AP Calculus AB Summer Assignment

Name: AP Calculus AB Summer Assignment Due Date: The beginning of class on the last class day of the first week of school. The purpose of this assignment is to have you practice the mathematical skills necessary to be successful in Calculus AB. All of the skills covered in this packet are skills from Algebra and Pre- Calculus. If you need to, you may use reference materials to assist you and refresh your memory (old notes, tetbooks, online resources, etc.). While the graphing calculators will be used in class, there are no calculators allowed on this packet. You should be able to do everything without a calculator. AP Calculus AB is a fast paced course that is taught at the college level. There is a lot of material in the curriculum that must be covered before the AP eam in May. Therefore, we cannot spend a lot of class time re- teaching prerequisite skills. This is why you have this packet. Spend some time with it and make sure you are clear on everything covered in the packet so that you will be successful in Calculus. Of course, you are always encouraged to seek help from your teacher if necessary. This assignment will be collected and graded as your first test, the last class day of the first week of school. Be sure to show all appropriate work to support your answers. In addition, there may be a quiz on this material during the first quarter. All questions must be complete with the correct work. You must return in September knowing how to do all the material in this packet. For assistant with the packet you may contact me at gnaem@roselleschools.org. Emails may take a few days for a response. Please be specific in your email what you need assistance with, include the section and the question number as well. Sample online websites are giving at the end of the packet. Good Luck ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 1

I. Lines The slope intercept form of a line is y m + b where m is the slope and b is the y- intercept. The point slope form of a line is y y m( 1) ( 1, where m is the slope and 1 y1) is a point on the line. You should be very comfortable using the point slope form of the line. Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes. Eample 1: Find the equations of (a) line parallel and (b) perpendicular to 1 y + 5 that contains the point (-,1) Solution: Write an equation of a line through the point (a) parallel to the given line and (b) perpendicular to the given line: 1. Point: (1,5) line: 6- y8. Point: (-,) line: +5y8 Part a (using slope from eample above) 1 1 ( ) + b Using the slope- intercept form with 1 m and point (-,1) 1 + b Multiply - 1and - 1 b Subtract from both sides Find the slope and y- intercept of the line:. - y 1 b Get a common denominator of 1 b Combine like terms 1 1 y + This is the equation of the line parallel to the given line that contains (-,1) 4. 8. 4+y1 Part b (using slope from eample above) 1 ( ) + b Using the slope- intercept form with m and point (-,1) 5. y 1 6 + b Multiply and - 1 6 b Subtract - 6 from both sides 7 b Subtract y + 7 This is the equation of the line perpendicular to the given line that contains (-,1) 6. - 51 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page

Eample : Find the slope and y- intercept of 6 5y 15 Find the equation of a line in slope intercept form: 7. Contains (1,) and (-,4) Solution: First you must get the line in slope- intercept form. 5y 15 6 Subtract 6 form both sides 15 6 y Divide by - 5 5 6 y Simplify 5 8. Contains (,1) and m4 The slope is m 5 6 and the y- intercept is - Eample : Find the equation of the line that passes through the point (1,- ) and has slope m -. Solution: Since we are given a point and slope it is easier to use the point slope form of a line. 9. Contains (1,7) and m0 y ( 1) Substitute into point slope form for ) and m ( 1, y1 y + + 1 Minus a negative makes + and distribute - y 1 Subtract from both sides 10. Contains (6,5) and m is undefined ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page

Eample 4: Find the equation of the line that passes through (- 1,) and (4,5). Write the following equations in point slope form 11. contains (- 1,4) and (,8) Solution: You will need to find slope using m 5 4 1 5 m y y1 1 Choose one point to substitute back into either the point slope or slope- intercept form of a line. 5 (4) + b Using the slope- intercept form with 5 and point (4,5) m 5 8 5 + b Multiply and 4 5 1. Passes through (6,) and had a y- intercept of 5 8 8 5 b Subtract from both sides 5 5 5 8 b 5 5 Get a common denominator of 5 17 b 5 Combine like terms 17 y 5 5 Equation of the line in slope intercepts form. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 4

Eample 5: Graph the following equation: Solution: y + Sketch a graph of the equation: 1. y- + First you must get the equation in slope- intercept form: y 1 y 1 The slope 1/ and the y- intercept - 1 14. 1. 4 y Plot the point (0,- 1). From the first point go up 1 and over to the right to get a second point 15. y- 1+1 Now connect the two points to get the line. 16. 7 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 5

