Ch18 radian弧度

1、18.1 Radian & 18.2 Length of arc and area of sectorThe radian is useful to distinguish between quantities of different nature but the same dimension. For example angular velocity can be measured in radians per second (rad/s). The angle subtended at the center of a circle by an arcof circumferenc

2、e equal in length to the radius of the circleis one radian. Angle measures in radians are often given without any explicit unit. When a unit is given, sometimesthe abbreviation rad is usedThere are 2 (approximately 6.28318531) radians in a complete circle.How can we convert betweenradians and degree

3、s?A complete circle 3602 rador180 radUsing this conversion, find:1 degree in radians = 1 radian in degrees = 0.0174557.2958.sr22srThe ratio of the arclength to the circumferenceis the same as the ratioof the angle to 360 deg.srCan we use the same idea to find the area A ?A22Ar212ArRemember theseform

4、ulae butonly with radians!Why measure angles in radians?Here is one reason (there are others ).Suppose we want to calculate the first derivativeof y = sinx at x = 0.We dont have a simple formula, so lets use the definitionof derivative (limits):0sin()sin( )limxdyxxxdxxPut x = 0. Whathappens?0sinlimx

5、xx00.511.522.5311.522.533.544.5Why measure angles in radians?rrsinrAs 0, sinrr“arc”“chord”P(pán)QAOWhy measure angles in radians?1dydxNow you should read carefully Example 18.2.1 on p. 266and then do:Exercise 18A Q1 a,e,i Q2 a,e,i Q5, Q8, Q10Later, we will discuss the formula:(sin )cosdxxdxwhich works on

6、ly in radians.Example 18.2.1Find the perimeter and the area of the segment cut off by a chord of length 8cm from a circle center and radius 6cm. Give your answer correct to 3 significant figures.PQOThe perimeter of the segment consists of two parts, the straight part of the length 8cm,and the curved

7、 part;to calculate the length of the curved part you need to know the angle POQCall this angle , As triangle is isosceles, a perpendicular drawn from to bisects both and angle POQPOQOPQPQThen the perimeter cm is given by ,666. 06421sinSo and 7297. 021459. 1d756.1668dThe perimeter is 16.8cm,correct t

8、o 3 significant figures.To find the area of the segment,you need to find the area of the sector ,and then subtract the area of the triangle . Using the formula for the area of a triangle, the area of the triangle is given by . Thus the area in of the shaded region is OPQOPQAbcsin21sin212rPOQ2cm459.

9、1sin621459. 1621sin21212222rr381. 8The area is 8.38 ,correct to 3 significant figures.2cmExample 18.2.2A chord of a circle which subtends an angle of at the center of the circle cuts off a segment equal in area to of the area of the whole circle.31(a) Let rcmbe the radius of the circle. Using a meth

10、of similar to the on in example 18.2.1, the area of the segment is sin212122rrThis is of the area of the whole circle if 3122231sin2121rrrMultiplying by 2 and dividing by you find 2r32sin(b) If you substitute in the equation you get 61. 2 sinf103. 261. 2sin61. 261. 2fwhich is very close to094. 232Th

11、is suggest that is close to 2.61, but it is not enough to show that it is 2.61 correct to 2 decimal places. To do that you need to show that lies between 2.605 and 2.615.It is obvious form Fig.18.5 that lies between 0 and ,and that the shaded area gets larger as increases. So you have to show that t

12、he area is too small when and too large when 615. 2605. 2The first of these is smaller, and the second larger, than andf,093. 2605. 2sin605. 2605. 2.112. 2615. 2sin615. 2615. 2f094. 232It follows that the root of the equation is between 2.605 and 2.615; that is, the root is 2.61, correct to 2 decima

13、l places.18.3 Graphs of the trigonometric functionsThe graph of when the angle is measured in radians have a similar shape to those for which are drawn in Figs.10.3,10.4 and 10.5 on pages 139 and 140. The only change is the scale along the tan sin,cosyandyytan sin,cosyandyy.axisyxy = sin xyxy = cos

14、x22yxy = tan x232223 The function have the following properties. Periodic property: Odd property: Translation property :sin cosandcos2cossin2sincoscossinsincoscoscoscossinsinsinsinIf you refer to the graph of ,and think in the same way as with the cosien and sine graphs,you can obtain similar result

15、s: tan The function have the following properties.Periodic property:Odd property:tantantantantantantanNote that the period of the graph of is 180.and that the translation property of is the same as the periodic propertytantan18.4 Inverse trigonometric functionsNone of the trigonometric functions sat

16、isfies the horizontal line test, so none of them has an inverse. The inverse trigonometric functions are defined to be the inverses of particular parts of the trigonometric functions; parts that do have inverses.Inverse Sine Functionyxy = sin xSin x has an inverse function on this interval. Recall t

