ࡱ> c5@ ?bjbj22 "^XX?ڤڤڤڤf<2NIKKKKKK$R joo@;;;I;I;&;ar}T I]ڤDV02ttt";6Joo$[\H1 \Trigonometry Trigonometry takes on two distinct forms: right triangle trig oscillating function trig Each one plays a huge role in basic physics, calculus and engineering right triangle trig: force diagrams, geometry, vector analysis oscillating functions: calculus, mass-spring systems, circuits, sound, and any vibrations Contents: 1. Right Triangle Trig 1.1 Pythagorean Theorem 1.2 SOH CAH TOA and similar triangles 1.3 Trig values for 0, 30,45,60, and 90 1.4 Degrees and Radians 1.5 Trig Identities 2. Oscillating Functions 2.1 basics of sine and cosine 2.2 period 2.3 frequency 2.4 phase shift 2.5 standing waves 1. Right Triangle Trigonometry The Pythagorean Theorem The Pythagorean Theorem is the cornerstone of trigonometry and analytic geometry. It relates the lengths of the sides of right triangles. First you need to remember the terminology:  If we denote the lengths of the three sides by a and b for the legs and c for the hypotenuse, then in a flat plane the Pythagorean Theorem states that c2 = a2 + b2 The most widely known right triangle is perhaps the 3-4-5 triangle shown below  where we note that 32 + 42 = 9 + 16 = 25 = 52. Other right triangles with integer sides are 5-12-13 and 20-21-29. The reader should check that the two sets of numbers really do satisfy the Pythagorean Theorem. Skill: Completing the Triangle One thing that one should be able to do is as follows: given two sides of a right triangle, be able to fill in the length of the third side. This is called completing the triangle. Example 1: Complete the following triangle  Solution: we must have x2 + 32 = 72 or x2 + 9 = 49 or x2 = 49 9 = 40 so x =  EMBED Equation.3  (most problems result in a square root appearing somewhere) Example 2: Complete the following triangle  Solution: We need to determine the hypotenuse this time. We have 92 + 132 = x2 or 81 + 169 = x2 or 250 = x2 so x =  EMBED Equation.3  Problem: why is it true that a + b > c ?? 1.2 SOH CAH TOA If one of the angles (not the 90 degree one) are denoted by  and we distinguish the legs as to being adjacent to the angle or opposite it then the triangle looks like  The Pythagorean Theorem in this case states Hyp2 = Adj2 + Opp2 The basic trig functions are defined as ratios of the sides of the triangle as the legs relate to  (either adjacent to it or opposite it).  EMBED Equation.3  You need to memorize these!!! They are pronounced: sin() =  sine theta cos() =  cosine theta tan() =  tangent theta For the right triangle shown below, we would then have  that sin() = 10/26 = 5/13, cos() = 24/26 = 12/13 and tan() = 10/24 = 5/12. Sometimes you have to complete the triangle first before you can figure out the values of the trig functions as shown by the following example Example For the triangle shown below, determine sin(), cos() and tan().  Solution: step 1  complete the triangle the adjacent side is found by solving x2 + 42 = 92 or x2 = 81  16 =65 so x =  EMBED Equation.3  step 2  read off the trig values  so that that sin() = 4/9 cos() =  EMBED Equation.3 /9 and tan() = 4/ EMBED Equation.3  Complementary Angles It is well known that for any triangle, the sum of all three angles is always 180. Now for a right triangle, one of those angles is 90, so the other two angles add up to 90. They are called complementary because of this. Thus if you have a right triangle and you know one angle, you really know all three. If, for example, one angle is 37 degrees than the other must be 63 degrees and the triangle might look like  Knowing this and SOH CAH TOA can save us some work. We start with the following triangle and base what follows on it:  Now, remembering SOH CAH TOA, we can write down all of the following: sin() = Opp/Hyp cos() = Adj/ Hyp as well as (check this) sin(c) = Adj/Hyp cos(c) = Opp/Hyp because of where c is located on the triangle. Putting this together, cos(c) = sin() and sin(c) = cos() or, equivalently sin() = cos(90 - ) or, put verbally: if you know sine of an angle then you know cosine of its complement, and if you know cosine of an angle, you know sine of its complement. For example, if we are given that sin(40) = .643 then we automatically know that cos(60) = .643 . If we know that cos(55) = .82 then we know from this that sin(35) = .82. Part of geometry and, in particular trigonometry, is getting all the possible mileage you can out of the given information in a problem. deductive reasoning. 