
Trig NotesFall SemesterTrig1ChapterP "Prerequisites"Trig2A "Trig Intro."
ANGLE SUM AND DIFFERENCE FORMULASChapter 4 Lesson 4Since 15 degrees are missing angles from the unit circle,let's try calculating sin 15 and cos 15.Use this rectangle to derive sin75 and cos75.Using a different diagram allows derivation of general formulasfor the sum and differences of angles.All we have to do is find the missing sides in terms of sines and cosines.Generalizing the diagram enables derivation of the formulas:From cosAcosB = cos(A+B) + sinA sinB we have:SUM: cos (A+B) = cosA cosB  sinA sinBSUM: sin (A+B) = sinA cosB + cosA sinBFrom adding and subtracting segments to congruent segments, we have:COSINE DIFFERENCE:cos(AB) = cosA cosB + sinA sinBSINE DIFFERENCE:sin(AB) = sinA cosB  cosA sinBYou can also derive the sine angle addition formula from Area = 1/2 bc sinA of a triangle.1/2 dc sin(A+B) = (d*sinB) (c*cosA) /2 + (c*sinA) (d*cosB) /2 givingsin(A+B) = sinB cosA + sinA cosB, which can be rearranged to become:sin(A+B) = sinA cosB + cosA sinBPYTHAGOREAN IDENTITIESChapter 4 Lesson 1Let a = sinθ, b = cosθ and c = 1. Then by the Pythagorean Theorem,Note the sine and cosine segments are perpendicular.We can obtain two other pythagorean identities by dividing each addend:to get
Here's how the related triangles appear on the unit circle:Notice the tangent touches the circle once and the secant will cut it twice.Placing all the triangles on the same unit circle yields:Note the tangent and cotangent segments are perpendicular.We can also rotate the triangles so that the right angle is tangent to the unit radius.This time the cosecant and secant segments are perpendicular.Find the segments lengths of theta θ is 45 degrees.Find the coordinates below if theta θ is 60 degrees.P( ?, ?); R( ?, ?); Q( ?, ?)The other segments formed have names as well.Here's a way to remember the identities:
SOLVING TRIANGLESUse Law of Cosines for SSS cases when given sides a, b, and c;and SAS when given a, ∠B, and c.Use Law of Sines for ASA (given ∠A, c, and ∠B with ∠C = 180 ∠A∠B)and AAS (∠A, ∠B, and a) cases.When solving triangles with the SSA case of Law of Sines,one of three things can happen.This ambiguity is therefore called "The Ambiguous Case."This is why there is no Angle Side Side triangle congruencetwo triangles.Note: in this case SSA becomes HL (HypotenuseLeg congruence.)
For a more detailed explanation with numbers, check this link out:In summary when doing a problem:If sine < 1 there are two solutions(or one if the angle you get is less than the given angle. You can check by seeing if the supplement violates the triangle sum of 180).If sine = 1, there is one solution.There are no solutions if sine > 1.(a) sine > 1 (b) sine = 1 (c) sine < 1, a < b (d) a > b, S.s.A. case => 1 triangleUnit 2B Lesson 1.7Arcsine and ArcCosSo for the range of the inverse trig functions we have:Now let's graph the data points for these inverse trig functions.Graphing these coordinates (t, y) and (y, t) yields:Graphing (t, x) and (x, t) gives us the arccos graph.The graphs of all six inverse trig functions follow:NOTE: The inverse cotangent pictured is wrong.
Unit 2B Lesson1.5sine graphUnit 3 Notes 3 (identities)
6.a) 3600 seconds b) 1/60 = 0.01666... c) 1/3600 = 0.0002777...d) 5 + 0.5 + 0.00166... = 5.501666...e) 89 + 0.3 + 0.011666... = 89.3116666...7. a) 57 degrees + (0.29578)60 minutes = 57 deg 17.746 minutes = 57 deg 17' 60(0.746) seconds= 57 deg 17' 45" (rounding 44.8 seconds)8. a) 400 grad / 360 deg = 10/9 = 1 1/9 = 1.111... b) 90 deg / 100 grad = 9/10 = 0.9c) 30 * (10/9) = 100 / 3 = 33 1/3 = 33.333... gradiansd) 66 2/3 * 90 deg / 100 grad = (200/3)*(9/10) = 180/30 = 60 degreese) 1 grad * 9/10 = 0.9 degrees = = (0.9) * 60' = 54 minutesf ) 25 grad * 54 min = 1350 min / 60 deg = 22.5 degrees = 22 degrees 30 secondsor 25*9/10 = 22.5
Let's define sine and cosine functions from their unit circle ratios.Using the angle measures in degrees, observe the Quadrants:Quadrant I: 0 < θ < 90, Quadrant II: 90 < θ < 180, Quadrant III: 180 < θ < 270, Quadrant IV: 270 < θ < 360.Note the Quadrants where both sine and cosine are positive (+, +), negative (,) or opposites (,+) & (+,).Also observe how sine starts at 0 and cycles through all the values ending in 0 again after 360 degrees.This is the period of the sine function. In radians, the period is 2π.Likewise the cosine starts at 1 and goes down to 1 before ending at 1 again after 2π radians. Its period is also 360°.Since the angle can rotate around the circle infinitely many times,both in positive counterclockwise directions and negative clockwise directions,the domain of each is the set of all Real numbers: ∞ < t < ∞ or (∞, ∞).As all the fractions are between 1 and 1, the range of each function [1 , 1].For y(t) = sin(t) the range is 1 ≤ y ≤ 1 and for x(t) = cos(t) the range is 1 ≤ x ≤ 1.Since they are right triangles, apply the Pythagorean Theorem.
