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# PRE Math 1

UCI

GPA 3.73

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This 53 page Class Notes was uploaded by Adam Crona on Saturday September 12, 2015. The Class Notes belongs to Math 1 at University of California - Irvine taught by Staff in Fall. Since its upload, it has received 31 views. For similar materials see /class/201868/math-1-university-of-california-irvine in Mathematics (M) at University of California - Irvine.

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674 ProduchuniandSuanr0ductIden tkm In this section we will learn following identities ProductSum Identities sinz cosy sinm y sinm 7 cosz siny sinm y 7 sinm 7 sinz siny cosm 7 y 7 cosm cosz cosy cosm y cosm 7 SumProduct Identities zy ziy 2 sinz siny 2 sin cos 2 sinz7siny 2cosmysinmiy 2 2 coszcosy 2cos3ycosmgy i z y i z 7 y cosz 7 cosy 72 sin sin 2 Proof of the product sum identities sinm y sinm 7 y sinz cosy cosz siny sinz cosy 7 cosz siny 2 sinz cosy Hence we have i 1 i i sinz cosy 5 s1nz y s1nm 7 Similarly we can obtain other product sum identities Start with a product sum identity sina cos sina B sina 7 3 1 Set 04 B m ai y Solving this system we have mty 2 2 Substituting 2 into equation 1 and simplifying7 we obtain the sum product identities m y cos m 7 y 2 2 Similarly7 we can obtain other product sum identities Example Find exact values 1 sin 1050 sin 150 2 sin 1050 7 sin 150 sinzsiny 2 sin Solution 1 sin 1050 sin 150 cos1050 7 150 7 cos1050 150 cos 90 7 cos 1200 1 1 1 7 0 7 21 lt 2 4 2 0 0 0 7 0 sin 105 7 sin 15quot 2 cos sin M 2 2 cos 600 sin 450 Q 7 2 2 7 239 Example Verify the identities sin 2t 1 sin 4t 7 cos 2t 7 cos 4t 7 sin 7 sin 2 7 Got cos x 7 cos y 2 cot t cosmy7zcosyz7zcoszz7y cosm y 1 3 cosz cosy cosz 1 Solution sin 2t 1 sin 4t 2 sin 2t cos 2 cos 2t 7 cos 4t 7 72 sin 2t sin 2 3 2 sin3t cost 72 sin3tsin7t cost Q cott sin z 7 sin y 2 sin L42 cos Lgy cos x 7 cos y 72 sin L42 sin Lgy 721 7 cos 2 7 7 m sm 2 m y 7 cot 7 2 3 COS COS y COS Z 1 cosm y cosm 7 cosz cosm y cosz cosm 7 y cos 2 cosz y z cosm y 7 cosz 7 y z cosm 7 y 7 icoszy7zcosyz7zcoszm7ycoszyz Example Verify the identities 1700s2y 2 1 17siny1siny itan y coszy 2 17 7 smy 1siny cot2m71 3 cot2zm 4 1 sin 2t sint cos t2 Solution 1700s2y 7 1700s2y 17siny1siny 7 17sin2y sinzy coszy tanzy 2 coszy 17sin2y 1siny 7 1siny 17siny1siny 1siny 17siny 5 5 Solving Right Triangles Solving right triangles Find all unknown parts of a right triangle given the measure of two sides or the measure of one acute angle and a side FIGURE 1 Basic relations between the elements of the right triangle 04 900 a2b202 Locate the right triangle in the rst quadrant of a rectangular coordinate system we have following relations C sin CSCB c b cos E sec E C a b a tan 7 COt a b 00lt6lt900 FIGURE Side 1 is often referred to as the side opposite angle 3 a as the side adjacent to angle 3 and c as the hypotenuse Using these designations for an arbitrary right triangle we have the relations Figure 2 Example Solve the right triangle with c 625 feet with B 3220 Solution Solve for 04 Oz 900 7 3220 5780 1 sin csc 7H Hyp OPP yp Opp cos 37 sec gig 0 Ad Adj tan A421 cot 0 lt B lt 900 FIGURE 3 625ft b I FIGURE 4 Solve for b sin C b 3220 7 M 625 b 625 sin3220 333feet Solve for a cos a C a 3220 7 COS 625 a 625 cos3220 529feet D Example Solve the right triangle with a 432 centimeters and b 262 centimeters 65 Trigonometric Equations Suggestions for solving trigonometric equations algebraically Page 486 of the textbook 1 Regard one particular trigonometric function as a variable and solve or it a Consider using algebraic manipulation such as factoring comb ing or separating fractions and so on b Consider using identities 2 After solving for a trigonometric function solve for the variable Example Find all solutions exactly for 2 coszzicosz0 Solution Step 1 Solve for cos x 2 cos2 x 7 cos x 0 cosz2 cosmil 0 cosz0 or 2cosz710 cos7 2 Step 2 Solve each equation for z over one period 0 2 7139 Step 3 Write an expression for all solutions 7r32k7r 7r22k7r 37r22k7r 57r32k7r z k any integer Example Find all real solutions for 3cos2z8sinm7 1 2 Solution Step 1 Solve sinz from the equation 300s2z8sinm 7 317sin2m8sinz 373sin2z8sinm77 3sin2z78sinm4 sinz723sinz72 0 sinz720 or 3sinz720 CDC sinz 2 sinz 3 Step 2 Solve each equation over one period 07 2 7139 The rst equation sinz 2 has no solution since 71 sinz 1 Solve the second equation sinz 2 3 sin l g 07297 x 7r 7 07297 24119 Step 3 Write an expression for all solutions 07297 2 k 7139 z 24119 2k k any integer Example Find 9 so that 5 sin20 7 5 730457 00 g 20 7 5 g 3600 Solution Step 1 Make a substitution Let u 2 t9 7 57 then u satis es 5 sinu 730457 00 g u g 3600 Step 2 Solve for sin u 73045 7 sin u 7 70609 Step 3 Solve u over 00 g u g 3600 Figure 1 Let Oz sin 170609 7375170 then u 04 3600 3224830 or u 1800 7a1800 375170 2175170 Step 4 Now solve for 0 u 2175170 u 3224830 20752175170 2075 322483quot 0 1112590 0 1637420 FIGURE 1 Example Find exact solutions for sinzzsin2z 0 xlt27r Solution 1 sinzz 5 sin2z l 2 sinm cosz 2 sinzzisinmcosz 0 sinzsinmicosz 0 sinz0 or sinzicosz0 9507 7139 sinzcosm sinz cosx tanm1 7r 57139 m7 7 47 4 Combining the solutions from both equations7 we have the complete set of solutions 5 7 4 39 Example Solve cos2z 4 cosi2 for all real z 4 Solution Step 1 Solve for cos m cos2m 4cosz72 2cos2z71 4cosz72 2cos2z74cosz1 0 4 i 167421 2 2 1707107 or 0292893 Step 2 Solve each equation over one period for z The rst equation cos x 1707107 has no solution The second equation cos x 0292893 cosm we have cosil 0292893 12735 or 27139 7 cos 1 0292893 50096 Step 3 Write an expression for all solutions 7 127352kr k At 35 7 50096 2 any 1n eger Example Find the exact solutions over the indicated intervals 1 coszcotz7 0 zlt27r 2cosm7sinz17 0 z 27r Solution cos x cot z cos x cos x sin x cos x cos x cosz 0 or cosx cosx sinz 1 sin z sinz 1 Find the solutions for z cosz0 sinz1 0 zlt27r z z 2 Combining the solutions7 we have the complete set of solutions 7139 37139 m E 3quot cot 37quot 0 1 7r 7 7r 7 Check cos 7 cot g 7 07 cos 2 Jim Lambers IVIath 1B Fall Quarter 2004 05 Lecture 9 Notes These notes correspond to Section in the text Angles and Their Measure It is to find countless examples of objects that conceptually contain straight lines that meet at a common point such the top and right edges of a piece of paper or the beams of a building To describe the relationship between these straight lines we introduce the following concept De nition 1 Angle Let aquot and s be two rays with a common endpoint P An angle a rotation