Calculus Summer TUTORIAL
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- Lenard Nichols
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1 Calculus Summer TUTORIAL The purpose of this tutorial is to have you practice the mathematical skills necessary to be successful in Calculus. All of the skills covered in this tutorial are from Pre-Calculus, Algebra, and Algebra. The material covered is from our district approved, Pre-calculus, tetbook. If you need to, you may use reference materials to refresh your memory (old notes, tetbooks, online resources, etc.). While graphing calculators will be used during a few tests and quizzes, the majority of in class assessments are non-calculator. You are encouraged to learn how to be calculator-independent. At the end of this page, there are links to some suggested online calculators. Calculus is a fast-paced course that is taught at the college level to prepare you for AP Calculus. There is a lot of material in the curriculum that must be covered before the end of the year. Therefore, we cannot spend a lot of class time re-teaching prerequisite skills. This is why you have this tutorial. Spend some time with it and make sure you are clear on everything covered in this tutorial so that you will be successful in Calculus. Of course, you are always welcomed to seek help from your teacher if necessary. For assistance with the assignment part, you may contact me at gnaem@roselleschools.org. s may take a few days during summer for a response. Please be specific in your for what you need assistance with, include the section and the question number as well. For each question in the assignment, there is an eample in this tutorial. The summer assignment packet is posted online on our school s website. Reference to the corresponding section is listed under each question. Calculators Links Online Calculator Emulator for Download Good Luck! CALCULUS SUMMER TUTORIAL JUNE, 08 Page
2 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( - ) where m is the slope and (, y ) 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: Find the equations of (a) line parallel and (b) perpendicular to y = that contains the point (,). 3 Part a = - - (-) + b Using the slope-intercept form with m = and point (,). 3 3 = 3 + b Multiply and. - 3 = b Subtract 3 from both sides = b Get a common denominator of 3. = b Combine like terms. 3 y = Part b This is the equation of the line parallel to the given line that contains (,) = 3(-) + b Using the slope-intercept form with m = 3 = -6 + b Multiply 3 and = b Subtract 6 from both sides. 7 = b Subtract. y = Eample : and point (,). This is the equation of the line perpendicular to the given line that contains (,). Find the slope and y-intercept of 6-5y =5. First you must get the line in slope-intercept form. -5y = 5-6 Subtract 6 form both sides. y = y = Divide by 5. Simplify. The slope is m= 6 and the y-intercept is 3 5 CALCULUS SUMMER TUTORIAL JUNE, 08 Page
3 Eample 3: Find the equation of the line that passes through the point (, ) and has slope m= 3. Since we are given a point and slope it is easier to use the point slope form of a line. y - - = -3( -) Substitute into point slope form for (, y ) and m. y + = -3 + Minus a negative makes + and distribute 3. y = -3 - Subtract from both sides. Eample 4: Find the equation of the line that passes through (,3) and (4,5). m = = 5 You will need to find slope using m = y - y -. 5 = (4) + b Choose one point to substitute back into either the point slope or slope-intercept 5 form of a line. Using the slope-intercept form with m = 5 and point (4,5). 5 = 8 + b Multiply and = b Subtract 8 5 from both sides = b 5 Get a common denominator of 5. 7 = b 5 Combine like terms. y = Equation of the line in slope intercepts form. Eample 5: Graph the following equation: 3y + 3= 3y = - 3 First you must get the equation in slope-intercept form by isolating y. y = 3 - The slope = 3 and the y-intercept = Plot the point (0, ). From the first point go up and over 3 to the right to get a second point. Now connect the two points to get the line. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 3