17、hat for a function to have an inverse, it must be a one-to-one function and pass the Horizontal Line Test.f(x) = sin x does not pass the Horizontal Line Testand must be restricted to find its inverse.Copyright by Houghton Mifflin Company, Inc. All rights reserved.21The inverse sine function is defin

18、ed byy = arcsin x if and only ifsin y = x. Angle whose sine is xThe domain of y = arcsin x is 1, 1.Example: 1a. arcsin261 is the angle whose sine is .6213b. sin233sin32This is another way to write arcsin x.The range of y = arcsin x is /2 , /2.Copyright by Houghton Mifflin Company, Inc. All rights re

19、served.23Inverse Cosine FunctionCos x has an inverse function on this interval.f(x) = cos x must be restricted to find its inverse.yxy = cos xCopyright by Houghton Mifflin Company, Inc. All rights reserved.24The inverse cosine function is defined byy = arccos x if and only ifcos y = x. Angle whose c

20、osine is xThe domain of y = arccos x is 1, 1.Example: 1a.) arccos231 is the angle whose cosine is .32135b.) cos2635cos62 This is another way to write arccos x.The range of y = arccos x is 0 , .Copyright by Houghton Mifflin Company, Inc. All rights reserved.26Inverse Tangent Functionf(x) = tan x must

21、 be restricted to find its inverse.Tan x has an inverse function on this interval.yxy = tan xCopyright by Houghton Mifflin Company, Inc. All rights reserved.27The inverse tangent function is defined byy = arctan x if and only iftan y = x. Angle whose tangent is xExample: 3a.) arctan363 is the angle

22、whose tangent is .631b.) tan33tan33This is another way to write arctan x.The domain of y = arctan x is .(, ) The range of y = arctan x is /2 , /2.Copyright by Houghton Mifflin Company, Inc. All rights reserved.29Composition of Functions:f(f 1(x) = x and (f 1(f(x) = x. If 1 x 1 and /2 y /2, thensin(a

23、rcsin x) = x and arcsin(sin y) = y. If 1 x 1 and 0 y , thencos(arccos x) = x and arccos(cos y) = y. If x is a real number and /2 y /2, thentan(arctan x) = x and arctan(tan y) = y. Example: tan(arctan 4) = 4Inverse Properties:Copyright by Houghton Mifflin Company, Inc. All rights reserved.30Example:a

24、. sin1(sin (/2) = /2 15b. sinsin353 does not lie in the range of the arcsine function, /2 y /2. yx5335233 However, it is coterminal with which does lie in the range of the arcsine function.115sinsinsinsin333 Copyright by Houghton Mifflin Company, Inc. All rights reserved.31Example:2Find the exact va

25、lue of tan arccos.3xy32adj22Let =arccos, thencos.3hyp3uu 22325opp52tan arccostan3adj2uuCopyright by Houghton Mifflin Company, Inc. All rights reserved.3218.5 Solving trigonometric equations using radiansWhen you have a trigonometric equation to solve, you will sometimes want to find an angle in radi

26、ans. The principles are similar to those that you used for working in degrees in section 10.5,but the function will now have the meanings that were assigned to them in section 18.4. 111tan sin,cosandExample 18.5.1Solve the equation ,giving all the roots in the interval correct to 2 decimal places. 7

27、 . 0cos20Step 1 This is one root in the interval 346. 27 . 0cos120Step 2 Use the symmetry property to show that -2.346. is another root. Note that -2.346. is not in the required interval. coscosStep 3 Use the periodic property, ,to say that is a root in the required interval. cos2cos936. 32346. 2The

28、 roots of the equation in are 2.35 and 3.94,correct to 2 decimal places.7 . 0cos20Example 18.5.2Solve the equation ,giving all the roots in the interval correct to 2 decimal places. 2 . 0sinStep 1 This is one root in the interval 201. 02 . 0sin1Step 2 Another root of the equation is , but this is no

29、t in the required interval.342. 3201. 0Step 3 Subtracting gives ,the other root in the interval 2940. 2Therefore the roots of in are -2.94 and -0.20,correct to 2 decimal places.2 . 0sinExample 18.5.3Solve the equation ,giving all the roots in the interval correct to 2 decimal places. 3 . 01 . 03cosL

30、et ,so that the equation becomes . As lies in the interval lies in the interval which is 1 . 033 . 0cos1 . 03,1 . 031 . 03324. 9524. 9The first part of the problem is to solve for 3 . 0cos324. 9524. 9Step 1 This is one root in the iterval .226. 13 . 0cos1324. 9524. 9Step 3 Adding to and subtracting from gives and as the other roots.22266. 1017. 5549. 7Since ,the roots of the original equation a