of 9 around P until it coincides with s The ray 9 called the initial side of the angle and the ray s called the terminal side of the angle The point P called the vertex of the angle An angle said to be positive if the rotation in the counterclockwise direction and negative if it in the clockwise direction Two angles are said to be coterminal if they have the same initial and terminal sides Degree and Radian Measure Because two rays that form an angle can conceivably point in any direction there are infinitely many angles that the rays can form Therefore we need some method of measuring angles so that w can distinguish them just we use length to measure line segments This leads to the following definition De nition 2 Degree Let 9 be a real number An angle said to have a measure of 9 degrees written as 9 if the rotation of the angle 9360 of a complete rotation The following terms are useful for classifying angles based on their degree measure De nition 3 Types of angles An angle said to be 0 straight if its measure equal to 180 0 right if its measure equal to 90 0 acute if its measure 9 satis es 0 lt 9 lt 90 0 obtuse if its measure 9 satis es 90 lt 9 lt 180 Two angles are said to be o complementary if the sum of their measures equal to 90 0 supplementary if the sum of their measures equal to 180 Examples of straight right acute and obtuse angles are shown in Figure 1 Straight angle 9 180 Right angle 9 90 444 Acute angle 06 lt e lt 909 obtuse angle 906 lt e lt190a Figure 1 Examples of straight right acute and obtuse angles For each angle the solid arrow is the initial side and the dashed arrow is the terminal side The vertex of each angle is marked with a circle Example 1 Degree measure is used to define the concepts of latitude and longitude which to gether define a point on the earth s surface The latitude of a point P on the surface is the degree measure of an angle which has its vertex located at the center of the earth The terminal side of the angle is the ray that passes through the point P and the initial side of the angle is the ray that passes through the equator but lies in the plane containing the terminal side and the line that passes through the north and south poles El CS 217 Light and Geometry in Vision Spring 2009 Lecture 1 7 April 1 Scribe Kuang Chih Liu Lecturer Deva Ramanan Note These lecture notes are still rough7 and have only have been mildly proofread 1 1 Image Formation o What determines where the image of a scene point will lie o What determines its brightness in the nal images 1 1 1 Light Light is an electromagnetic radiation exhibits properties of 0 wave quotW wavelength refraction wave can change direction in new medium di raction wave can bend when passing through small openings polarization wave oscillates in 1 direction eg polarized sunglass rectrict light passing through lens 0 particle model light transport using energy packets quantum called photons black body radiation hot metal glow 112 Measuring Light 1 We measure the amount of light hitting a surface with Irradiance power of electromagnetic radiation area wattmeter2 6P P z AA 0 6A AA as H 6A area of an in nitesimal surface patch power energy sec7 watt joules sec 1 1 CS 217 Lecture 1 7 April 1 Spring 2009 2 We measure the amount of light leaving a surface as Radiance power foreshortened areasolid angle watt meter2 sr 62F N P mm W AAAw Steradian surface area of an unit radians sphere cut out by a solid angle 0 N 47139 in 1D reduces to radian length of an unit radius circle cut out by a unit angle 0 N 27139 we need foreshortened area because a patch directly overhead 6A sees more of A 113 Imaging 21 pixel pixel intensity i i acAac A 1 7r 1 sensor response oc total irradiance fan f y ft0 f4 f0 El l t 19 Ch Ell51155195143 0 lt f0 43 lt 1 c gt will tend to l for 0 directly overhead patch Q Why do we not see an image of a scene on a paper A Restrict directions of incoming light with a pinhole Pinhole optics Right hand coordinate system place the scene at Z N eu CS 217 Lecture 1 7 April 1 Spring 2009 Special case Orthographic Projection Assume we image a plane at depth zo eg scene is constant depth Where z ZOV i y39 y39 y fl39i f i Zi 20 f ymy wheremi Zo lets choose pixel unit st m1 Orthographics an M Assume points in scene always lie in a plane z zo An AZ ltlt zo f f lt 7 g 7 V m V y Z0 Am 9 Z0 M 91 If the depth variation max AZ is small compared to the distance to the camera orthographic projection is a good mo e 1 assume Where zo I m I X m i 5 Scaled ortho ra hic ro39ection y m y z s p p J e de ne m1 adjusting pixel units E0 E Z i i 5 Orthographic projection 3 Scene as collection of planar objects e Z1 zm for m objects 7 Z 7 Z I i for j1 m 5 Periobject scaled orthographic projection 95 can fall in this category CS 217 Lecture 1 7 April 1 Spring 2009 114 Lens 0 Pinhole must have nite diameter 7 this blows collection of photons onto a region o If the pinhole is too small 1 Not enough photons hit scene 2 Diffraction effect occur wave interpretation 0 Solution gather light with a lens um ohm mm Real Image 0 Thin lens model ignore refraction of light from passing through thick lens Two rules for thin lens math 1 All rays pass through lens are unaltered 2 All rays parallel to z axis converge at a point called the focal length 7 E 1 lif 7 f plug 1 into 2 3 l i 1 thin lens formula 2 72 f Given lens with f7 if we place image plane at Z77 then all 3D scene point at depth z will be imaged as points Jim Lambers IVIath 1B Fall Quarter 2004 05 Lecture 5 Notes These notes correspond to Section 44 in the text Common and Natural Logarithms In the previous lecture we defined the logarithmic function with base I where l is any positive real number other than one In this lecture we focus on the two most frequently used bases De nition and Evaluation The most frequently used base for exponentiation across all applications areas in which mathemat ics pla s a significant role is certainly base 10 The logarithmic function of base 10 is often useful for solving equations involving exponentiation with base 10 Therefore this logarithmic function long is called the common logarithm and is often written simply log In Lecture 3 we introduced the natural exponential function which of base 6 Because of this function s importance we similarly define the natural logarithmic function or natural logarithm to be the logarithmic function with base 3 This function log J is written an where the n in the symbol ln means natural We will see that in a number of applications it is necessary to solve the equation logJ g or an y where one of J and g is known and