So there are three Pythagorean identities.Explore the relationship of the new functions: secant, cosecant, and cotangent.Not only are the cofunctions reciprocals of sine, secant, and tangent,they are related via complementary angles. Hence, the prefix "co":cosine(angle) = sine(complement),cotangent(angle) = tangent(complement) andcosecant(angle) = secant(complement.)Likewise sin(angle) = cos(complement), tan(angle) = cot(complement) and sec(angle)= csc(complement).Unit 2 Notes 3 (six trig ratios)Unit 2 Notes 2 (Unit Circle coordinates)Since dividing the unit circle into eight and twelve equal sectors results in angles that are multiples of 30, 45, and 60 degrees, the side of special right triangles provide the (x, y) coordinates on the unit circle.Recall the 306090 triangle and the 454590 triangle.We divide each side by the length of the hypotenuse so that the newer triangles can fit into the unit circle of radius 1.So the 306090 triangle gets divided by 2, and the 454590 triangles gets divided by √2.Now let's fit a 306090 triangle into the first Quadrant of the unit circle.Then fit each new special triangle all around the unit circle.By accounting for positive (+) and negative () signs in each quadrant, we can find the (x, y) coordinates all the way around the circle.Filling in the coordinates all the way around the circle we now have:Unit 2A Chapter 1.11.4Lesson 1.1Notes #1 Radian MeasureNotes #2 9/18/2012"Prerequisites"Click on the link below for each lesson's notes:A scene in the movie Smilla's Sense of Snow compares human life to the number system:Solving a linear equation:Factoring a quadratic with a leading coefficient:Solving a rational equation:Check for extraneous solutions. 2 makes the expression undefined (dividing by 0).Radical Equation (square both sides to eliminate the root)Once again, check the solutions.If there are two radicals, square both sides twice!and check the solutions."Cartesian Plane"Linear EquationsAugust 29, 2012The definition of a function is written in blue.A function can map numbers from one set to another via a diagram,or a table, or it can relate input and output values through an equation, for example y = x.You can tell the graph is of a function because it passes the vertical line test.You can tell from the table by checking if there are repeat values for x. In the case above, there are not.In the case below, there are repeat values of x.This circle is not a function because some xvalues are associated with two yvalues.So the graph fails the vertical line test and there are repeat values of x in the table.The domain and range are the same in this relation: {0, 3, 4, 5, 4, 3, 5}.These are discrete points since we can count them. Find the circle's equation to connect its dots.For the equation of the circle, the domain is 5 ≤ x ≤ 5 and the range is also y ≤ 5.If you are given equations instead of graphs or tables, ask these two questions:1) What values of x result in square rooting a negative number?2) For what values of x is there division by zero?P6 "Graphs of Functions""Function Transformations"Press Y= button. Press MATH button. Press right arrow to highlight "NUM" menu.Press 1 to select "abs(". Press X key, then ")". Press ZOOM key.Arrow to or press "6". Press GRAPH key to graph Y1. Press ZOOM and "6" keys.to get the absolute value graph to square up like a "V".Enter these equations into calculator:f(x) +2 moves f(x) = x up 2 units. f(x+2) moves f(x) left 2 units.How would f(x) get moved down? How can f(x) be moved right 2 units?f(x) reflects f(x) = x+2 over xaxis. f(x) reflects f(x) = (x+2) over yaxis.Why are x and x the same graph?A vertical stretch multiplies f(x)'s yvalues, pulling it up. A compression pushes it down by dividing.A horizontal compression pushes the "V" graph toward the center. A horizontal stretch pulls it out to both sides.On the absolute value graph x, 2f(x) = 2x = 2x = 2x = f(2x) so that the horizontal compression appears as a vertical stretch.Likewise for horizontal stretches and vertical compressions. This is true for all linear functions. So let's change the parent function f(x).Can you identify each function's graph?Vertical Stretch = 4f(x) Vertical Compress = f(x)/4 Horizontal Stretch = f(x/4) Horizontal Compress= f(4x)Multiplying outside parentheses performs a vertical stretch, while dividing outside is a vertical compression.Multiplying inside parentheses is a horizontal compression, while