Trigonometry Chapter 3 Lecture Notes

Transcription

1 Ch Notes Morrison Trigonometry Chapter Lecture Notes Section. Radian Measure I. Radian Measure A. Terminology When a central angle (θ) intercepts the circumference of a circle, the length of the piece subtended (cut off) is called the arc length (s). B. The radian measure of an angle (θ) is the ratio of the arc length (s) to the radius of the circle (r), i.e. s θ r C. If the arc length subtended by angle θ is equal to the radius (i.e., when s = r), then θ has a measure of radian. D. Since a radian is defined as a ratio of two lengths, the units cancel and the measure is considered unit-less. Therefore, if an angle measure is written with no degree symbol, it is assumed to be in radians. Though it is not essential, it is often customary to write radians or rads after an input measured in radians, especially when doing conversions and canceling units. E. In many applications of trigonometry, radian measure is preferred over degree measure because it simplifies calculation and allows us to use the set of real numbers as the domain of the trig functions rather than just angles. II. Converting Standard Angles Between Degrees and Radians A. One full rotation measures or approximately.8 radians. Thus radians = 0 ; radians = 80 ; radians 90 ; radians 0 ; radians 5 ; and radians 0. B. It is absolutely essential that you know the equivalent degree and radian measures for all standard angles. This is not difficult if you think of every standard angle as a multiple of one of our four special angles - 0, 5, 0, and 90. For example, if 0, then 0, 90, etc. If 5, then 90 ; 5 ; and so on. Example : Convert the following angles from degree measure to radian measure. 0 (#8) b. 5 c c = 0 b

2 Example : Convert the following inputs from radian measure to degree measure. 7 rads (#) b. radians c. Ch Notes Morrison 5 radians b c C. You should memorize the following diagram. D. You should memorize the following table. Degree-Radian Conversion Factors Multiples of 0 & Multiples of 5 & Multiples of 0 & Multiples of 90 & 0 = 5 = 0 = 90 = 0 = 90 = 0 = 80 = 90 = 5 = 80 = 70 = 0 = 80 = 0 = 0 =

3 Ch Notes Morrison 5 50 = 80 = 7 0 = 8 0 = 9 70 = 0 00 = 0 = 0 = 5 5 = 70 = 7 5 = 8 0 = 5 00 = 0 = III. Example Converting Nonstandard Angles Between Degrees and Radians A. To convert a non-standard angle from degrees to radians, multiply by Convert each nonstandard angle from degree measure to radians. 7 50' (#0) b (#5) 80 and simplify. b. Example B. To convert a nonstandard angle from radians to degrees, multiply by 80 and simplify. Convert each radian measure to degrees. In part b, round to the nearest minute. 7 (#) b..0 (#50) 0 b. b.

4 Ch Notes Morrison Example 5 Angle θ is an integer when measured in radians. Give the radian measure of the angle. (#) Example Find the exact value of each expression without using a calculator. 5 cot (#) b. sin c. 5 sec b. terminates in Quadrant II and has a reference angle of. cot A O 5 7 is coterminal with which terminates in Quadrant III. 5 7 O It has a reference angle of. sin = sin H c. 5 7 is coterminal with which terminates in quadrant IV. 5 7 H It has a reference angle of. sec = sec A ************************************************************************************ Section. I. Arc Length Applications of Radian Measure The length (s) of the arc intercepted on a circle of radius (r) by a central angle θ (measured in radians) is given by the product of the radius and the angle, i.e., s = r θ. Caution: θ must be in radians to use this formul Example Find the length of the arc intercepted by a central angle r = 7.9 cm. (#0) θ 5 in a circle of radius

5 Ch Notes Morrison Example Find the distance in kilometers between Farmersville, California, N, and Penticton, British Columbia, 9 N, assuming they lie on the same north-south line. The radius of the earth is 00 km. (#) Example A small gear and a large gear are meshed. An 80.0 rotation of the smaller gear causes the larger gear to rotate Find the radius of the larger gear if the smaller gear has a radius of.7 cm. (#) II. Area of a Sector of a Circle A. A sector of a circle is the portion of the interior of the circle intercepted by a central angle. B. The area (A) of a sector of a circle of radius r with central angle θ (measured in radians) is given by A r θ. Caution: θ must be in radians to use this formul 5