1.3 Trig values for 0, 30,45,60, and 90 You will frequently need to know the values of sine, cosine and tangent for the angles listed above. What follows will try to cut down on the amount of memorization and also explain where the values come from. For starters, we will focus only on sine and cosine and then fall back on the fact that tan() = sin()/ cos(). Our goal is to fill in the following table based on what we know: sin()cos()030456090 45 is a good place to begin. The triangle may look like  if we make the hypotenuse 1. Note the second 45 because they have to add up to 90. This makes the triangle isosceles and hence Opp = Adj . By the Pythagorean Theorem, we then have 12 = Adj2 + Opp2 = Adj2 + Adj2 = 2Adj2 so Adj2 = and therefore Adj =  EMBED Equation.3  Therefore the triangle looks like  and thus cos(45)= sin(45)=  EMBED Equation.3 . The table now looks like sin()cos()03045 EMBED Equation.3  EMBED Equation.3 6090 Let s set up an equilateral *triangle with sides 1 unit long:  *equilateral means that all three sides are the same length. Isosceles means two equal sides Now, you point out, this is not a right triangle and we were doing trigonometry, the study of right triangles. True but lets add a line from the left vertex to the midpoint of the vertical side. Now things look like  and we note that the green, horizontal line is perpendicular to the vertical, blue line. We have created two, congruent, right triangles, called 30-60-90 triangles. We have 2 of the three sides of each. If we let x denote the length of the green, horizontal side then one of them looks like  where, from the Pythagorean Theorem, we have x2 + (1/2)2 = 12 so x =  EMBED Equation.3  and the triangle, completed, is  From SOH CAH TOA, we get sin(30)= EMBED Equation.3  = cos(60) and cos(30) =  EMBED Equation.3  = sin(60) The table now looks like sin()cos()030 EMBED Equation.3  EMBED Equation.3 45 EMBED Equation.3  EMBED Equation.3 60 EMBED Equation.3  EMBED Equation.3 90 and we note that because of the property of complementary angles discussed earlier, every time we determine 2 trig values, we effectively get 2 more free of charge. Thus if we can figure out sin and cosine of 0 we are really done. For 0 we can start with a very, very slim triangle with hypotenuse of 1. This might appear as  Remembering that sine is opposite divided by 1 and cosine is adjacent divided by 1, we then imagine  getting smaller and smaller and becoming 0. The adjacent and the hypotenuse both become the same thing so cos(0) = 1. By similar thinking, the opposite side becomes 0 as the angle collapses, so sin(0) = 0. The table now looks like sin()cos()00130 EMBED Equation.3  EMBED Equation.3 45 EMBED Equation.3  EMBED Equation.3 60 EMBED Equation.3  EMBED Equation.3 90 For 90 degrees the role of sine and cosine are reversed so we can fill in the last row as sin()cos()00130 EMBED Equation.3  EMBED Equation.3 45 EMBED Equation.3  EMBED Equation.3 60 EMBED Equation.3  EMBED Equation.3 9010 The property of complementary angles has saved us half the work. It can save you half of the memorization by the following pattern of duplication of values (indicated by corresponding colors): sin()cos()00130 EMBED Equation.3  EMBED Equation.3 45 EMBED Equation.3  EMBED Equation.3 60 EMBED Equation.3  EMBED Equation.3 9010 where each pairing occurs for the simple reason that sin() = cos(90 - ) Completing Right Triangles Using Trig Functions This is a skill where, given only the hypotenuse and an angle in a right triangle, one can fill in all three sides of the triangle. For example, consider the following problem: Problem: determine all three sides of the triangle shown:  Solution: Lets find the adjacent (horizontal) side first. Call its value x. Then by definition, the ratio of adjacent to hypotenuse is the cosine of 30 degrees, or cos(30) = x/20 =  EMBED Equation.3  so x = 20 EMBED Equation.3  = 10"3 To next find the opposite (vertical) side, we have two ways to do it: use the Pythagorean Theorem or use trig functions as we just did. We ll go with the latter to illustrate it as you already know how to use the Pythagorean Theorem Call this unknown side y. Then by definition sin(30) = y/20 = 1/2 so y = 20(1/2) = 10 The triangle, now completed, looks like  Another approach: we developed the trig values for 30 degrees using a triangle with a hypotenuse of 1 unit and adjacent angle of 30. What we needed in this problem was a triangle 20 times as big and similar (in the mathematical sense) to the original so we just multiply or scale all three sides by 20 and we are done  Comment: to do these kinds of problems , you need to know SOH CAH TOA cold as well as the values from the trig table. There is no way around this. Problem: complete the following triangle  Solution: lets get the hypotenuse first. Call it H. By definition sin(30) = 8/H We know sin(30) = so = 8/H or H = 8/(1/2) = 16 To get the horizontal side, which we call A, we know that cos(30) = A/H = A/16 and, from the table, that cos(30) =  EMBED Equation.3  so A/16 =  EMBED Equation.3  and A = 16 EMBED Equation.3  = 8  EMBED Equation.3  The completed triangle now looks like  Comment: we can check our work by seeing if our three sides satisfy the Pythagorean Theorem. If not, we made a mistake. In this problem,  EMBED Equation.3  so we are ok. Problem: complete the following triangle  Solution: lets call the vertical side y and the hypotenuse h. Then we are looking for values to complete  Lets find h first. By definition, cos(15) = 12/h so h = 12/cos(15). To finish, we need the cosine of 15 degrees which requires a calculator. This tells us cos(15) is about .966 so h = 12/.966 = 12.42 We now have two out of three sides. We can get y either with the Pythagorean Theorem or Trig functions. 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