4 II. Intercepts The -intercept is where the graph crosses the -ais. You can find the -intercept by setting y = 0. 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 = ( + 3) - 4 X-intercept 0 = ( + 3) - 4 Set y = 0 4 = ( + 3) Add 4 to both sides ± = ( + 3) Take square root of both sides - = ( + 3) or = ( + 3) Write as equations -= Subtract 3 from both sides -5 = Y-intercept y = (0 + 3) - 4 Set = 0 y = 3-4 Add 0+3 y = 9-4 Square 3 y = 5 Add 4 to both sides CALCULUS SUMMER TUTORIAL JUNE, 08 Page 4
5 III. System of Equations Eample: Use substitution or elimination method to solve the system of equations. ìï + y = 0 í îï - y - 9 = 0 Elimination Method: = 0 Add the two equations and combine like terms = 0 Divide by. ( - 3)( - 5) = 0 Factor the trinomial. = 3 and = 5 Use zero product property to find the values of. 3 - y - 9 = y - 9 = 0 Plug =3 and = 5 into oneof theoriginal equations.. -y = 0 y = 0 6 = y y = ±4 Solve Points of Intersection (5,4), (5,-4) and (3,0) Substitution Method: y = Solve one equation for one variable. (st equation solved for y) - ( ) - 9 = = 0 Plug what y is equal to into second equation. Combine like terms = 0 Divide by. ( - 3)( - 5) = 0 Factor the trinomial = 3 and = 5 Use zero product property to find the values of. 3 - y - 9 = y - 9 = 0 Plug =3 and = 5 into oneof theoriginal equations.. -y = 0 6 = y Solve y = 0 y = ±4 Points of Intersection (5,4), (5,-4) and (3,0) CALCULUS SUMMER TUTORIAL JUNE, 08 Page 5
6 IV. Functions To evaluate a function for a given value, simply plug the value into the function for. Reads 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: Given f () = + and g() = - 4 find f(g()). f (g()) = f ( - 4) = ( - 4) + = ( ) + = f (g()) = Eample : Given: f() = and g() =. Find: f(g()), g(f()) and f(g()) To find f(g()) we must first find g(): g() = () = 4 = 3 Since g() = 3 we can find f(g()) = f(3) = 3(3) + 5 = = 4 To find g(f()) we must first find f(): f() = 3() + 5 = = Since f() = we can find g(f()) = () = () = = To find f(g()) we must put the function g() into f() equation in place of each. f(g()) = f( ) = 3( ) + 5 = = 6 + 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. If the domain were - < 7 then in interval notation the domain would be (-,7]. Eample 3: Find the domain and range for f () = - 3 Since we can only take the square root of positive numbers 3 0 which means that 3. So we would say the domain is [3, ). Note that we have used a [ to indicate that 3 is included. If 3 was not to be included we would have used (3, ). The smallest y value that the function can return is 0 so the range is (0, ). CALCULUS SUMMER TUTORIAL JUNE, 08 Page 6
7 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. 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 : Is - y = a function? The graph is below: When a vertical line is drawn it will cross the graph more than one time so it is NOT a function. Eample : Test for symmetry with respect to each ais and the origin. Given equation: y = 0 -ais (change all y to y): (- y) = 0 -y = 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 (-) = 0 -y = 0 Since there is no way to make this look like the original it is NOT symmetric to the y-ais. Origin: (change all to and change all y to y ) -(- y) (-) = 0 y = 0 Since this does look like the original it is symmetric to the origin. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 7
8 Eample 3: The figure to the right shows the graph of y = 0. It is symmetric only to the origin. Eample 4: Determine algebraically whether f() = is even, odd, or neither. 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 true for 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 as follows: f( ) = 3( ) + 4 Plug in for. = 3( ) + 4 Square. = Multiply by 3. The final epression is the same thing I'd started with, which means that f() is even. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 8