the other needs to be computed To solve logJ g or an g for t given the value of J one can use a typical scientific calculator that will be able to evaluate both of these logarithmic functions computing logarithms in any other way is beyond the scope of this course On the other hand to solve logJ g for J given the value of y we can convert the equation from logarithmic form to exponential form and compute J 109 using a calculator if g is not an integer Similarly we can solve an g for J by converting the equation to exponential form and computing J all using a calculator Applications We now discuss some applications of the common logarithm logJ In general the common loga rithm is particularly useful for measuring quantities that tend to over an extremely wide range of values spanning several orders of magnitude such the range of real numbers from 10 10 to 1010 For such a quantity the common logarithm provides a reasonably r ed number that indi cates the magnitude of the quantity s value because it represents the values exponent the power to which 10 must be raised to obtain this value 5 2 Circular Functions De nition of the Circular Functions sine function sinz b cosine function cosz a a b i b tangent function tanm 7 a 7S 0 m Wrapping 170 U Circular 01 Function Functions cosecant function cscz F b 7 0 i 1 secant function sec z 7 a 7 0 a cotangent function cot z b 7 0 FIGURE 1 Circular Functions If x is a real number and 11 are the coordinates ofthe circular point Wm7 then sinzb cscz7 by O cosza secz77 ay O a tanz a7 0 cotzb7 0 a The domain and range of circular functions are given at Figure 1 Exact Values for Particular Real Numbers Example 1 Evaluate each circular function exactly for z 7r3 Solution From de nition of wrapping function7 we have 7139 1 W3 va 7 39 Function De nition domain range sine sinz 1 00 lt z lt 00 11 cosine cosz a 00 lt z lt 00 11 tangent tanm ab such that a 7 0 00 lt tanm lt 00 a cotangent cotz a 1 such that b 7 0 00 lt cotz lt 00 1 secant secz ab such that a 7 0 1g lseczl lt 00 a 1 cosecant cscz g a 1 such that 1 3A 0 1 lcsczl lt 00 TABLE 1 Thus mid 686li 3 2 3 b 2 cos 7r 1 sec 7r 1 1 2 7 a 7 7 7 7 3 2 3 a 12 Wibi Qi wiailgil 3772 3 1 Wr D Example 2 Evaluate exactly A sin 57r6 B cot7r C sec27r3 D tan77r4 Solution A W 5 7 Su B W7r 710 cot7r 71 Not de ned 0 WltJgt 7 773 secs 2 71 2 D We 77 tame lg 71 a Sign Properties As a circular point moves form quadrant to quadrant its coordi nates a b undergo sign changes Hence the circular functions also undergo sign changes Figure 2 Basic Identities The circular functions are de ned through a b of a b and sinmb cosma Sign in quadrant Circular U Function 1 II III IV sin x b cscz 11 cosz a secz 10 tanm 90 cot z ab FIGURE 2 Hence we can obtain other circular function from sinz and cosz as follow ing 1 1 7 01 cscm b sinz 1 1 secz 7 02 a cosx a 1 1 cotz Eb7a 03 tanm 8mm 04 a cosx a cosx t 7 05 CO m b sinz Because the circular points and W7z are symmetric with respect to the horizontal axis we have the following sign properties sin7z 7b 7 sinz cos7m a cosz 7b b tan7z 7 if itanz a a Finally because a b cos m sin z is on the unit circle 742 l v2 1 it follows that cos z2 1 sin z2 1 which is usually written is sin2 z cos2 x 1 06 where sin z and coszz are concise ways of writing sin z2 and cos m2 respectively Caution sin z2 7 sin m2 cos z2 7 cosz 2 2 Theorem 1 Basic Trigonometric Identities For x any real number in all restricted so that both sides of an equation are de ned Reciprocal Identities cscm secz 7 cot z 7 sin z cos x tan z Quotient Identities sin z cos x cot z i cos no sin 9c Identities for Negatives sin7z 7 sinz cos7m cosz tan7z itanm Pythagorean Identity sin2 z coszz 1 The basic identities are true for all replacements ofx by real numbers for which both sides of an equation are de ned These basic identities must be memorized along with the de nitions of sicc circular functions Example 3 Use the basic identities to nd the values of the other ve circular functions given sinz 7 and tanm gt 0 Solution Since sinz lt 0 and tanm gt 07 the circular point is in quad rant lll We next nd cosz using the identities sin2 z cos2 x 1 1 75 cos2 x 1 3 2 7 7 cos 9c 7 4 3 cosz 7 Since Wmis in quadrant lll Jim Lambers IVIath 1B Fall Quarter 2004 05 Lecture 4 Notes These notes correspond to Sections 26 and 43 in the text Logarithmic Functions Inverse Functions In Lecture 1 we briefly discussed the notion of an inverse function If f is a one to one function then it has an inverse function denoted by f l such that f lfo ll 1 and ff 1yy 1 for any J in the domain of and any g in the range of In words these equations which are known the cancellation equation show that f and f 1 describe proce that undo one another That is if f is applied to an input J to produce an output g and f is applied to y then the result is the original input Similarly if f 1 is applied to an input g to obtain an output J and f is applied to J then the result is g The cancellation equations can be used to obtain an equivalent characterization or definition of the inverse function that is actually more useful for the task of finding the inverse of a given function We now state this definition De nition 1 Inverse Function Let f be a onetoone function with domain D and range R The inverse function of f denoted by f l the function with domain R and range D de ned by the statement J f 1y if and only if y fJ39 2 where J belongs to D and g belongs to R The reason why this definition of the inverse function is useful for finding the inverse of a given function f is that we can solve the equation 1 f x for J and then use this definition to conclude that the expression we obtain for J is equal to f 1 g and therefore we will have found the inverse function The following result provides an way of determining whether a function is oneto one and therefore would have an inverse Recall that a function f is increasing if fJ gt fg whenever J gt g and decreasing if fJ lt y whenever J gt g Theorem 1 If a function f increasing on its entire domain or decreasing on its entire domain then f onetoone Proof Suppose that is increasing Then if J lt y then lt Let J and y be any two real numbers in the domain of such that J 7 9 Then we must have J lt y or J gt g If J lt y then fJ lt y and if J gt y then fJ gt y Therefore fJ y Since this is true for any J and y in the domain of f we conclude that f is oneto one The proof for the case where f is decreasing is similar El De nition of the Logarithmic Function In this lecture we discuss the inverse function of the exponential function with base I f J b As discussed in lecture 