II. Intercepts Find the intercepts for each of the following. The - intercept is where the graph crosses the - ais. You can find the - intercept by setting y0. 1. y + The y- intercept is where the graph crosses the y- ais. You can find the y- intercept by setting 0. Eample: Find the intercepts for y ( + ) 4 Solution:. y + X- intercept 0 ( + ) 4 Set y0 4 + ( ) Add 4 to both sides ± ( + ) Take square root of both sides ( + ) or ( + ) Write as equations 5 or 1 Subtract from both sides. + y ( + 1) Y- intercept y (0 + ) 4 Set 0 y 4 Add 0+ y 9 4 Square y 5 Add 4 to both sides 4. y 4 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 6

III. System of Equations Use substitution or elimination method to solve the system of equations. Find the point(s) of intersection of the graphs for the given equations. Eample: + y 16 + 9 0 y 9 0 1. + y 8 4 y 7 Elimination Method 16 + 0 0 8 +15 0 ( )( 5) 0 and 5 Plug and 5 into one original y 9 0 5 y 9 0 y 0 16 y y 0 y ±4 Points of Intersection (5, 4), (5,4) and (, 0). + y 6 + y 4 Substitution Method Solve one equation for one variable. y +16 9 (1st equation solved for y) ( +16 9) 9 0 Plug what y is equal to into second equation. 16 + 0 0 (The rest is the same as 8 +15 0 previous eample) ( )( 5) 0 or 5. 4y 0 64y 17 0 16 + 4y 0 + 64y + 1600 0 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 7

IV. Functions Let f()+ and g()1+ find each of the following: To evaluate a function for a given value, simply plug the value into the function for. ( f g)() f (g()) OR f [g()] read f of g of Means to plug the inside function (in this case g() ) in for in the outside function (in this case, f()). Eample 1: Given f () + 1 and g() 4 find f(g()). 1. f(g(0)). g(g()). g(f()) Solution: f (g()) f ( 4) ( 4) + 1 ( 8 + 16) + 1 16 + + 1 f (g()) 16 + Eample : Given: f()+5 and g()- 1 Find: f(g()), g(f()) and f(g()) 4. f(g()) If, 1. "# h, find the following: 5. f(h(- 1)) Solution: To find f(g()) we must first find g(): g()()- 1 4-1 For f() 8, find Since g() we can find f(g())f()()+59+514 To find g(f()) we must first find f(): f()()+56+511 Since f()11 we can find g(f())g(11)(11)- 1-11 6. () To find f(g()) we must put the function g() into f() equation in place of each. For f(), find 7. () f(g())f(- 1)(- 1)+56- +56+ ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 8

The domain of a function is the set of values for which the function is defined. The range of a function is the set of y values that a function can return. In Calculus we usually write domains and ranges in interval notation. Find the domain and range for each function give your answer using interval notation: 8. h ( ) 9 If the domain were - 1 < 7 then in interval notation the domain would be (- 1,7]. 9. h() sin Eample : Find the domain and range for 10. f( ) + Solution: Since we can only take the square root of positive numbers - 0 which means that. So we would say the domain is [, ). Note that we have used a [ to indicate that is included. If was not to be included we would have used (, ). The smallest y value that the function can return is 0 so the range is (0, ). 11. + 1, < 0 f( ) +, 0 1. 1. 14. " ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 9

V. Symmetry: - ais substitute in y for y into the equation. If this yields an equivalent equation then the graph has - ais symmetry. If this is the case, this is not a function as it would fail the vertical line test. Test for symmetry with respect to each ais and the origin. 1. + y- ais substitute in for into the equation. If this yields an equivalent equation then the graph has y- ais symmetry. A function that has y- ais symmetry is called an even function. Origin substitute in for into the equation and substitute in y for y into the equation. If this yields an equivalent equation then the graph has origin symmetry. If a function has origin symmetry, it is called an odd function. In order for a graph to represent a function it must be true that for every value in the domain there is eactly one y value. To test to see if an equation is a function we can graph it and then do the vertical lines test. Eample 1: Is y a function? The graph is below:. 6 ) Solution: When a vertical line is drawn it will cross the graph more than one time so it is NOT a function. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 10

