Analytic Trigonometry - Ougouag

Review of Trigonometry for CalculusAnalytic Trigonometry7

Basic Trigonometric Identities2

Example 1 – Simplifying a Trigonometric ExpressionSimplify the expression cos t tan t sin t.Solution:We start by rewriting the expression in terms of sine andcosine.cos t tan t sin t cos t sin tReciprocal identity Common denominator Pythagorean identity sec tReciprocal identity3

Addition and Subtraction FormulasWe now derive identities for trigonometric functions ofsums and differences.4

Example 2– Simplifying an Expression Involving InverseTrigonometric FunctionsWrite sin(cos–1 x tan–1 y) as an algebraic expression in xand y, where –1 x 1 and y is any real number.Solution:Let cos–1x and tan–1y. We sketch triangles withangles and such that cos x and tan y (seeFigure 2).cos xtan yFigure 25

Example 2 – Solutioncont’dFrom the triangles we havesin cos sin From the Addition Formula for Sine we havesin(cos–1 x tan–1 y) sin( ) sin cos cos sin Addition Formula for Sine6

Example 2 – Solutioncont’dFrom trianglesFactor7

Double-Angle FormulasThe formulas in the following box are immediateconsequences of the addition formulas.8

Half-Angle FormulasThe following formulas allow us to write any trigonometricexpression involving even powers of sine and cosine interms of the first power of cosine only.This technique is important in calculus. The Half-AngleFormulas are immediate consequences of these formulas.9

Example 3 – Lowering Powers in a Trigonometric Expressionusing Double Angle and Half Angle formulasExpress sin2 x cos2 x in terms of the first power of cosine.Solution:We use the formulas for lowering powers repeatedly.10

Example 3 – Solutioncont’dAnother way to obtain this identity is to use theDouble-Angle Formula for Sine in the formsin x cos x sin 2x. Thus11

Example 4 – Evaluating an Expression Involving InverseTrigonometric FunctionsEvaluate sin 2 , where cos with in Quadrant II.Solution :We first sketch the angle in standard position withterminal side in Quadrant II:Since cos x/r ,we can label a side and thehypotenuse of the triangleTo find the remaining side, weuse the Pythagorean Theorem.12

Example 4 – Solutionx2 y2 r2(–2)2 y2 52cont’dPythagorean Theoremx –2, r 5y Solve for y2y Because y 0We can now use the Double-Angle Formula for Sine.Double-Angle FormulaFrom the triangleSimplify13

Basic Trigonometric Equations14

Example 5: Basic Trigonometric EquationsFind all solutions of the equations:(a) 2 sin – 1 0(b) tan2 – 3 0Solution:(a) We start by isolating sin .2 sin – 1 02 sin 1sin Given equationAdd 1Divide by 215

Example 5 - Solutioncont’dThe solutions are 2k 2k where k is any integer.(b) We start by isolating tan .tan2 – 3 0tan2 3tan Given equationAdd 3Take the square root16

Example 5 - Solutioncont’dBecause tangent has period , we first find the solutionsin any interval of length . In the interval (– /2, /2) thesolutions are /3 and – /3.To get all solutions, we add integer multiples of tothese solutions: k – k where k is any integer.17

Solving Trigonometric Equationsby Factoring18

Example 6 – A Trigonometric Equation of Quadratic TypeSolve the equation 2 cos2 – 7 cos 3 0.Solution:We factor the left-hand side of the equation.2 cos2 – 7 cos 3 0(2 cos – 1)(cos – 3) 02 cos – 1 0 or cos – 3 0cos orcos 3Given equationFactorSet each factor equal to 0Solve for cos 19

Example 6 – Solutioncont’dBecause cosine has period 2 , we first find the solutions inthe interval [0, 2 ). For the first equation the solutions are /3 and 5 /320

Example 6 – Solutioncont’dThe second equation has no solution because cos isnever greater than 1.Thus the solutions are 2k 2k where k is any integer.21

Example 7 – Solving a Trigonometric Equation by FactoringSolve the equation 5 sin cos 4 cos 0.Solution:We factor the left-hand side of the equation.5 sin cos 2 cos 0cos (5 sin 2) 0cos 0or 5 sin 4 0sin –0.8Given equationFactorSet each factor equal to 0Solve for sin 22

Example 7 – Solutioncont’dBecause sine and cosine have period 2 , we first find thesolutions of these equations in an interval of length 2 .For the first equation the solutions in the interval [0, 2 ) are /2 and 3 /2 . To solve the second equation, wetake sin–1 of each side.sin –0.80 sin–1(–0.80)Second equationTake sin–1 of each side23

Example 7 – Solution –0.93cont’dCalculator (in radian mode)So the solutions in an interval of length 2 are –0.93and 0.93 4.0724

Example 7 – Solutioncont’dWe get all the solutions of the equation by adding integermultiples of 2 to these solutions. 2k –0.93 2k 2k 4.07 2k where k is any integer.25

Solving Trigonometric Equationsby Using Identities26

Example 8 – Solving by Using a Trigonometric IdentitySolve the equation 1 sin 2 cos2 .Solution:We first need to rewrite this equation so that it containsonly one trigonometric function. To do this, we use atrigonometric identity:1 sin 2 cos2 Given equation1 sin 2(1 – sin2 )Pythagorean identity2 sin2 sin – 1 0(2 sin – 1)(sin 1) 0Put all terms on one sideFactor27

Example 8 – Solution2 sin – 1 0sin cont’dorororsin 1 0sin –1 Set each factorequal to 0Solve for sin Solve for in theinterval [0, 2 )Because sine has period 2 , we get all the solutions of theequation by adding integer multiples of 2 to thesesolutions.28

Example 8 – Solutioncont’dThus the solutions are 2k 2k 2k where k is any integer.29

Solving Trigonometric Equationsby Graphs on calculator30

Example 9 – Solving Graphically by Finding Intersection PointsFind the values of x for which the graphs of f(x) sin x andg(x) cos x intersect.Solution :The graphs intersect where f(x) g(x).We graph y1 sin x and y2 cos x on the same screen, forx between 0 and 2 .(a)(b)31

Example 9 – Solutioncont’dUsingor the intersect command on the graphingcalculator, we see that the two points of intersection in thisinterval occur where x 0.785 and x 3.927.Since sine and cosine are periodic with period 2 , theintersection points occur wherex 0.785 2k andx 3.927 2k where k is any integer.32

Equations with Trigonometric Functionsof Multiples of Angles33

Example 10 – A Trigonometric Equation Involvinga Multiple of an AngleConsider the equation 2 sin 3 – 1 0.(a) Find all solutions of the equation.(b) Find the solutions in the interval [0, 2 ).Solution:(a) We first isolate sin 3 and then solve for the angle 3 .2 sin 3 – 1 02 sin 3 1sin 3 Given equationAdd 1Divide by 234

Example 10 – Solutioncont’d3 Solve for 3 in the interval[0, 2 )35

Example 10 – Solutioncont’dTo get all solutions, we add integer multiples of 2 tothese solutions. So the solutions are of the form3 2k 3 2k To solve for , we divide by 3 to get the solutionswhere k is any integer.36

Example 10 – Solutioncont’d(b) The solutions from part (a) that are in the interval [0, 2 )correspond to k 0, 1, and 2. For all other values of kthe corresponding values of lie outside this interval.So the solutions in the interval [0, 2 ) are37

Analytic Trigonometry