2 b is an increasing function if b gt 1 and it is a decreasing function if 0 lt b lt 1 In either case by Theorem 1 b is oneto one It follows that b has an inverse function We now precisely define this inverse function De nition 2 Logarithmic Function Let b gt 0 and assume that b 7 1 Let y by be the exponential function with base b The logarithmic function with base b the inverse function of f and denoted by f lJ logb That y long if and only if by J 3 where J gt 0 and y a real number The statement 9 long in equation 5 called the logarithmic form and the statement by J in equation 5 called the exponential form The number 9 in equation 5 called the logarithm or log to base I of It is important to realize that the value of log J is an exponent Specifically if i long then 9 is the exponent to which I must be raised in order to obtain The definition provides a verv useful statement for solving equations involving exponential and logarithmic functions the equivalence of the exponential form by J and the logarithmic form 9 log It is important to be able to convert between these forms in order to solve such equations Properties of Logarithmic Functions The properties of exponential functions introduced in Lecture 2 can be used to derive similar properties for logarithmic functions We will see that these properties are extremely useful in solving problems involving logarithmic functions In the statement of the following theorem that lists these properties we use the symbol ltgt which means if and only if That is the statement p ltgt 1 means that the statements p and q are true under exactly the same circumstances so in effect they are equivalent statements Theorem 2 Properties of Logarithmic Functions Let b i and r be real numbers where b J and y are positive and b 7 1 Then log 1 0 2 log I 1 33 logbbr 4 biogquot T r where aquot gt 0 5 log m 2 long logb y 2 J 10315 10gw 40319 7 logb J aquot logb J 8 long logby ltgt J 9 Proof 1 Since I gt 0 b0 1 Rewriting this equation in logarithmic form immediately yields log 1 0 2 Since I gt 0 1 I Rewriting this equation in logarithmic form immediately yields log I 1 3 Let z If If we rewrite this equation in logarithmic form we obtain logb z 9 Substituting bf for 2 yields logb bf 3 Alternatively let f r N Then the inverse function of f is f 1 z log z Because f and 1 are inverse functions of one another 1f 9 However by the definition of these functions we also have fquotf 3 f 1 N log N Therefore logb bf r 5 Let 2 logb aquot Note that the definition of 2 makes sense only if r gt 0 which we assume in this case Then rewriting the equation z log 9quot in exponential form yields bz Substituting z log 9quot into the exponential form yields biog r Alternativel 7 let z M Then the inverse function of is 1 log 9quot Because and 1 are inve e functions of one another 10 9 However by the definition of these functions we also have f f 10 flogb r biog Therefore logb bf r Jim Lambers IVIath 1B Fall Quarter 2004 05 Lecture 16 Notes These notes correspond to Section 61 in the text Basic Identities and Their Use Basic Identities The following identities which were introduced in Lecture 8 provide useful relationships among the trigonometric functions that allow values of these functions to be computed using limited information and allow problems involving these functions to be transformed in a number of so that they can be simplified o Reciprocal Identities From the definitions of the circular functions we observe that the following pairs of functions are reciprocals of one another 1 39 t 7 1 w J tanx39 1 secJ 7 2 cosJ cst39 1 3 f sinJ o Quotient Identities Let WJ 11 Since sinJ I and cosJ a it follows from the definition of tanJ 00 and cotJ 11 that sinJ tanJ 4 cosJ cosJ cotJ a sinJ o Identities for Negatives A function f 1 is an even function if f J f 1 and it is an odd function if f J f 1 The functions sinJ and tanJ are odd functions and cosJ is an even function That is sin J sin 1 COS J cos 1 7 tan J tan 8 Jim Lambers IVIath 1B Fall Quarter 2004 05 Lecture 11 Notes These notes correspond to Section in the text Solving Right Triangles A right triangle is a triangle that has a right angle A triangle has three angles whose measures sum to 180 which implies that a right triangle has two acute angles Suppose that a right triangle is placed so that one of the acute angles which we denote by 9 is in standard position The sides of the right triangle are 0 The side that is adjacent to the angle 9 We denote its length by a This side coincides with the initial side of 9 which is the positive o The side that is opposite the angle 9 We denote its length by b o The hypotenuse which is the side opposite the right angle We denote its length by c This side coincides with the terminal side of 9 This triangle is illustrated in Figure 1 Consider the point cos 9 sin 9 This point lies on the terminal side of 9 Therefore this point and the origin form a line segment of length one that overlaps the hypotenuse This line segment along with the line segment joining the origin to cos 9 0 and the line semgment joining cos 90 to cos 9 sin 9 form a right triangle one of whose acute angles is 9 Therefore this right triangle is similar to our original right triangle which means that the ratios of the lengths of corresponding sides are equal that is c a I 7 7 7 lt1 1 cos9 sin9 From the equality of these ratios it follows that sin9 2 cos9 3 From the quotient and reciprocal identities introduced in Lecture 8 it follows that cot9 csc9 tan 9 sec9 2 52 c elfle 31b cosesine 1 sine 9 case a Figure 1 Right triangle with its acute angle 9 in standard position The adjacent side has length a the opposite side has length I and the hypotenuse has length 0 The overlapping similar triangle has a hypotenuse of length 1 an adjacent side of length cos 9 and an opposite side of length sin 9 These relationships between the values of the trigonometric functions at 9 and the sides of the right triangle are most easily remembered by using the intuitive names associated with the sides of the right triangle to refer to their respective lengths opposite 088 adjacent 8mquot hypmenuse opposite 1 7 W 8 a Jacent Low OPPOSW hvpotenui hVPOtenuse A 39 o 39A 39 41 5 8 adJacent L 8 OPPOSW Using these relationships well the fact that the two acute angles of a right triangle are complementary one can use limited information about the sides or angles of a right triangle to compute the lengths of the remaining sides or angles This process is known solving a right triangle Specifically one can use the lengths of any two sides to determine the length of the third