Eample : Test for symmetry with respect to each ais and the origin. Given equation: y 4 0 Test for symmetry with respect to each ais and the origin.. Solution: - ais (change all y to y): ( y) 4 0 y 4 0 since there is no way to make this look like the original it is NOT symmetric to the - ais y- ais: (change all to ) y 4 ( ) 0 y 4 0 since there is no way to make this look like the original it is NOT symmetric to the y- ais 4. origin: (change all to and change all y to y ) ( y) 4 ( ) 0 y 4 0 since this does look like the original it is symmetric to the origin. Eample : The figure to the right shows the graph of y 4 0. It is symmetric only to the origin. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 11

Eample 4: Determine algebraically whether f() + 4 is even, odd, or neither. Solution: Show work to determine if the relation is even, odd, or neither: 5. 7 If I graph this, I will see that this is "symmetric about the y-ais"; in other words, whatever the graph is doing on one side of the y-ais is mirrored on the other side: This mirroring about the ais is a hallmark of even functions. But the question asks me to make the determination algebraically, which means that I need to do it with algebra, not with graphs. So I'll plug in for, and simplify: 6. 4 f( ) ( ) + 4 ( ) + 4 + 4 My final epression is the same thing I'd started with, which means that f() is even. 7. 4 4 + 4 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 1

VI. Asymptotes and Holes Given a rational function if a number causes the denominator and the numerator to be 0 then both the numerator and denominator can be factored and the common zero can be cancelled out. This means there is a hole in the function at this point. For each function below list all holes, vertical asymptotes and - intercepts 1. ( )( + ) ( )( + 1) Eample 1: Find the holes in the following function Solution: When is substituted into the function the denominator and numerator both are 0. Factoring and canceling ( + 1)( ) 1 but ( ) this restriction is from the original ( + 1) function before canceling. The graph of the function f() will look identical to 1 f( ) ecept for the hole at. ( + 1). 1 y + 1 1 f( ) ( + 1) note the hole at Given a rational function if a number causes the denominator to be 0 but not the numerator to be 0 then there is a vertical asymptote at that value. Eample : Find the vertical asymptotes for the function Solution: When - 1 is substituted into f() then the numerator is - 1 and the denominator is 0 therefore there is an asymptote at 1. See the graphs above. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 1

Given a rational function if a number causes the numerator to be 0 but not the denominator to be 0 then the value is an - intercept for the rational function.. 1 + 8 Eample : Discuss the zeroes in the numerator and denominator + Solution: When - is substituted into the function the numerator is 0 and the denominator is - 6 so the value of the function is f(- )0 and the graph crosses the - ais at -. Also note that for 0 the numerator is and the denominator is 0 so there is a vertical asymptote at 0. The graph is to the right. Eample 4: Find the holes, vertical asymptotes and - intercepts for the given function: f ) + 6 ( 4. 9 + 14 g ( ) + + Solution: First we must factor to find all the zeroes for both the numerator and denominator: ( ) ( + ) Numerator has zeroes 0 and Denominator has zeroes 0 and -. 0 is a hole - is a vertical asymptote is a - intercept ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 14

VII. Inverses To find the inverse of a function, simply switch the and the y and solve for the new y value. Find the inverse for each function. 1. f () + 1 Eample 1: f () + 1 y + 1 Rewrite f() as y Switch and y y + 1 Solve for your new y ( ) Cube both sides ( ) y + 1 y + 1 y 1 f 1 () 1 Simplify Solve for y Rewrite in inverse notation. f () Also, recall that to PROVE one function is an inverse of another function, you need to show that: Prove f and g are inverses of each other. f (g()) g( f ()). f () g() Eample : If: f () 9 and g() 4 + 9 show f() and g() are 4 inverses of each other. ( ) 9 " f (g()) 4 9 % $ '+ 9 g( f ()) 4 + 9 # 4 & 4 9 + 9 4 + 9 9 4 4 4 f (g()) g( f ()) therefore they are inverses of each other. 4. f () 9, " 0 g() 9 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 15

VIII. Finding Solutions A. Solve each equation: A. Factoring or using the quadratic formula to solve equations. 1. 7 0. 4 5 1 B. Solving Inequalities by factoring, creating a number line, and checking regions. + 6 + 4 0 C. Solve by finding the common denominator. 4. + 0 5. + 1 6. + " B. Solve each inequality: 7. 16 > 0 8. + 6 16 > 0 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 16

