LESSON 6: TRIGONOMETRIC IDENTITIES by Thomas E. Price ...

1 THE UNIVERSITY OF AKROND epartment of Theoretical and Applied MathematicsLESSON 6: TRIGONOMETRIC IDENTITIESbyThomas E. PriceDirectory Table of Contents BeginLessonCopyrightc Revision Date: August 17, 2001 Table of Elementary sum and di erence double and half angle IDENTITIES and Factor to Exercises1. IntroductionAnidentityis an equality relationship between two mathematical expressions. Forexample, in basic algebra students are expected to master variousalgbriac factoringidentitiessuch asa2 b2=(a b)(a+b)ora3+b3=(a+b)(a2 ab+b2): IDENTITIES such as these are used to simplifly algebriac expressions and to solve alge-briac equations. For example, using the third identity above, the expressiona3+b3a+bsimpliflies toa2 ab+b2:The rst identiy veri es that the equation (a2 b2)=0istrue precisely whena= b:The formulas or TRIGONOMETRIC IDENTITIES introduced inthis LESSON constitute an integral part of the study and applications of IDENTITIES can be used to simplifly complicated TRIGONOMETRIC expressions.

2 Thislesson contains several examples and exercises to demonstrate this type of IDENTITIES can also used solve TRIGONOMETRIC equations. Equations ofthis type are introduced in this LESSON and examined in more detail inLesson student's convenience, the IDENTITIES presented in this LESSON are sumarized inAppendix A2. The Elementary IdentitiesLet (x; y) be the point on the unit circle centered at (0;0) that determines the angletrad:Recall that the de nitions of the TRIGONOMETRIC functions for this angle aresint=ytant=yxsect=1ycost=xcott=xycsct =1x:These de nitions readily establish the rst of theelementaryorfundamentalidentitiesgive n in the table below. For obvious reasons these are often referred toas thereciprocalandquotientidentities. These and other IDENTITIES presented inthis section were introduced in LESSON 2 :Table : Reciprocal and Quotient 2: The Elementary Identities5 Example 1 Use the reciprocal andquotient formulas to verifysectcott= csct:Solution: Sincesect=1costandcott=costsintwe havesectcott=1costcostsint=1sint= csct:Example 2 Use the reciprocal andquotient formulas to verifysintcott= cost:Solution: We havesintcott= sintcostsint= cost:Section 2: The Elementary Identities6xy),(yxt t),(yx Several fundamental IDENTITIES follow from the sym-metry of the unit circle centered at (0;0).

3 As indicatedin the gure, if (x; y) is the point on this circle thatdetermines the angletrad;then (x; y) is the pointthat determines the angle ( t) rad:This suggests thatsin( t)= y= sintand cos( t)=x= cost. Suchfunctions are calledoddandevenrespectively1. Sim-ilar reasoning veri es that the tangent, cotangent, andsecant functions are odd while the cosecant function iseven. For example, tan( t)= yx= yx= tant:Identi-ties of this type, often called thesymmetry IDENTITIES ,are listed in the following functionfisoddiff( x)= f(x) andeven iff( x)=f(x) for allxin its domain. (Seesection 2insection 5for more information about these two properties of 2: The Elementary Identities7sin ( t)= sintcos ( t) = costtan ( t)= tantcsc ( t)= csctsec ( t) = sectcot ( t)= cottTable : The Symmetry next example illustrates an alternate method of proving that the tangentfunction is odd:Example 3 Using the symmetry IDENTITIES for the sine and cosine functions verifythe symmetry identitytan( t)= tant:Solution: Armed with theTable havetan( t)=sin( t)cos( t)= sintcost= tant:This strategy can be used to establish other symmetry IDENTITIES as illustrated inthe following example and inExercise 1.)

4 Example 4 The symmetry identity for the tangent function provides an easy methodfor verifying the symmetry identity for the cotnagent function. Indeed,cot( t)=1tan( t)=1 tant= 1tant= cott:Section 2: The Elementary Identities8 The last of the elementary IDENTITIES covered in this LESSON are thePythagoreanidentities2given inTable let (x; y) be the point on the unit circlewith center (0;0) that determines the angletrad. Replacingxandyby costandsintrespectively in the equationx2+y2= 1 of the unit circle yields the identity3sin2t+ cos2t= 1. This is the rst of the Pythagorean IDENTITIES . Dividing this lastequality through by cos2tgivessin2tcos2t+cos2tcos2t=1cos2twh ich suggest the second Pythagorean identity tan2t+ 1 = sec2t. The proof of thelast identity is left to the reader. (SeeExercise 2.)