side and the measure of both acute angles or use the length of one side and the measure of one acute angle to compute the length of the other two sides and the measure of the other acute angle We now illustrate this process with some examples Example 1 Suppose that we know that one of the acute angles of a right triangle has a measure of 30 and that the length of its adjacent side is 2 units What are the lengths of the other two sides Solution Using the relation o osite tan 30 gt 4 adjacent it follows that the opposite side has length 2 tan 30 2 2x33 Similarly we can use the relation adj acent cos 30 gt F hypotenuse J to conclude that the hypotenuse has length 2 2 403 6 cos30 g 3 Finally we can use the fact that the acute angles of the triangle are complementary to conclude that the other acute angle has measure 90 30 2 60 This right triangle is an example of a 300090 triangle in which the hypotenuse is twice the length of one of the sides El Example 2 Suppose that a hexagon is inscribed in a circle of radius 4 units What is the length of each side Solution Consider the triangle formed by two adjacent vertices of the hexagon and the center of the circle Then because a hexagon has sides the angle whose vertex is the center has a measure of 360 6 60 Also because the sides that form this angle have the same length that is the triangle is an isosceles triangle we can bisect this triangle into two congruent right triangles that is their sides are the same length using a line segment that begins at the center of the circle and is perpendicular to the side of the hexagon we are considering This is illustrated in Figure 2 Each of these triangles has an angle whose vertex is at the center and whose measure is 60 2 30 In other words each triangle is a 30 60 90 triangle Because each of these triangles has a hypotenuse with one endpoint at the center of the circle and the other endpoint being a vertex of the hexagon its length is 4 units Therefore we can use the relation Sin 300 opposite hypotenuse A l V to obtain 1 opposite 4sin 30 4E 2 8 7 1 Law of Sines Acute triangle Obtuse triangle a b FIGURE 1 Oblique triangles Triangles without a right angle Every oblique tri angles is either acute all angles between 00 and 900 or obtuse one angle between 900 and 1800 Note how the sides and angles of the oblique triangles in Figure 1 have been labeled Side a is opposite angle a side I is opposite angle Band c is opposite angle 39y Also note that the largest side of a triangle is opposite the largest angle Solving the triangle Given any three of the six quantities indicated in Figure 1 we are interested in nding the remaining three if they exist Basic relations quantities of a triangle a y 1800 11 gt c bc gt a ca gt b laibl lt c lbicl lt a lcial lt 1 Method to solve the triangle law of sines and law of cosines Law of Sines Theorem 1 Law of Sines ln words the ratio of the sine of an angles to its opposite side is the same as the ratio of the sine of either of the other angles to its opposite side 1 sin 04 sin 3 sin 39y a b c FIGURE 2 From the law of sines7 we can solve triangles given 1 Two sides and an angle opposite one of them SSA 2 Two angels and any side ASA or AAS Example Solve the triangle with 04 2800 7 B 45020 7 c 120meters Solution Solve for 39y 04 B 39y 1800 39y 1800 7 04 B 1800 7 2800 4590 106040 Solve for a sin 04 sin 39y a C c sina sin 120 sin 2800 sin 106040 588 meters Solve for 1 sin 3 sin 39y T T 0 sin 3 sin 39y 120 sin 45020 sin 106040 891 meters 3 Example Solve the triangles with 04 123071 23 centimeters and a 47 centimeters Solution From a rough sketch Figure 37 we see that there is only one triangle 47 cm 23 c A C FIGURE 3 Solve for B sin 7 sina b 7 a sing 7 b sina 7 23 sin 1230 7 a 7 47 23 sin 1230 71 0 24 3 sin lt 47 gt Solve for 39y a y 1800 39y 1800712307240330 Solve for 0 sin 04 sin 39y a C 47 330 a isan 31 centimeters sin 04 s1n1230 D Example Solve the triangles with 04 2607a 10 meter and b 18 meters Solution There are two possible triangles l and H Figure 4 18 m 10 m FIGURE 4 Solve for B and B We start by nding 3 and B using the law of sines sin 7 sina b 7 a i i 0 sing b sina 18 s1n26 07891 a 10 Angle B can be either obtuse or acute 7 1800 7 sin 1 07891 or 6 sin 1 07891 0 1800 7 520 1280 We next nd 39y and 39y 39y 180 7 260 128quot 260 39y 180 7 260 5201020 Finally7 we solve for c and c Solve for 39y and 39y Solve for c and c sina 7 1 sina 7 sin39y a 7 ci a 7 cl asin39y asin39y c c s s 7 151311260 7 10nsin1020 7 sin260 7 sin260 10 meter 22 meter In summary Triangle l B 1280 Triangle ll 3 520 39y 1020 39y 260 c 10 meter 0 22 meters Jim Lambers IVIath 1B Fall Quarter 2004 05 Lecture 19 Notes These notes correspond to Section 64 in the text ProductSum and Sum Product Identities The sum and difference identities for sine and cosine can be used to derive new identities that relate sums and products of sines and cosines These identities are useful for solving various problems in calculus and for the study of ProductSum Identities Recall the sum and difference identities for cosine cosJ y cosxcostJ sinxsian cosJ y cosJcostJ sinxsian If we add these two equations we obtain the identity cosJ y cosJ y 2 cosJcos J On the other hand if we subtract equation 1 from equation 2 we obtain cosJ y cosJ y 2 sinJsin J Similarly consider the sum and difference identities for sine sinJ y sinJ cos y cosJ sin y SlnJ y sinJ cos y cosJ sin J If we add these two equations we obtain the identity sinJ y SlllJ y 2 sinJcos J On the other hand if we subtract equation 6 from equation 5 we obtain sinJ y sinJ y 2 cosJ sin 9 AA cum VV 44 Common and Natural Logarithms Common and Natural Logarithms Common logarithms are logarithms with base 107 and denoted by log Natural logarithms are logarithms with base 57 and denoted by In Logarithm Notation logz log10 z Common logarithm lnz loge z Natural logarithm FIGURE 1 LogarithmExponential Relationships logz y is equivalent to z 10y lnz y is equivalent to z ey Example 1 Use a calculator to evaluate each to six decimal places A log3184 ln0000349 Proof log 3184 3502973 Verify 1033952973 318399957 B ln0000349 77960439 Verify 10 7960439 0000349 B Example 2 Use a calculator to evaluate each to six decimal places log2 2 A 131 7 01 271 11 lt gt low lt gt ogL1 lt gt og og Proof log2 030103 log 11 00413927 1 log2 7 030103 39 log11 00413927 B log 7 log 181818 0260 11 C log2 7 log 11 030103 7 00413927 0260 7273 Example Simplify using the properties of logarithmic functions A ln1 log 10 lne 1 1010g7 Solution ln1 0 log 10 1 C ln52m12m1 lne1 101 g77 D Example If ln3 110 and ln7 195 nd A 1mg B In an