Chapter 6. Trigonometric Functions of Angles 6.1 Angle Measure Radian Measure 1 radians 180º Therefore, o 180 π 1 rad, or π 1º 180 rad Angle Measure Conversions π 1. To convert degrees to radians, multiply by. 180 2. To convert radians to degrees, multiply by 180. π : (a) 90º / 2 rad (b) 60º / 3 rad (c) 5 / 4 rad 225º
An angle is in if it is drawn on the -plane with its vertex at the origin and its initial side on the positive (i.e., the right half of) -axis. Coterminal Angles Two angles are if their sides coincide. Coterminal angles differ from each other by an integer multiple of 360º (respectively, by a multiple of 2 radians). Conversely, every pair of angles that differ by an integer multiple of 360º or 2 radians are coterminal. : 30º, 390º, 750º, 330º are coterminal (they are all in the form of 30º ± 360 º). Similarly, / 2, 5 / 2, 3 / 2, and -11 / 2 radians are all coterminal (they are all in the form of / 2 ± 2 radians). Area of a Circular Sector
The area of a circular sector with a central angle is the fraction / (angle of one revolution) of the area of the entire circle. It is found by the formulas: In a circle of radius, the area of a sector with a central angle, is θ ( π 2π θ ( π 360 2 θ 2 ) ( in radians), or 2 2 θπ 2 ) ( in degrees). 360 Length of a Circular Arc Similarly, the length of the arc subtending a central angle is the fraction / (angle of one revolution) of the circumference of the circle. It is found by the formulas: In a circle of radius, the length of the arc subtending a central angle, is θ (2π ) θ 2π ( in radians), or θ θπ (2π ) ( in degrees). 360 180 : A circular sector has radius 10 and central angle / 3. Find its area and the subtending arc. A [( / 3) / 2] 10 2 100 / 6 50 / 3 ( / 3) 10 10 / 3
Circular Motion motion along a circular path Linear Speed vs. Angular Speed: Linear speed distance traveled / elapsed time Angular speed change in central angle / elapsed time Suppose a point moves along a circle of radius and the ray from the center of the circle to the point traverses (in radians) in time. Let be the distance the point travels along the arc in time. Then the speed of the object is given by Angular speed θ ω (radians per unit ) Linear speed If a point moves along a circle of radius with angular speed linear speed is given by, then its : [#79] A circular saw has a blade with a 6-in radius. Suppose the blade spins at 1000 rpm. (a) Find the angular speed of the blade in rad/min. (b) Find the linear speed of the blade in ft/sec.
6.2 Trigonometry of Right Triangles [See Trig. function summary handout.] Special Triangles 45º - 45º - 90º triangle 30º - 60º - 90º triangle Special Angles, in, in degrees radians cos tan csc sec cot 0º 0 0 1 0-1 - 30º / 6 1 / 2 3 / 2 3 / 3 2 2 3 / 3 3 45º / 4 2 / 2 2 / 2 1 2 2 1 60º / 3 3 / 2 1 / 2 3 2 3 / 3 2 3 / 3 90º / 2 1 0-1 - 0
Solving a Right Triangle Consider the right triangle: We have 2 2 + 2 cos : Let 4 and 7, find and derive the values of the six trigonometric functions for. : [#48] From the top of a 200-ft lighthouse, the to a ship in the ocean is 23º. How far is the ship from the base of the lighthouse? : [#49] A 20-ft ladder leans against a building so that the angle between the ground and the ladder is 72º. How high does the ladder reach on the building?
6.3 Trigonometric Functions of Angles Let be a right triangle with an acute angle θ shown below Then + 2 2 θ cos θ tan θ sec θ, 0 cot θ csc θ, 0
Signs of the Trigonometric Functions (vs. Quadrants) Quadrant I: Quadrant II: Quadrant III: Quadrant IV: All positive Sin and csc positive Tan and cot positive Cos and sec positive Reference Angle Let be an angle in standard position. The associated with is the acute angle formed by the terminal side of and the -axis. Therefore, the reference angle is always measured against the nearer half of the -axis. Evaluating Trigonometric Functions for Any Angle For any angle : 1. Find the reference angle associated with the angle. 2. Determine the sign of the trigonometric function of by noting the quadrant in which lies. 3. The value of the trigonometric function at the desired angle is equal to its value at the reference angle times the sign found in step 2. : (a) Evaluate cos(210º), tan(210º), and csc(210º). (b) Evaluate (315º), cot(315º), and sce(315º).