5 Sin2t+ cos2t=1tan2t+ 1 = sec2t1 + cot2t= csc2tTable : Pythagorean IDENTITIES are so named because angles formed using the unit circle also describe a right tri-angle with hypotenuse 1 and sides of lengthxandy:These IDENTITIES are an immediate consequenceof the Pythagorean expression sin2tis used to represent (sint)2and should not be confused with the quantitysint2:Section 2: The Elementary Identities9 The successful use of trigonometry often requires the simpli cation of complicatedtrigonometric expressions. As illustrated in the next example, this is frequently doneby applying TRIGONOMETRIC IDENTITIES and algebraic 5 Verify the following identity and indicate where the equality is valid:cos2t1 sint= 1 + sint:Solution:By rst using the Pythagorean identitysin2t+ cos2t=1and then thefactorization1 sin2t= (1 + sint)(1 sint);the following sequence of equalities canbe established:cos2t1 sint=1 sin2t1 sint=(1 + sint)(1 sint)1 sint= 1 + sint;1 sint6=0:As indicated, the formula is valid as long as1 sint6=0orsint6=1.

6 Sincesint=1only whent= 2+2k wherekdenotes any integer, the identity is valid on the set< ft:t= 2+2k wherekis an integerg:Section 2: The Elementary Identities10 The process of using TRIGONOMETRIC IDENTITIES to convert a complex expression toa simpler one is an intuitive mathematical strategy for most people. Sometimes,however, problems are solved by initially replacing a simple expression with a morecomplicated one. For example, in some applications the expression 1+sintis replacedby the more complex quantitycos2t1 sint. This essentially involves redoing the stepsinExample 5in reverse order as indicated in the following calculations:1 + sint=(1 + sint)(1 sint)1 sint=1 sin2t1 sint=cos2t1 sint:In particular, the rst step would be to multiply 1 + sintby the fraction1 sint1 sint(which has value one as long as 1 sint6= 0) to obtain the quantity(1 + sint)(1 sint)1 sint:The reader is advised to review the calculatons above while keeping in mind the in-sights required to perform the steps.

7 The strategy of replacing seemingly simpleexpressions by more sophisticated ones is a rather unnatural and confusing , with practice the strategy can be mastered and understood. The nextexample further illustrates this type of 2: The Elementary Identities11 Example 6 Determine the values oftsuch that2 sint+ cos2t=2:Solution: Equations such as these are usually solved by rewriting the expression interms of one TRIGONOMETRIC function. In this case it is reasonable to use the rstidentity inTable changecos2tto the more complicated expression1 sin2t:This will produce the following equation involving only the sine function:2 sint+1 sin2t=2:This last equation should remind the reader of the corresponding quadratic equation2x+1 x2=2which can be solved by factoring . That is what we will do here.

8 First,subtract2from both sides of the above equation and then multiply through by( 1)toobtainsin2t 2 sint+1=0: factoring this expression yields(sint 1)2=0:The only solution to this last expression is given bysint=1ort= 2+2k wherekis any The sum and di erence formulasThis section begins with the veri cation of thedi erence formulafor the cosinefunction:cos( ) = cos cos + sin sin where denotes the measure of the di erence of the two angles and :Oncethis identity is established it can be used to easily derive other important veri cation of this formula is somewhat complicated. Perhaps the most di cultpart of the proof is the complexity of the notation. A drawing (Figure )shouldprovide insight and assist the reader overcome this obstacle. Before presenting theargument, two points should be reviewed.

9 First, recall the formula for the distancebetween two points in the plane. Speci cally, if (a; b) and (c; d) are planer points,then the distance between them is given byp(a c)2+(b d)2:(1)Second, the argument given below conveniently assumes that 0< <2 . Theassumption that <2 is justi ed because complete wrappings of angles (integermultiples of 2 ) can be ignored since the cosine function has period 2 :The assump-tion that the angle is positive is justi ed because theSymmetry Identitiesguarantee that cos( ) = cos( ):We can now derive the 3: The sum and di erence formulas13xy (x1,y1)(x2,y2)xy (w, z)abFigure : Di erences of , observe that the angle appears inFigure (a)and (b)and is in standard positioninFigure This angle deter-mines achordor line segment ineach drawing (not shown), one con-necting (x1;y1)to(x2;y2) (inFig-ure ) and one connecting (w; z)to (1;0) (inFigure ).

10 Thesechords have the same length sincethey subtend angles of equal mea-sure on circles of equal radii. (SeeLesson 3 Section 6.) This observa-tion and the distance formula (Equation 1) permit the equalityp(x2 x1)2+(y2 y1)2=p(w 1)2+z2.(2)Section 3: The sum and di erence formulas14 Squaring both sides ofEquation 2removes the radicals resulting in(x2 x1)2+(y2 y1)2=(w 1)2+ the squares of the binomials suggests thatx22 2x1x2+x21+y22 2y1y2+y21=w2 2w+1+z2.(3)Since (x1;y1) is a point on the unit circle so that the equalityx21+y21= 1 holds, thesum ofx21andy21inEquation 3can be replaced by 1:Similar statements hold for thepoints (x2;y2);and (w; z). These replacements yield2 2x1x2 2y1y2=2 2w,or, after dividing by 2 and solving forw,w=x1x2+ desired formulacos( ) = cos cos + sin sin (4)follows from the observations that (Refer toFigure )w= cos( );x2= cos ;x1= cos ; y2= sin ;andy1= sin.