Solution lng ln7 7 ln3 195 7110 085 B Ins ln2113 1n3 x 7 ln3 1n7 110 195 102 D Applications Sound Intensity The sound intensity can be measure by decibel as following D 1OlogIi Decibel scale 0 where 1 D The decibel level of the sound 2 I The intensity of the sound measured in watts per square meter Wmz 3 0 The intensity of the least audible sound that an average healthy young person can hear standardized to be 0 10 12 watt per square meter Sound intensity Wmz Sound Decibel level 10 Threshold of hearing 0 52 x 10 10 Whisper 2716 32 x 10 6 Normal conversation 6505 85 x 10 4 Heavy traffic 8929 32 x 10 3 Jackhammer 9505 10 x 100 Threshold of pain 120 83 x 102 Jet plane with afterburner 14919 TABLE 11 Typical Sound Intensities and Decibel Level Earthquake Intensity 3 The energy released by the earthquake can be represented by the mag nitude M on the Richter scale as following 2 E M g log0 Richter scale where 1 M The Magnitude on Richter scale 2 E The energy released by the earthquake measure in joules 3 E0 The energy released by a very small reference earthquake stan dardized to be E0 1043940 joules Magnitude on Richter scale Destructive power M lt 45 Small 45 lt M lt 55 Moderate 55 lt M lt 65 Large 65ltMlt75 Major 75 lt M Greatest TABLE 2 The Richter Scale Example 6 If the energy release of one earthquake is 1000 times that of another how much larger is the Richter scale reading of the larger than the smaller Solution Let E1 and E2 the energy releases of smaller and larger earthquake respectively Let 2 E1 M771 7 1 3 OgEO and 2 E2 M771 7 2 3 OgEO be the Richter equations for the smaller and larger earthquakes respectively Substituting E2 1000 E1 into the second equation we obtain 1000 E1 E0 E lt10g103 log jg M2 og who who who 2M1 Thus an earthquake with 1000 times the energy of another has a Richter scale reading of 2 more than the other 62 Sum Difference and Cofunction Identities In this section we will learn following identities Summary of Identities Sum Identities cosm y cosz cosy 7 sinz siny sinm y sinz cosy cosz siny tan tan tanz y m y 1 7 tan z tan y Difference Identities cosm 7 y cosz cosy sinz siny sinm 7 y sinz cosy 7 cosz siny tan 7 tan tamm y 1 tan z tan y Cofunction Identities Replace 7r2 with 900 if z is in degrees 7139 i i 7f 7139 cos7 7 z sinz s1n 7 z cosz tan 7 z cotz 2 We will prove difference identity for cosine cosm 7 y cosz cosy sinz siny 1 And other identities can be readily veri ed from this particular one At rst we assume that z and y are in the interval 027r and z gt y gt 0 We associate z and y with arcs and angles on the unit circle as indicated in Figure 1 We form a triangle AOB with A cos y sin y B cos msiny Now we rotate the triangle AOB clockwise about the origin until the terminal point A coincides with D1 0 then terminal B will be at C cosz 7 y sinm 7 Since the rotation preserves lengths we have dAB dCD c7agt2ltd7bgt2 7 lt17egt2lt07fgt2 Cia2db2 1762f2 0272aca2d272dbb2 172552f2 02d2a2b272ac72db 172552f2 Sincepoints ABandCare on unit circles we have 02d21a2b21 and 52 f2 1 and we have eacbd 1 a b 5 f Acosy sin y Ccosz 7 y sinm 7 o D10 1 D10 C d Bcosmsinz FIGURE 1 Replacing e a c b d with cosm7y cos y cos x sin y and sin m respectivelyFigure 1 we obtain cosm 7 y cosy cosz siny sinz cosz cosy sinz siny Thus we prove the difference identity for cosine for z and y are in the interval 0 2 7r and z gt y gt 0 It then follows easily by periodicity and basic identities that the identity holds for all real numbers and angles in radian or degree measure Verify other identities Sum identity for cosine Replace y with 7y in equation 1 we obtain the sum identity for cosine cosm y cosz cosy 7 sinz siny 2 Cofunction identities Set z g in the identity 1 we have 7T 7139 i 7139 1 cos 573 cos cosys1n siny 0 cos y 1 sin y siny Hence we have the cofunction identity for cosine cos 7 y siny 3 Cofunction identities for sine Now let y 7r2 7 z in equation 3 we have This is the cofunction identity for sine sin 7 z cosz 4 Cofunction identity for tangent The cofunction identity for tangent can be obtained as following sin 7 z cosz 7r tanlt77mgt cotm 2 cosg7m sinz Thus we have the cofunction identity for tangent tan 7 z cotz 5 Difference and sum identity for sine sinm 7 y cos 7 m 7 7 cosiltgzgtltygti cos 7 z cos7y sin sin7y sinz cosy 7 cosx siny Hence we have the difference identity for sine sinm 7 y sinz cosy 7 cosz siny 6 Replace y in equation 6 with 7y we obtain the sum identity for sine sinm y sinz cosy cosz siny 7 Difference and sum identity for tangent mm 7 E sin z cos y 7 cos x siny cos x cos y sin z siny sinm cosy 7 cosm siny cosm cosy cos x cosy cosm cosy sinm siny cosm cosy cos x cosy sinm 7 sinm cos x cosy sinm siny 1 cosm cosy tan z 7 tan y 1 tan z tan y Thus we have the difference identity for tangent tan z 7 tan y t 7 8 anw y 1tanztany 63 DoubleAngle and HalfAngle Identities In this section7 we will learn following identities Doubleangle and halfangle identities DoubleAngle Identities sin 22 2 sinz cosz 7 2 2 7 2 7 2 cos2z 7 cos m7s1n 7172s1n zi2cos 71 2tanm 2cotz 2 tan2m 7 17tanzz cotzz71icotz7tanz HalfAngle Identities z 1 7 cos x i 7 i 7 sin 2 2 z 1 cos x 7 i 7 cos 2 4 2 z 17cosm sinz 17cosm tan 7 i 7 7 7 2 1cosm 1cosm sinz where the sign is determined by the quadrant in which 22 lies Verify the identities DoubleAngle Identities We start with the sum identities sinzy sinz cosy cosz siny 1 cosm y cosz cosy 7 sinz siny 2 tanm tan tanz y W 3 Doubleangle identity for sine sin 2 z sinm z sinz cosz cosz sinz 2 sinz cosz Doubleangle identity for cosine cos 2 z cosm z cosz cosz 7 sinz sinz cos2 x 7 sin2 2 First double 7 angle identity for cosine 17 sin2 2 7 sin2 2 1 7 2 sin2 2 Second double 7 angle identity for cosine coszz 7 17 cos2 x 2 cos2 x 7 1 Third double 7 angle identity for cosine 1 Doubleangle identity for tangent tan2 z tanz z tan z 1 tan z 1 7 tan z tan z 2 t First double 7 angle identity for tangent 1 2 m 1 2 2 calm cot2 z 7 609960 7 C0102 2 t Second double 7 angle identity for tangent cot z 7 1 2 cot z tanm cot2 z 7 1 tanm 2 Third double 7 angle identity for tangent cot z 7 tan z HalfAngle Identities Half angle identities are simply doubleangle identities stated in an alter nate form We start from the doubleangle identity for cosine cos2m172 sinzm2 cos2m71 Replace m With z2 and solve for sinm2 or cosz2 i 1 cosz 172s1n27 2 i 2m 17cosm sin 7 7 2 2 1 7 sin i Half 7 angleidentityforsine where the sign is determined by the quadrant in which m2 lies cosz 2 cos2 3 7 1 6082 g 7 1 cosz 2 7 2 1 cos i Half7angleidentityforcosine where the sign is determined by the quadrant in which m2 lies 3 Finally7 