Trigonometric Functions on the Unit Circle: the Pythagorean Identities For any angle : cos 2 θ + 2 θ 1 1 + tan 2 θ sec 2 θ 1 + cot 2 θ csc 2 θ : (a) If cos 1 / 3, and is in quadrant IV, find and cot. (b) If tan 3 / 5, and is in quadrant III, find and cos. Area of a Triangle The area A of a triangle with sides of lengths angle is 1 A θ 2 and, and with included
6.4 The Law of Sines The Law of Sines In any triangle,. That is, in any triangle, the lengths of the sides are directly proportional to the es of the corresponding opposite angles. Solving a Oblique Triangle / Determining Congruence : [#4] Suppose the triangle has sides 56.3 and 80.2, and angle 67º. Find side and angle. 56.3.921 80.2 0.6462 40.25º. 180 72.75º (.921) (80.2) (72.75º) 83.16
At the minimum, to solve a triangle we need to know 3 of the 6 sides/angles, at least one side s length must be known. Not every combination is equally useful for example, angle-angle-angle (AAA) does not provide enough information to enable us to solve the triangle. If 4 of the 6 pieces of information are known, then a triangle can always be solved, as in the previous example. Congruent Triangle - ASA Suppose 2 angles and the included side are known, say angles, and side. Then immediately the third angle can be found by 180º. Therefore, both and will be known, giving us a complete ratio, namely, /. Then just apply the Law of Sines: Cross-multiply and solve, we will have ( ) / and ( ) /. : [#8] Solve the triangle given 30º, 2, 100º. Congruent Triangle - SAA Suppose now one side, an adjacent angle and the angle opposite of the given side are known, say side and angles,. Here, again, 180º. As well, we have a complete ratio, this time /. Apply the Law of Sines and solve: ( ) / and ( ) /.
You might recall that side-side-angle (SSA) is not, in general, a conclusive demonstration of congruence. Sometimes it works, more often it doesn t. Why is it the case? Congruent Triangle SSA (one-solution case) : [#24] Solve the triangle given 135º, 100, 80. 135 100 80 80 135º / 100 0.5657 34.45º (Use a calculator s 1 function.) Is there another possible value of? Why or why not? 180º 135º 34.45º 10.55º 135 100 10.55 100 10.55º / 135º 25.9
The Ambiguous Case: Indeterminate Triangle SSA (two-solutions case) : [#29] Solve the triangle given 40º, 15, 20. 40 15 20 20 40º / 15 0.8571 59º (Again, use a calculator s 1 function.) But 59º is not the only possible angle! There is another one in the second quadrant, namely its supplement angle 121º. Both angles are valid possibilities, given what we know. (Why?) Therefore, we have 2 triangles satisfying the given conditions: 40º, 59º, 81º, or 40º, 121º, 19º. The third side is, respectively, 23 and 7.6. The No-Solution Case: : [#19] Solve the triangle given 125º, 20, 45. 125 20 45 45 125º / 20 1.8431 1 (1.8431)??? Therefore, no triangle can satisfy the given conditions.
6.5 The Law of Coes The Law of Coes In any triangle, 2 2 + 2 2 cos 2 2 + 2 2 cos 2 2 + 2 2 cos Congruent Triangle - SAS Suppose 2 sides and the included angle are known, say angle, sides and. Then the misg side can be found by the corresponding equation given by the Law of Coes. In this case, 2 + 2 cos. With all 3 sides now known, angles and can then be found ug the other 2 equations of the Law. : Unlike the arce function, arccoe (the inverse coe function) is one-to-one on [0º, 180º], which is exactly its range. Therefore, each angle can be solved uniquely.
: [#12] Solve the triangle given 60, 30, 70º. 2 60 2 + 30 2 2(60)(30) cos 70º 2 4500 3600 cos 70º 3268.73 57.17 2 3600 57.17 2 + 30 2 2(57.17)(30) cos 3600 57.17 2 30 2 3430.2 cos cos 0.1657 80.46º 180º 70º 80.46º 29.54º Congruent Triangle - SSS If all 3 sides,, and are known, then clearly we can apply them to each of the 3 equations of the Law of Coes to find the coes of the angles,, and. For instance, 2 2 + 2 2 2 2 cos 2 + 2 cos arccos( 2 2 2 + 2 ) : Solve the triangle given 10, 16, 24.
Another Formula for Finding the Area of a Triangle: Heron s Formula The area A of triangle is given by A ( )( )( ) 1 Where ( + + ) is half of the perimeter of the triangle. 2 : Find the area of the triangle whose sides are 5, 8, 11. 1 2 ( + + ) 1 (5+ 8+ 11) 12 2 A ( )( )( ) 12(7)(4)(1) 336 4 21