9. 10 10. + 4 11. + 4 4 1. sin sin, 0 < C. Solve for : 1. 14. + 5 15. 1 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 17

IX. Absolute Value and Piecewise Functions A. In order to remove the absolute value sign from a function you must: 1. Find the zeroes of the epression inside of the absolute value. A. Write the following absolute value epressions as piecewise epressions (by remove the absolute value): 1. 4. Make sign chart of the epression inside the absolute value.. Rewrite the equation without the absolute value as a piecewise function. For each interval where the epression is positive we can write that interval by just dropping the absolute value. For each interval that is negative we must take the opposite sign. Eample 1: Rewrite the following equation without using absolute value.. 6 + + 1 + 4 Solution: +40 Find where the epression is 0-4 Subtract 4-4/ Divide by - Simplify _- - + Put in any value less than - into +4 and you get a negative. Put in any value more than - into +4 and you get a positive.. 4 + 1 + 4 + 4 < Write as a piecewise function. Be sure to change the signs of each term for any part of the graph that was negative on the sign chart. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 18

Eample : Rewrite the following equation without using absolute value. + 5 B. Solve the following absolute value inequalities: 4. > 1 Solution: +5-0 Find where the epression is 0 (- 1)(+)0 factor - 10 or +0 Set each factor equal to 0 X1/ or - Solve each equation + _- + Put in any value less than - into (- 1)(+) and you get a positive number. - 1/ Put in any value more than - and less than ½ into (- 1)(+) and you get a negative number. 5. 4 Put in any value more than 1/ into (- 1)(+) and you get a positive number. Write as a piecewise function. Be sure to change the signs of each term for any part of the graph that was negative on the sign chart. f( ) + 5 + 5 < and > 1/ 1/ 6. 10 + 8 > ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 19

Eample : Rewrite the following equation without using absolute value. 9 + 7. 4 > Solution: - 90 Find where the epression is 0 (For the part in the absolute value only.) 9 Add 9 Divide by _- + Put in any value less than into - 9 and you get a negative. Put in any value more than into - 9 and you get a positive. 8. 6 > 8 + 9 + 9 + < Write as a piecewise function. Be sure to change the signs of each term that is inside the absolute value for any part of the graph that was negative on the sign chart. + 11 7 < Simplify B. Absolute value inequalities require you to write two separate inequalities. You were probably taught to Keep Flip Change. One inequality will be identical to the inequality, just without the absolute value sign. The second inequality will have a flipped inequality sign and the opposite value. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 0

X. Eponents Write without fractional eponents: A fractional eponent means you are taking a root. For eample 1/ is the same as. 1.. 1/ y (16 4 ) 1/ Eample 1: Write without fractional eponent: / y. y 1/ / 4 7 Solution: y Notice that the root is the bottom number in the fraction and the power is the top number in the fraction. 4. 9 Negative eponents mean that you need to take the reciprocal. For eample. means 1 and means 5. 64 Eample : Write with positive eponents: y 4 5 Solution: y 4 5 6. 8 7. 7 Eample : Write with positive eponents and without 1/ fractional eponents: ( + 1) ( ) 1/ ( ) Solution: ( + 1) Write with positive eponents: 8. 9. y (4 ) 10. When factoring, always factor out the lowest eponent for each term. 11. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 1

Eample 4: y 1 + 6 Factor then simplify: Solution: The lowest eponent for is - so can be factored from each term. Leaving y (1 + 11). Notice that for the eponent for the 6 term we take 1- (- ) and get. For the new eponent. 1 term we take - 1- (- ) and get 1 as our 1. 4 + 18 When dividing two terms with the same base, we subtract the eponents (numerator eponent- denominator eponent). If the difference is negative then the term goes in the denominator. If the difference is positive then the term goes in the numerator. Eample 5: Simplify () 8 1/ 1. 5 ( ) 1/ + ( ) Solution: First you must distribute the eponent. 8. Then since we have two terms with as the 8 base we can subtract the eponents. Since - 8 results in - 5 we know that we will have 5 in the denominator. 8. 5 Eample 6: Simplify + 1 ( 1) 1 14. 6( 1) 4( 1) Solution: First we must factor both the numerator and denominator. ( 1). Then we can see that ( + 1)( 1) we have the term (- 1) in both the numerator and denominator. Subtracting eponents we get - 11 so the term will go in the numerator with 1 as it s eponent. ( 1). ( + 1) ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page