dividing inside is a horizontal stretch.Note that multiplying inside () has the opposite effect horizontally than it does vertically.Vertical transformations match up with operations outside the ()'s.Horizontal transformations involve operations inside ()'s.Why are operations inside the parentheses opposite those from outside?If y = f(x) + k, then y  k = f(x) moves the function down k units.Likewise, if y = k*f(x), then y/k = f(x) is a vertical stretch despite dividing by k.Professor Pi approximating Pi with inscribed and circumscribed polygons using sine and tangent.Table of ContentsLesson 2.1 Trig IdentitiesCosine is an even function.Secant is an even function.All four trig functions with sine in them are odd.Combining these three triangles into Quadrant I of the unit circle shows why ......the tangent is named (it's the height of the segment that touches the circle at one point)and the secant (extend the purple hypotenuse to QIII where it will cross unit circle a second time.)Note here also that the cotangent and tangent segments are perpendicular(right angle→90 degrees→complementary angles→tangent (complement) = cotangent(θ).)The cosecant is the hypotenuse of the red triangle, the secant the hypotenuse of the purple.The height of the brown triangle is the sine, and the base of it is the cosine.All six trig ratios are lengths of sides of these three triangles drawn onto the unit circle.EXTENSION: So if θ is 30 degrees, or π/6 radians, find all six lengths.Lesson 2.4 Part ICosine Difference FormulaLesson 2.4 Part IISum and Difference Formulas for Cosine(αβ) and Sine (α±β)Lesson 2.4 Part IIIDouble Angle Formulas for cos(2θ), sin(2θ), and tan(2θ)Lesson 2.4 Part IVHalf Angle Formulas for cos(x/2) and sin(x/2)Lesson 2.4 Part Vtan(x/2)Lesson 2.4 Part VICotangent Formulas for Cot(α±β) and Cot(2θ)
Lesson 1.8A "Solving Right Triangles"
Lesson 1.8B "Harmonic Motion"
Lesson 3.1A "Law of Sines"
Lesson 3.1B "Area of SAS Triangle"
Lesson 3.2 "Law of Cosines"
Lesson 3.2B "Area of SSS Triangle"Lesson 1.8 "Harmonic Motion"October 24th, 2011Lesson 1.8 "Solving Right Triangles"October 22nd, 2011Lesson 3.1B"Area of Triangle (SAS formula)"October 27th, 2011Lesson 3.1A
"Law of Sines"
October 25th 2011[ModuleInstance>Description]
You need at least version 7 of Flash player. Download latest version here!
Lesson 3.2
"Law of Cosines"
October 31st, 2011[ModuleInstance>Description]
You need at least version 7 of Flash player. Download latest version here!
Unit 2B Chapter 1.51.7 "Trig Graphs"Notes 1 Lesson 1.5A "Amplitude and Frequency"
Notes 2 Lesson 1.5B "Phase Shifts"
Notes 3 Lesson 1.6 "Graphs of tan, cot, sec, csc"
Notes 4 Lesson 1.7A "Inverse intro"Notes #0 (optional)September 27, 2011Graphing Calculator Keystrokes.At any time, press 2nd MODE to return to the calculator Home Screen.Press Y= to enter equations to graph. Press GRAPH to see the graph.Press ZOOM and 7 ): ZTrig to set the window to [2π < x < 2π] by [4 < y < 4],or press the down arrow cursor to "7" and press ENTER .To enter unit circle values into a LIST, press STAT then ENTER.If there are any numbers in the columns, move the cursor to hilite L1, etc., and press CLEAR .Then place the cursor under L1 and type each number and ENTER in between each value.(i.e. where the commas are in this list {0, 30, 45, 60, 90, 135, 180, 270, 360})When done, quit to the Home Screen (see above.) Enter the following keystrokes:2nd 1 X 2nd ^ / 180 ENTER, where "/" means divide. You should see "L1* π/180" on the screen.This converts all entries in list L1 to decimal radians. 2nd ^ is the π key. (π = pi)Press STO 2nd 2 to store this into list L2. Or combine all steps:2nd , 1 , X , 2nd , ^, / , 180 , STO , 2nd , 2 , ENTER . You should see L1*π/180 > L2 on your home screen.You can manually enter the unit circle yvalues for sine into L3 as before but SIN(L1) STO L3 ENTER is faster.(2nd, 3 is L3). The first "(" is automatic but you have to enter the right " ) ".The screen should read sin(L1) > L2 followed by decimals in the brackets {0 ...