6 Ch Notes Morrison Example Find the area of a sector of a circle with radius 8. cm and central angle θ = 5. Round to the nearest tenth. (#) Example 5 The Ford Model A, built from 98 to 9, had a single windshield wiper on the driver s side. The total arm and blade was 0 inches long and rotated back and forth through an angle of 95. If the wiper blade was 7 inches, how many square inches of the windshield did the blade clean? (#) ************************************************************************************ Section. The Unit Circle and Circular Functions I. Introduction In the 00s, scientists began using trigonometry to solve problems in physics and engineering. Such applications necessitated extending the domains of the trigonometric functions to include all real numbers, not just a set of angles. This extension was accomplished by using a correspondence between an angle and the length of an arc on a unit circle (a circle with a radius of, centered on the origin, with equation x y ). II. The Unit Circle Imagine that the real number line is wrapped around a unit circle. Zero is at the point (, 0); the positive numbers wrap in a counterclockwise direction; and the negative numbers wrap in a clockwise direction. Each real number t corresponds to a point (x, y) on the circle. If central angle θ, in standard position, measured in radians, subtends an arc length of t, then according to the arc length formula (s = r θ), t = θ.

7 Ch Notes Morrison Thus, on a unit circle, the measure of a central angle and the length of its arc can both be represented by the same real number, t. III. The Unit Circle Definitions of the Trigonometric Functions If t is a real number and (x, y) is the point on the unit circle corresponding to t, then y cos t = x sin t = y tan t =, x 0 x x sec t =, x 0 csc t =, y 0 cot t =, y 0 x y y Example : Use the diagram below to evaluate the six circular values of θ. (#) 5 cos θ 7 sin θ tan θ sec θ csc θ cot θ IV. Domains of the Circular Functions Since x = 0 when t and t, sec t and tan t, which both have x in the denominator, are not defined there or at any odd multiple of. Thus the domain of sec t and tan t is t t (n ). Since y = 0 at t 0, t, and t, csc t and cot t, which both have y in the denominator, are undefined at any integer multiple of. Thus the domain of cot t and csc t is t t n. Sine and cosine do not have any restrictions on their domain, thus their domain is,. V. Finding Exact Values of Circular Functions by Using Special Points on the Unit Circle Angle 0 or Point (, 0),,, (0, ) (-, 0) (0, -) The x-value of a point on the unit circle tells you cos t and the y-value tells you sin t. The sign of the function values depends on where the angle terminates: Quadrant I: Quadrant II: Quadrant III: Quadrant IV: x and y are both positive, therefore all function values are positive. x is negative and y is positive, therefore only sine and cosecant are positive. x and y are both negative, therefore only tangent and cotangent are positive. x is positive and y is negative, therefore only cosine and secant are positive. 7

8 Ch Notes Morrison Example : Example : Find the exact value of (a) cos θ, (b) sin θ, and (c) tan θ for θ. (#) The point associated with is (0, ). Therefore, cos θ = 0; sin θ = ; tan θ is undefined; cot θ = 0; csc θ = ; sec θ is undefined. Find the exact value of (a) csc (#); (b) cos (#); (c) 0 cot. terminates in quadrant IV. Its reference angle is which is associated with the point cos ;,. In quadrant IV, only cosine and secant are positive. sin ; sec ; csc ; cot tan ; b. terminates in quadrant II. It has a reference angle of which is associated with the point. In quadrant II, only sine and cosecant are positive. cos ; sec, sin tan ; csc ; cot 8

9 Ch Notes Morrison c. 0 is larger than, so we must first find its smallest positive coterminal angle. 0 which terminates in quadrant III and has a reference angle of. The point associated with is cotangent are positive. cos ; sec ;, sin ; tan csc ; cot and in quadrant III only tangent and VI. Finding Approximate Values of Circular Functions To find the approximate value of a circular function, we use a calculator set on RADIAN mode. For secant, cosecant, and cotangent, we use sec θ, csc θ, and cot θ. cos θ sin θ tan θ Example : Find the approximate value of sec (-8.9). (#) sec (-8.9) = cos VII. Determining a Number with a Given Circular Function Value A. To determine, without a calculator, what input generates a specified output, we must be able to recognize our special point values in all their alternate forms, for example, ; ; and. And we must know which functions are positive in each quadrant (A S T C). Example 5 Find the exact value of t if sin t in the interval,. Do not use a calculator (# 58). A sine value of indicates the reference angle, t, is and since the given interval is quadrant III, t must be. B. To determine, with a calculator, what input generates a specified output, we use an inverse trig function and the fact that if fxnt y ; then fxn y t. 9