Eample 7: Factor and simplify 1/ 4( ) + ( ) 1/ Simplify rational epressions: (4 ) 15. Solution: The common terms are and (- ). The lowest eponent for is 1. The lowest eponent for (- ) is - 1/. So factor out. 1/ ( ) and obtain ( ) 1./ [4( ) + ] ( ) 1/ [4 1 + ] (5 1).. This will simplify to. Leaving a final solution of 16. ( + 1)( ) y 4 ( ) ( 1) 17. f 16 ) 4 ( 8 + 1 + 1 18. y 5 10 + 5 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page

XI. Rational Epressions When simplifying comple fractions, multiply by a fraction equal to 1, which has a numerator and denominator composed of the common denominator of all the denominators in the comple fraction. 1.. + Eample 1: 7 6 +1 5 +1 7 6 +1 i +1 5 +1 +1 7 7 6 5 7 1 5. 4. 5. Eample : + 4 5 1 4 + 4 5 1 4 ( 4) + () 5()( 4) 1() i ( 4) ( 4) 6. "" " +8+ 5 0 +8 5 1 7. ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 4

Eample : Simplify, using factoring of binomial epressions. Leave answers in factored form. 8. ( +1) (4 9) (16 + 9)( +1) ( +1) ( +1)(4 9) (16 + 9) # " $ ( 6)( +1) ( 6)( +1) Eample 4: ( +1) (4 5 9 16 9) ( 6)( +1) ( +1) (4 1 18) ( 6)( +1) ( +1) (4 + )( 6) ( 6)( +1) ( +1)(4 + ) 9. 10. Simplify by rationalizing the numerator. + 4 + 4 + 4 + + 4 + + 4 4 ( + 4 + ) ( + 4 + ) 1 + 4 + 11. 1. + 9 1. + h h ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 5

XII. Natural Logarithms y ln and y e (eponential function) are Recall that ( ) inverse to each other Properties of the Natural Log: Epress as a single logarithm: 1. ln + ln 4 ln ( AB) ( A) + ( B) ln ln ln Eample 1: ln( ) + ln( 5) ln( 10) ln A ln ln B ( A) ( B) 6 Eample : ln ( 6) ln ( ) ln ln ( ). ln 1 ln p ( A ) pln( A) 4 Eample : ln( ) 4ln( ) ( ) ( ) ( ) ln ln ln 8 ln ( e ) Eample 4:, ln( ) 1 and e, ( ) ln 1 0, 0 e 1. 7 Use the properties of natural logs to solve for : 5 11 7 5 11 7 5 11 ln ln 7 ( ) ( ) ( ) ( ) ln 5 ln 7 ln 11 ln ( ) ( ) ( ) ( ) ( ) ( ) ln ( 11) ln ( ) ln ( 5) ln ( 7) ln 5 ln 7 ln 11 ln ( ln 5 ln 7 ) ln ( 11) ln ( ) ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 6

XIII. Graphing Trig Functions Graph two complete periods of the function. f( ) sin ( ) f( ) cos ( ) 1. f () 5sin -5 5-5 5 - - y sin and y cos have a period of and an amplitude of 1. Use the parent graphs above to help you sketch a graph of the functions below. For f () Asin(B + C) + K, A amplitude, B period, C Phase Shift (positive C/B shift left, negative C/B shift B right) and K vertical shift.. f () sin. # f () cos " & $ % 4 ' ( 4. f () cos ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 7

XIV. Trig. Equations and Special Values 1. You are epected to know the special values for trigonometric functions. Fill in the table to the right and study it. Use 180 radians to get rid of radians and convert to degrees. Use radians 180 to get rid of degrees and convert to radians. You can determine the sine or cosine of a quadrantal angle by using the unit circle. The - coordinate of the circle is the cosine and the y- coordinate is the sine of the angle. (-1,0) (0,1) - (0,-1) (1,0) Degrees 0 0 45 60 90 10 15 150 180 10 5 40 70 00 15 0 60 Radian s) Cos Sin Quadrant Eample 1: sin90 1 cos 0 You should study the following trig identities and memorize them before school starts: Reciprocal identities sin 1 csc cos 1 sec tan 1 cot Find all solutions to the equations. You should not need a calculator. (Hint: one of these has NO solution.). 4cos 4cos 1 csc 1 sin sec 1 cos cot 1 tan Tangent Identities tan sin cos cos cot sin Pythagorean Identities sin + cos 1 tan + 1 sec cot + 1 csc. sin + sin + 1 0 Double angle Identities sin sin cos cos cos sin cos 1 1 sin ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 8