} on the next line.Before moving on press 2nd, STAT, ENTER to confirm the entries in lists L1, L2, and L3.We're going to make a STAT PLOT of these values. Press 2nd, Y=, ENTER to go into the STAT PLOT menu.Hilite Plot1 and press ENTER, hilite "On" ENTER, then select the first type of stat plot (scatter).Arrow down to XList: and set it to L2 by pressing 2nd, 2. Select L3 for YList.Leave the marker as the square (not the dot or cross.)Make sure your calculator is in radian mode by pressing MODE, hilite "Radian" or "Rad" and press ENTER.Then press GRAPH . You should see little squares on the graph.If you got a DIM: MISMATCH error, your L2 and L3 lists are of different sizes (you missed an entry.) Fix it and try again.If there are other lines on the graph you must CLEAR the equations in the Y= menu.Press Y=, CLEAR all equations, and enter sin(x) into Y1 by pressing SIN and X,T,O,n , and GRAPH. (O is supposed to be Theta) If your graph does not appear, be sure the equation is turned on by hiliting the = sign after Y1.You can press WINDOW to adjust the viewing settings of the graph.Repeat the process for the Cosine on the back of the Sine and Cosine graphs worksheet. Print this page for 1 point extra credit in Notes packet 2B and for future reference.
Notes #5 (October 11th, 2011)First, enter these angles into List L1 and make these lists for sin, cos, and tan.Turn Plot1 "On" and graph (ZOOM 7) the sine graph for 90 < x < 180.Then swap the Domain and Range with XList : L2 and YList: L1.Before graphing, set WINDOW to 4 < x < 4 by 360 < y < 360. Then press GRAPH.The plotted points fail the vertical line test and so the inverse relation is not a function.However, enter y = sin^(1)(x) by pressing. Y=, 2nd, SIN, X to get the inverse sine function.For cosine, we'll fist convert to radians, then repeat the process.Again, the inverse cosine relation is not a function. However,if we restrict the domain of the angle to run from 0 to π radians (180 degrees),we run through all the xvalues on the unit circle for cosine.This restricted cosine is called Cos(x) and its inverse function is ArcCos(x).(It's called an 'arc' because it returns the angle, or arc measure, on the unit circle.)If the angle theta runs from 0 to 180 degrees (π radians), all the cosine values are covered from 1 to 1.To cover all the yvalues of sine, the angle must go from 90 degrees to 90 degrees (or π/2 to π/2 rad).For tangent, what angles must be covered so that the line through unit circle takes on every slope value?Restrict the domain to just graph one branch of tangent. Call new function Tan(x) with capital T.Then its inverse ArcTan(x) is graphed from interchanging the x and yvalues.Domain of [π/2, π/2] of Tan becomes Range of ArcTan andRange [∞, ∞] of Tan becomes Domain of ArcTan.Notes #4 (October 10, 2011)Now that we can find trig values of given angles, let's do it backwards.soThis method of "undoing" what sin, cos, or tan does is called an "inverse."In the notation the "1" means inverse, not an exponent for a reciprocal.(You wouldn't undo addition by division, and we won't undo sine by dividing for the same reason.They are different levels of inversesadd and subtract, multiply and divide, sine and inverse sine, etc.)and yetBoth sin(π/4) and sin (π/4) have a value of 0.7071. So what angle will inverse sine of 0.7071 output?Simplifying radicals with Mr. Happy Face!Notes #3 (October 6, 2011)For using a graphing calculator with the tancot/seccsc graphing worksheet:This is the graph of the tangent function. y = tan(x)Note the intercepts where sin(x) = 0 and the asymptotes where cos(x) = 0.What are these xvalues? To graph y = cot(x) enter 1/tan(x) or cos(x)/sin(x).Label the xintercepts and the asymptotes. Where are they?Finish the cosecant graph on the worksheet.Let's recap the domain and range of all six trig functions:Notes #2 (October 4, 2011)Notes #1 (October 3, 2011)Amplitude and FrequencyGraph y = sin(x). Then graph y = 2sin(x) with a vertical stretch factor of 2.Then graph y = sin(x). A stretch factor of 1 reflects over the xaxis.The original range of sin(x) of [1, 1] is not affected by reflecting over the xaxis.However, 2sin(x) has a range of 2 < y < 2. What about y = 2sin(x)?The Amplitude is the absolute value of the stretch factor. A = a if y = asin(x) and 2 = 2.When dividing by 3, a vertical compression results the same as multiplying by 1/3.So the cosine graph will be sandwiched between y = 1/3 and y = 1/3.Now let's look at f(kx) , a horizontal compression factor.For y = cos(2x), the cosine cycled through its wave twice over 2π on the xaxis.Normally, cosine has a Period of 2π, , but here P = 2π/2 = π.If y = asin(bx), then Amplitude A = a, frequency = b, and its period = 2π/f.[ModuleInstance>Description]
You need at least version 7 of Flash player. Download latest version here!