9 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. Eample: Find the holes in the following function f () = - f () = ( +)( - ) f () = ( +) Factoring and canceling. When = is substituted into the function the denominator and numerator both are 0. But ( ) this restriction is from the original function before canceling. The graph of the function f() will look identical to f () = ecept for the hole at =. ( +) f () = f () = ( +) 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 f () = - -. When = is substituted into f() then the numerator is and the denominator is 0 therefore there is an asymptote at =. See the graphs above. 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. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 9
10 Eample 3: Discuss the zeroes in the numerator and denominator f () = + 3 See the graph is to the right. When = 3 is substituted into the function the numerator is 0 and the denominator is 6 so the value of the function is f( 3)=0 and the graph crosses the -ais at = 3. Also note that for = 0 the numerator is 3 and the denominator is 0 so there is a vertical asymptote at = 0. Eample 4: Find the holes, vertical asymptotes and -intercepts for the function f () = ( - 3) f () = 3( + ) = 0 and = 3 = 0 and = = 0 = = 3 Factor to find all the zeroes for both the numerator and denominator. Numerator has zeroes. Denominator has zeroes. Is a hole. Is a vertical asymptote. Is a -intercept. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 0
11 VII. Inverses To find the inverse of a function, simply switch the and the y and solve for the new y value. To PROVE one function is an inverse of another function, you need to show that: f (g()) = g( f ()) = Eample : 3 Find the inverse function for f () = + 3 f () = + 3 y = + 3 = y + Rewrite f() as y. Switch and y. Solve for the new y. ( ) 3 Cube both sides. ( ) 3 3 = y + 3 = y + Simplify. y = 3 - Solve for y. f - () = 3 - Rewrite in inverse notation. Eample : If: f () = and g() = show f() and g() are inverses of each other. æ f (g()) = 4-9 ö ç è + 9 g( f ()) = ( 4 + 9) ø 4 = = = = 4 4 f (g()) = g( f ()) = Therefore they are inverses of each other. CALCULUS SUMMER TUTORIAL JUNE, 08 Page
12 VIII. Finding Solutions Factoring or using the quadratic formula to solve equations of the form a + b + c = 0. = -b ± b - 4ac a Solving Inequalities by factoring, creating a number line, and checking regions. Simplify Rational epressions by finding the common denominator. Eample : Solve - = 0. - = 0 Add 7 to both sides to get all terms to one side of equation = = 0 Determine a, b and c. a =, b =, and c = 0. = - - ( ) ± (-) - 4( ) (-0) ( ) ( ) = 4 ± = 4 ± = 4 ± 44 = 4 ± 44 = ± = 4 ± = 4 ± Use quadratic formula. Simplify your answer. = 4 ± = ± These are the Solutions. Eample : Solving Inequalities y The two -intercepts are =.69 and =.9. To find out if your Solutions are in between or outside, we need a test point. Choose a point in between. If the inequality is true for your test point, the solution is all the numbers in between. If the inequality is false for your test point, the solution is all the numbers outside. (0) ²+(0) 4 ³0 Let s pick = 0. 4 ³0 False! So the solutions are outside or ³.9 Solutions. CALCULUS SUMMER TUTORIAL JUNE, 08 Page
13 Eample 3: Simplify: LCD: 5 since 3 * 5 = 5. Multiply each fraction by missing part of denominator over itself to get a common denominator. Multiply numerators and denominators of each fraction. Add numerators and keep the same denominator. Eample 4: Simplify: ( + 3) æ ö ( + 3) ç è - 3 ø LCD: ( 3)(+3) Factored form of 9. Multiply each fraction by missing part of denominator over itself (only need to do this on right fraction) to get a common denominator. Multiply numerators and denominators of each fraction as needed. Add numerators and keep the same denominator. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 3