Reduction Identities 4. sin sin 0 sin( ) sin cos( ) cos tan( ) tan We use these special values and identities to solve equations involving trig functions. 5. sin cos Solve each of the equations for 0 < ". Isolate the variable, sketch a reference triangle, find all the solutions within the given domain, 0 < ". Remember to double the domain when solving for a double angle. Use trig identities, if needed, to rewrite the trig functions. 6. sin + sin Eample : Find all solutions to sin + sin 1 Solution: sin + sin 1 Original Problem 7. cos + 5cos (hint: use double angle identity) sin + sin 1 0 Get one side equal to 0. ( sin 1)(sin + 1) 0 Factor ( sin 1) 0and (sin + 1) 0 Set each factor equal to 0 8. sin(cos ) 1 1 sin and sin 1 Get the trig function by itself π + πk 6 and 5π + πk 6 π + πk Solve for (these are special values) 9. sin sin 0 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 9

XV. Inverse Trigonometric Functions For each of the following, epress the value for y in radians. Inverse Trig Functions can be written in one of ways: arcsin ( ) sin 1 ( ) 1. y arcsin Inverse trig functions are defined only in the quadrants as indicated below due to their restricted domains. sin - 1 >0 cos - 1 < 0 cos - 1 >0 tan - 1 >0 sin - 1 <0 tan - 1 <0. y arccos (1) Eample 1: Epress the value of y in radians. y arctan 1 Solution: Draw a reference triangle. - 1 This means the reference angle is 0 or 6. So, y 6 so. y arctan(1) that it falls in the interval from " < y < " Answer: y 6 ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 0

Eample : Find the value without a calculator. cos arctan 5 $ " # 6% & Solution: For each of the following give the value without a calculator. 4. tan arccos $ " # % & Draw the reference triangle in the correct quadrant first. Find the missing side using Pythagorean Thm. 5 Find the ratio of the cosine of the reference triangle. 6 cos 6 61 " 1% 5. sec sin 1 # $ 1& ' 6. sin arctan 1 $ " # 5 % & " 7% 7. sin sin 1 # $ 8& ' ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page 1

WEBSITES: Using the TI- 8/84 Graphing Calculator [highly recommended site] http://www.prenhall.com/divisions/esm/app/graphing/ti8/ Just Math Tutoring (click on Algebra/SAT videos on the left and scroll down to Solving equations and inequalities ) http://www.justmathtutoring.com/ Purple Math http://www.purplemath.com/modules/compfrac.htm Solving by the Substitution method: http://www.purplemath.com/modules/systlin4.htm Solving by the Elimination method: http://www.purplemath.com/modules/systlin5.htm LAWS OF EXPONENTS http://www.mathsisfun.com/algebra/eponent- laws.html http://www.algebralab.org/practice/practice.asp SLOPE OF A LINE http://www.ck1.org/algebra/slope- of- a- Line- Using- Two- Points/ WRITE THE EQUATION OF A LINE IN SLOPE- INTERCEPT FORM GIVEN TWO POINTS http://www.ck1.org/algebra/standard- Form- of- Linear- Equations/ GRAPH THE EQUATION OF A LINE http://www.ck1.org/algebra/graphs- Using- Slope- Intercept- Form/ OPERATIONS WITH FRACTIONS http://www.aaamath.com/fra.html http://www.ck1.org/arithmetic/equivalent- Fractions/ http://www.ck1.org/arithmetic/fractions- in- Simplest- Form/ http://www.ck1.org/arithmetic/sums- of- Fractions- with- Like- Denominators/ http://www.ck1.org/arithmetic/differences- of- Fractions- with- Different- Denominators/ http://www.ck1.org/arithmetic/products- of- Two- Fractions/ http://www.ck1.org/arithmetic/quotients- of- Fractions/ Unit Circle and Trig Equations http://tutorial.math.lamar.edu/classes/calci/trigequations.asp http://www.analyzemath.com/trigeqeplore/trigeqeplore.html ACHS AP CALCULUS SUMMER PACKET MR. NAEM 01-014 Page