we use quotient identity to obtain a half angle identity for tan gent i licosm t x sing iXT i licosm ani 7 2 cos i 1icosm 1 cosag7 2 where the sign is determined by the quadrant in which m2 lies Simpler version of half angle identities for tangent can be obtained as following z sin g tan 7 I 2 cos i 12 12 2 cos SID g g 2 cos2 cos2 sinlt2 e 2 coszg sinz 1 cosz z sin tan 7 I 2 cos i 2 sing sing 2 sing cosg A 2 2 s1n s1n2 17 cosz sinz Example Verify the identities litanzz A 2 cos x 1tan2 tanmisinm 2 7 B s1n g7 2tanz 4 Solution 17 tan2 z 1 tan2 z A 2 sin 1 1 cos2z A 2 sin 1 1 cos2z A 2 COSZm 7 sin2mgt cos 1 A 2 2 sin 1 cos x 1 605 coszz 7 sinzz cos2 x sin2 z cos2 7 sin2 z cos 2x B Start from the left side7 we have z I 1 7 cos x sin 7 i 7 2 2 sing E 7 17 cosz 2 7 2 Start from the right side7 we have tanm 7 sinz 7 sinz 2 tan z 7 2 5 C0551 cos x 7 sin z 7 cos x sinz 7 cosz sinz 7 2 sin z sin z17 cos x 2 sin z 7 1 7 cosz 7 2 Hence7 we have sinzz 717cosz tanm7sinz 2 7 2 2 tanm D Example Find exact values A Find the exact values of sin 2x and cos 2x if tanm 7 and z is a quadrant lV angle B Compute the exact values of sin 1650 using a half angle identity C Find the exact values of cosm2 and sinm2 is sinz 75 7139 lt mlt37r2 Solution From the reference triangle for m Figure 27 we have T x73242 5 and i 3 4 Sln7 cosm7 5 5 Now use doubleangle identities for sine and cosine FIGURE 1 sin 2x 2 sinz cosz 7 4 7 cos2z 2 cos2z7123271 7 0 sin1650 sin330 I 1 7 cos 3300 2 C From the reference triangle for m Figure 3937 we have 17 52773 74 and 4 cos x if 5 If7r lt z lt 37r27 then 53 Angles and Their Measure 1 ANGLES AND THEIR MEASURE 11 Angles An angle is formed by rotating a ray about its endpoint The starting position of the ray is called the initial side and the nal position of the ray is the terminal side The common endpoint V of the initial side and terminal side is called the vertex A counterclockwise rotation produces a positive angle7 and a clockwise rotation produces a negative angle See Figure 1a b7 where the Greek letter 9 has been used to denote an angle Two different angles may have the same initial and terminal sides Such angles are said to be coterminal angles Figure 2 Terminal Side Vertex Initial Side erminal side 0 Vertex Initial Side Terminal Side Vertex Initial side a b c FIGURE 1 a A Positive Angle t9 A Negative angle 9 Standard Position An angle in a rectangular coordinate system is said to be standard po sition if its vertex is at the origin and the initial side is along the positive z axis Figure 1c If the terminal side of an angle in standard position lies along a coordinate axis7 the angle is said to be a quadrantal angle If the terminal side does not lie along a coordinate axis7 then the angle is often referred to in terms of the quadrant in which the terminal side lies Degree and Radian Measure There are two common systems to measure the size of an angles degree measure and radian measure De nition 1 Degree Measure An angle formed by one complete rotation is said to have a measure of 360 degrees 3600 An angle formed by 361370 of a complete rotation is said to have a measure of 1 degree 10 The symbol 0 denotes degrees see Figure 3 Certain angles have special namesFigure 4 Right angle right angle has measure 90quot Straight angle straight angle has measure 1800 Acute angle The measure of an acute angle is greater than 00 and less than 900 1 y 2400 8300 00 O 7750 m 0 m 0 m 28 FIGURE 2 Coterminal Angles 71200 3600 450 300 A FIGURE 3 Degree Measure Obtuse angle The measure of an obtuse angle is greater than 900 but less than 1800 We will use the Greek letters 04 B 39y and 9 to denote angles For simplicity we sometimes refer to an angle 9 having measure 450 as a 45 angle or an angle of 450 This may be expressed as 9 450 Two positive angles are complementary angles if their sum equals 900 and are supplementary angles is their sum is 1800 5 A degree can be divided further using decimal notation Fractions of a de gree may be measured using minutes and seconds Each degree is divided into 60 equal parts called minutes and each minute is divided into 60 equal parts call seconds Symbolically minutes are represented by and seconds by For example the measurement 25045 30 represents 25 degrees 45 minutes 30 second Express in decimal degrees this measurement is 0 7 0 0 24 45 30 7 24 60 3600 Example Convert 2104712 to decimal degrees B Convert 1051830 to degree minute second form 0 2475830 900 1800 600 1300 W Right Angle Straight Angle Acute Angle Obtuse Angle FIGURE 4 Types of Angles Complementary Angles Supplementary Angles FIGURE 5 Complementary angles and supplementary angles Solution A 21047 12 21 0 217870 B 1051830 1050 0183 x 60 1050 1092 1050 10 098 x 60 1050 10 59 105010 59 D Example Find angles that are complementary and supplementary to a 34019 42 4 Solution If angle 3 is complementary to a then B 90 7 04 90 7 34 19 42quot 89 59 60quot 7 34 19 42quot 90 89 59 60quot 55 40 18quot An supplementary angle to 04 is given by 39y 180 7 04 39y 180 7 34 19 42quot 179 59 60quot 7 34 19 42quot 180 179 59 60quot 145 40 18quot D Radian Measure A second unit of angle measure is radians radian measure is a common unit of measurement in many technical including calculus Angle 9 in Figure 6 has a measure of one radian The vertex of 9 is located at the center of the circle and its initial and terminal sides intercept an arc Whose length is equal to the radius of the circle FIGURE 6 One radian Radian measure An angle that has its vertex at the center of a circle and intercepts an arc on the circle equal in length to the radius of the circle has a measure of one radian From geometry we know that the arc length 3 on a circle is proportional to the measure of the central angle 0 A central angle of 27139 radians corresponds to an arc length that equals the circumference C 27139 r Using proportions s 27139 0 271quot 5 which simpli es to s r0 Hence the radian measure of 9 can also be given by s 0 r And when 8 r then 051 r Note 3 and r must be measured in the same units Also note that 0 is being used in two ways as the name of an angle and as the measure of the angle The context determines the choice Thus when we write 0 sr we mean the radian measure of angle 0 is sr From Degrees to Radians and Vice Versa The circumference of a circle with radius r is C 27139 r Therefore one rotation contains 27139 radians and so 3600 is equivalent to 27139 radians Ra dian measure can be compared to degree measure using proportions Since 1800 is equivalent to 7139 radians it follows that radian measure 7 7139 degree measure 7 180039 Solving for