine and cosine of the original angles A and B without knowing the actual values of the angles A and B Negative Angles Because angles A and A create the same x7 coordinates on the unit circle but negative y x y coordinates we have cosA cos A and x y sin A sinA Complimentary Angles x Because the x and y values are switched for y complimentary angles such as 30quot and 60quot we have 90 A Xay sin90 A cosA and A cos900 A sinA Supplementary Angles Because supplementary angles such as 30quot o and 150 have the same yicoordinates but 180 A negative xicoordinates we have x y Ly 39 A sinA sin180o A and coslSOO A cosA Dr Neal Spring 2009 AdditionSubtraction Formulas sinA B sinA cosB cosA sinB cosA B cosA cosB sinA sinB sinA B sinA cosB cosA sinB cosA B cosA cosB sinA sinB Double Angle Formulas sin2A 2sinA cosA cos2A cos2 A sin2 A HalfiAngle Formulas B 1 005B B 1 cosB s1n cos i 2 2 2 2 2 5 Example Assume tanA 7 with angle A in Quadrant Ill and cscB 2 with angle B in Quadrant IV a Use rightitriangle trig to determine the sines and cosines of angles A and B b Find sinA B cosA B sinA B cosA B 0 Determine what quadrants the angles A B and A B lie in d Find sin2A and cos2A Determine what quadrant the angle 2A lies in e Find sinB 2 and cosB 2 Determine what quadrant the angle B 2 lies in Solution With A in 111 and tanA both x and y are negative So y 2 x 7 and zl2272 5 For B in IV y is negative but x is positive Also 4 sin B csc B 5 Angle A in Quadrant lll Angle B in Quadrant IV 2 4 3 sinA and cosA sinB and cosB J53 J53 5 5 Dr Neal Spring 2009 sinA B sinA cosB cosA sinB cosA B cosA cosB sinA sinB 2 3 7 4 7 3 2 4 z x5 x5 ZEXTEX5 628 22 21 8 29 5J5 5 J5 5J5 sinA B sinA cosB cosA sinB cosA B cosA cosB sinA sinB 2 3 7 4 7 3 2 4 Z X gtlt gtlt gtlt J5 5 J5 5 J5 5 J5 5 6 28 34 218 13 5J5 15J5 A B must be in Quadrant 11 cos is neg sin is pos A B must be in Quadrant III sin cos are both neg sin2A2sinA cosA 0052ACOSZASin2A 2 7 2 2 2E JE J 7 2 1 53 53 53 2A must be in the 1st Quadrant because both sin and cos are positive For angle B2 first note that angle B is in Quadrant IV Thus 270 SBS360quot B Dividing by Z we obtain 1350 S3 S1800 Thus B 2 is in Quadrant 11 So the sine will be positive but the cosine will be negative 3 l cosB 1g B 1cosB 1g sm cos quot 2 2 2 2 2 2 lez z X1 i2 3 5 2 5 5 5 2 5 5 5 Dr Neal Spring 2009 15 and 75 Angles The 150 and 750 reference angles througout the quadrants create many examples for using the Sum and Difference Angle Formulas Each angle can be written as a sum or difference involving the 30quot 45quot and 600 reference angles for which we know the sines and cosines For example we can use 150 600 7 450 or 150 450 7 30quot 105 750 165 150 V 195 345 255 285 Example Compute the exact value of sin15 and cos15 Then compute sin75 and cos75 Solution Using 150 450 7 30quot we have sin15 sin45 30 cos15 cos45 30 sin45o cos30quot cos45o sin30o cos45quotcos30quotsin45o sin30o J5 J3 J2 1 and 5 Ji 5 1 gtlt gtlt gtlt gtlt 2 2 2 2 2 2 2 2 Ja J2 J6 47 4 4 Because 750 is the complement of 15quot we should have that sin75 cos15 and cos75 sin15 But using 750 450 30quot we obtain sin75 sin45 30 cos75 cos45 30 sin45o cos30quotcos45o sin30o 005450 cos30quot sin45o sin30o J2 J2 J2 1 J2 J2 J2 1 gtlt gtlt and gtlt gtlt 2 2 2 2 2 2 2 2 cos15 w sin15 Dr Neal Spring 2009 Exercises 5 7 1 Suppose secA Z w1th A 1n Quadll and cotB E w1th B 1n Quad lll Draw the information on right triangles Then find the exact values of a sinA B and sinA B b cosA B and cosA B A A c s1n2B and cos2B d 81113 and 0053 e Explain what quadrants the angles A B A B 2B and A 2 are in Z Use the sum addition formulas to find the exact values of a sin105 and coslOSquot b sinl65 and cosl65quot c sin255 and cos255quot d sin345 and cos345quot Angle A in II sinA B sinA cosB cosA sinB 3 7 4 10 X X 5 J149 5 J149 21 40 19 5J149 5J149 sinA B sinA cosB cosA sinB 3 7 4 10 TW TW 21 40 61 5 W Dr Neal Spring 2009 Answer Angle B in 111 cosA B cos A cosB sinA sinB 4 7 3 10 X X 5 J149 5 J149 2830 58 5J149 5J149 cosA B cosA cosB sinA sinB 4 7 3 10 gtlt gtlt 5 J149 5 J149 28 30 2 5J149 5J149 A B must be in Quadrant 1 cos is pos sin is pos A B must be in Quadrant lll sin cos are both neg sin2B 2sin B 005B 10 7 J 2 J149 J149 140 149 cos2B cos2 B sin2 B 4 JZ l fr 2 149 149 49 100 i 2149 149 149 2B must be in Quadrant ll because cos is neg and sin is pos For angle A 2 first note that angle A is in Quadrant ll Thus 90 S A S 180 A Dividing by Z we obtain 450 S E S 90 Thus A 2 is in Quadrant I So the sine will be positive and the cosine will be positive A 1cosA cos quot 2 2 1 X 2 Lan a l

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