10 Ch Notes Morrison Example Find the values of t in 0, such that (a) cos t =.78. (#50) and (b) csc t =.09 (#5). VIII. Applying Circular Functions Since the given interval is in radians, we set the mode to RADIANS and type cos b. Again, the mode should be radians, but to find t = csc.09 we must first say csc t =.09 implies sin t and sin t Example 7 The temperature in Fairbanks, Alaska is modeled by Tx 7sin x where T(x) is the temperature in degrees Fahrenheit on day x, with x = corresponding to January and x = 5 corresponding to December. Use the model to estimate the temperature on March. (#7) 5. Since March corresponds to day 0, T0 7sin F ************************************************************************************ Section. Linear and Angular Speed I. Arc Length s r θ Alternate form: s r ω t For a circle with radius r and central angle θ measured in radians, if θ intercepts an arc length of s, then s r θ. Hint: For two pulleys connected by a belt or for two intermeshed gears, s must equal s, so r θ r. θ II. Linear Speed s v t Alternate forms: r θ v and v r ω t Consider a fly sitting on the tire of a bicycle. If we measure the linear distance (s) the fly travels per unit of time (t), we are describing its linear speed (v). Linear speed is measures in units such as miles per hour and feet per second. III. Angular Speed θ ω t Alternate form: s ω r t If we measure the angle (θ) the fly travels through per unit of time (t), we are describing its angular speed (ω). Angular speed is measured in units such as radians per second and degrees per minute. To convert revolutions per minute to an angular speed, we multiply by since there are radians in one revolution. Note: The symbol for angular speed, ω, is the Greek letter omeg 0

11 Ch Notes Morrison IV. The Relationship between Linear Speed and Angular Speed v r ω The linear speed (v) and the angular speed (ω) of a point moving in a circular path with radius r are related by the formula v r ω. Hint: For two pulleys connected by a belt or for two intermeshed gears, v must equal v, so r ω r ω. Example Example Suppose that point P is on a circle with radius r, and ray OP is rotating with angular speed ω. If r = 0 cm, ω radians per sec, and t = sec, find the following. (#) 0 the angle generated by P in time t ω θ t θ ω t θ 0 5 radians b. the distance traveled by P along the circle in time t s r ω t s 0 cm 0 c. the linear speed of P v r ω v 0 cm per sec 0 Given θ ω t θ radians and ω radians per min, find t. (#0) 8 θ t θ t 9 min ω ω 8 Example Given r = 8 cm and v r ω v 9 ω radians per sec, find v. (#) cm per sec 5 5 Example Given s m, r m, and 5 s r ω t ω, find t. (#) 5 s 5 t t 5 r ω 5 5 sec Example 5 Find ω for a line from the center to the edge of a CD revolving at 00 times per min. (#) 00 rev ω min rad rev 00 radians per min

12 Ch Notes Morrison Example Find v for a point on the tread of a tire of radius 8 cm, rotating 5 times per min. (#) Example 7 5 rotations v 8cm min radians rotation =0 cm per min The earth revolves on its axis once every hours. Assuming that Earth s radius is 00 km, find the following. (#8) angular speed of Earth in radians per day and radians per hour rotation radians ω radians per day day rotation radians day ω radians per hour day hours b. linear speed at the North Pole or South Pole Since r = 0 at the North or South Pole, v = 0 at the North or South Pole. c. linear speed at Quito, Ecuador, a city on the equator At the equator, r = 00 km, so v or v 00 5 km per hour km per day d. linear speed at Salem, Oregon (halfway from the equator to the North Pole) Since Salem is halfway between the equator and the North Pole, if we draw a radius from the center of the earth to Salem, it forms a 5 angle with the equator. Thus r sin 5 r 00 sin 5 00 r km Therefore, v km per day 800 or v km per hr