radian measure results in i 7139 radian measure degree measure x 1800 and solving for degree measure results in 1800 degree measure radian measure x RadianDegree Conversion Formulas 901i 7 9T Basic ro ortion 1800 arsed p p 0deg r m 0md Radians to degrees 7rra 0W1 1800 0deg Degrees to radians llegrees lool 300 450 600 900 1800 3600 lRadianlelglI 1 lwl27rl 4 3 3 TABLE 1 Equivalents measures in degrees and radians ExampleConvert the degree measures to radian measures and radian measures to degree measures a 750 5 radian 41012 1 H dewn 2 g 55g13 Emw m7rx57i6y Chapter 5 Trigonometric Function The trigonometric functions were originally de ned as the functions of angles and their applications were restricted to the indirect measurement of angles and distances And now the domain of trigonometric functions are not restricted to angles and applications involve sound light and electrical waves business cycles planetary motion etc In this chapter we will rst introduce trigonometric functions with real number domains then we de ne trigonometric functions with angle domains 51 The Wrapping Function De nition of the Wrapping Function Wrapping function denoted by Wm is a function that map real numbers to the points on the unit circle Unit circle means a circle of radius 1 with center at the origin of a rectangular coordinate system For example in the rectangular coordinate system 1412 the unit circle is given by the set of points 1412 that satisfy the equation FIGURE 1i Unit Circle The wrapping function is de ned as following we wrap a real num ber line with origin at 10 around the unit circlethe positive real axis is wrapped counterclockwise and the negative real axis is wrapped clockwise In this wayeach real number on the real line is paired with a unique point called a circular point on the unit circle see Figure of Section 51 Example Find the coordinate of following points A Wm B We 0 We D We E WW 1 2 Solution wlt0gt 10 We 071 mm 710 W37 0771 Wm 10 D An equivalent way of the de nition of wrapping function To nd the circular point P associated with the real number m we start at 1417 07 and move units along the unit circle7 counterclockwise if z is positive and clockwise if z is negative The length of arc AP is xlt0 0 FIGURE 2 The wrapping function and arc length Exact Values of Wrapping Functions for Particular Real Num bers We 42 We 7 mg 9 W7r4 can be obtained as following Assume that W7r4 01717 since 7r4 is onehalf the arc joining 17 0 and 07 17 the point 11 must lie on Jim Lambers IVIath 1B Fall Quarter 2004 05 Lecture 6 Notes These notes correspond to Section 45 in the text Exponential and Logarithmic Equations Now that we have discussed exponential functions and well their inverses logarithmic functions we are able to solve equations that involve either logarithms or exponents In this lecture we discuss some techniques for solving such equations Exponential Equations To solve an exponential equation for an unknown 1 where J occurs within exponents a useful approach is to rearrange the equation so that it has the form 9 am 1 where f 1 is some function of Then we can take the logarithm to some base I of both sides and use the properties of logarithms to obtain 1031 y 2 f J 10gb a Then we can try to solve this equation which is likely to be much simpler than the original equation Any positive number Z other than one can be used for the base typically the choice is dictated by convenience such setting I a so that log a log a 1 Example 1 Suppose that a sample of a radioactive substance decays from 100 g to 10 g in days We will compute the half life of the substance Recall that radioactive decay is modeled by the equation 1 an AAo 3 where AD is the initial amount of the substance A is the amount after time t has elapsed and Il is the halflife Substituting Ao 100 A 10 and t 5 we have 10 1002 4 where we have used the fact that 1 2 1 1 2 2 Rearranrving yields the equation O O y 10 1 2 5 z L41 V Since the base on the left side is 10 we take the common logarithm of both sides and use the properties of logarithms to obtain log 10 log2 Rearranging we obtain It 5log2 a 1505 7 since log 10 1 We conclude that the halflife of the substance is approximately 15 Another approach is useful in cases where the equation can be written in such a way that it contains expression of the form bf by by etc with no other occurrences of J in the equation Then we can make a substitution u bf 1 which yields I 1 uz I 1 13 and so on in View of the property a 0 9 The result is a polynomial in equation in it which we then solve for u The final step consists of solving the equation it f 1 for Example 2 We will solve the equation a 6 2 8 for If we multiply through by we obtain 621 1 26 9 Now we use the substitution u which yields the quadratic equation u 2a 1 0 10 The left side of this equation factors into u 12 which yields u 1 To obtain 1 we solve the equation it a for Since it 1 we have a 1 Taking the natural logarithm of both sides we obtain lnczx ln1 11 By the properties of logarithms this simplifies to Jln 0 12 and since lncz 1 we conclude that J 0 El Logarithmic Equations Logarithmic equations in which the unknown J is contained within a logarithm can be solved using techniques similar to those used to solve exponential equations If the equation can be written in the form logt f0 y 13 then we can apply the exponential function with base I to both sides Because I is the inverse function of log 1 we obtain f0 by 14 which we can hopefully solve to obtain Example 3 We will compute the sound intensi v in watts per square meter that has a decibel level of 60 Recall that the decibel level D is defined by I D 10log 7 1 3 10 where I is the sound intensity and 10 10 12 watts per square meter Substituting D 60 and 10 10 we obtain I 60 10logm 16 Simplifying and using the properties of logarithms yields 6 log log10 12 log 12 log 10 log 12 17 Rearranging we obtain an equation in logarithmic form log 6 18 To solve this equation for I w can convert to exponential form applying the exponential function of base 10 to both sides This yields the solution I 10 watts per square meter El Change of Base Although a logarithmic function can have any positive number except one a base it is not neces ry to use different methods to compute logarithms of different bases It is possible to express logb y in terms of logarithms of any other base We will show how logb y can be written in terms of natural logarithms It will follow that if we can compute natural logarithms we can easily compute logarithms of any base

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