Notes #6, Lesson 24September 26th, 2011Three rules for remembering unit circle coordinates:Make a table of reference angles versus sine, cosine and tangent.(Note: This table is transposed from the one in Unit Circle PDF and the notes below.)In QII, the reference angle = 180  θ or π  θ.In QIII, the formula for reference angle = θ  180 or θ  π.In QI, the reference angle is θ. In QIV, it's 2π  θ or (360  θ) for positive theta.For example, if θ = 330, then ref angle = 360330 = 30 degrees.For a negative theta angle, simply take the absolute value: e.g. 45 = 45 degrees.In this case, we have a negative θ in QIII, so ref angle = 180  135 = 45 degrees.Cosine is negative because so are the xcoordinates in Quadrant III.Period 3's NotesThis time, I color coordinated the table to be consistent with the Unit Circle from the Notes on 2.2.Degrees, Radians, Coordinates , with scratch work in purple and the table in brown and trig ratios in black.Also, with this arrangement, the tan ratios of sin/cos are easy to generate becausethe sine values are actually right on top of the cosine values. Note the simple radical formfor tan(30) as well as how the 1/2's cancel in tan(60). And yes, root(3) = tan(60) = 1.732...Lesson 2.4 "Right Triangle Identities"September 24th, 2011Apply the Pythagorean Theorem to a triangle withsides cosine(θ), sine(θ), and hypotenuse of 1.Then make a similar triangle by dividing all three sides by cos(θ) so the adjacent base is 1.Divide all three sides of first triangle by sine(θ) so that height opposite angle theta is 1.The five reference angles in degrees and radians for the three trig functions of sine, cosine, and tangent.Lesson 2.3 "Trig Functions"There are six trigonometric functions because there are six ways to choose two sides from three.Since order matters in a ratio, we use a permutation of choosing k objects from n.On an xygrid, the three sides of the triangle become r = hypotenuse, y = opposite, x = adjacent and t = θ.Including the reciprocals of the three trig ratios of sine, cosine, and tangent from geometry provide the new ratios of cosecant (= csc), secant (= sec), and cotangent (= cot).Each of the three "new" trig functions is a reciprocal of one of the original three. So tan = 1/cot.What ratio would result from cos/sin ?Lesson 2.2 "The Unit Circle"Combining the 8 and 12slice circles from the last lesson:Since dividing the circle into eight and twelve sectors results in angles that are multiples of 30 and 45, we can fit the 306090 and 454590 triangles from geometry into the unit circle. Scale each triangle so that each hypotenuse is 1 unit in length. This means dividing by the original hypotenuse of 2 or √2.Convert each triangle leg into simple radical form by rationalizing each denominator.As needed, multiply the top and bottom of the fraction by √2.Then the coordinates (x,y) are obtained by imbedding special right triangles into the circle:and inserting + and  signs for x and ycoordinates depending on the Quadrant:to get the completed unit circle.Lesson 2.1 "Radians and Degrees"September 12th, 2011First divide a circle of radius 1 centered on the origin into 8 equal slices. List each degree measure:One complete revolution is both 2 pi radians and 360 degrees. So the conversion factor is:2pi/360 = pi/180 to go from degrees to radians. Multiply by 180/pi to convert back to degrees.Divide another unit circle into twelve equal slices. The first quadrant (QI) will look like:And the whole thing looks likeHW: Combine both circles onto one giant circle so we can find the (x,y) ordered pair coordinates.Checking the conversion factor:Period 3's NotesEighthsTwelfthsUnit 1 Chapter "Prerequisites"Unit 4 Chapter 3 "Solving Triangles"
STAR TRIG
Lesson 31 "Law of Sines" (11/23/2010)[ModuleInstance>Description]
You need at least version 7 of Flash player. Download latest version here!
Lesson 32 "Law of Cosines" (11/30/2010)[ModuleInstance>Description]
You need at least version 7 of Flash player. Download latest version here!
Last Modified on November 28, 2018