14 IX. Absolute Value and Piecewise Functions In order to remove the absolute value sign from a function you must: Find the zeroes of the epression inside of the absolute value. 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: Rewrite the following equation without using absolute value. f () = = 0 = -4 = -4 Find where the epression is 0. Subtract 4. Divide by. = - Simplify. _- - + ì- - 4, < - f () = íî + 4, ³ - Plug in any value less than into + 4, you will get a negative value. Plug in any value more than into + 4, you will get a positive value. Write as a piecewise function. Be sure to change signs of each term for any part of the graph that was negative on the sign chart. Eample : Rewrite the following equation without using absolute value. f () = = 0 Find where the epression is 0. ( -)( + 3) = 0 Factor. -= 0 or + 3= 0 Set each factor equal to 0. = or = -3 Solve each equation _- + / Plug in any value less than 3 into ( -)( + 3), you will get a positive value. Plug in any value more than 3 and less than into ( -)( + 3), you will get a negative number. Plug in any value more than number. into ( -)( + 3), you will get a positive ì + 5-3, (-,-3)U( ï f () =, ) í Write as a piecewise function. Be sure to change signs of each term for any part of the graph that was negative on the sign chart. ï , - 3 ï î CALCULUS SUMMER TUTORIAL JUNE, 08 Page 4
15 Eample 3: Rewrite the following equation without using absolute value. f () = = 0 3 = 9 Add 9 = 3 Divide by 3 Find where the epression is 0 (For the part in the absolute value only.) _- + 3 ì , < 3 f () = í î3-9 +, ³ 3 Plug in any value less than 3 into 3-9, you will get a negative value. Plug in any value more than 3 into 3-9, you will get a positive value. Write as a piecewise function. Be sure to change signs of each term for any part of the graph that was negative on the sign chart. f () = -3 + ì íî 3-7 < 3 ³ 3ü ýþ Simplify 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. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 5
16 X. Eponents A fractional eponent means you are taking a root. For eample / is the same as. Negative eponents mean that you need to take the reciprocal. For eample - means / and / -3 means 3. When factoring, always factor out the lowest eponent for each term. When dividing two terms with the same base, we subtract the eponents (numerator eponentdenominator 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: Write without fractional eponent: y = /3 y = 3 Notice that the root is the bottom number in the fraction and the power is the top number in the fraction. Eample : Write with positive eponents: y = 5-4 y = 4 5 Eample 3: Write with positive eponents and without fractional eponents: f () = ( +)- ( - 3) f () = - 3 * - 3 ( +) Eample 4: Factor: y = ( - 3) - The lowest eponent for is so 3 - can be factored from each term. Leaving y = 3 - (+ 3 -) Notice that for the eponent for the 6 term we take ( ) and get 3. For the 33 - term we take ( ) and get as our new eponent. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 6
17 Eample 5: Simplify f () = ()3 8 First you must distribute the eponent. f () = Then since we have two terms with as the base we can subtract the eponents. Since 3-8 results in -5 we know that we will have 5 in the denominator. f () = 8 5. Eample 6: Simplify $$ f () = - + ( -) ( -) First we must factor both the numerator and denominator. f () =. Then we ( +)( -) can see that we have the term (-) in both the numerator and denominator. Subtracting eponents ( -) we get -= so the term will go in the numerator with as it s eponent. f () = ( +). Eample 7: Factor and simplify f () = 4( - 3) / + ( - 3) -/ The common terms are and (-3). The lowest eponent for is. The lowest eponent for (-3) is - /. So factor out ( - 3) -/ and obtain f () = ( - 3) -/ [4( - 3) + ]. This will simplify to f () = ( - 3) -/ [4 - + ]. Leaving a final solution of (5 -) - 3. CALCULUS SUMMER TUTORIAL JUNE, 08 Page 7
18 XI. Rational Epressions When simplifying comple fractions, multiply by a fraction equal to, which has a numerator and denominator composed of the common denominator of all the denominators in the comple fraction. Eample: = = = i + + Eample : = = = i -( - 4) + 3() 5()( - 4) - () = Eample 3: Simplify, using factoring of binomial epressions. Leave answers in factored form. é ù ë û ( +) 3 (4-9)- (6 +9)( +) ( -6)( +) = ( +) ( +)(4-9)-(6 +9) ( - 6)( +) = ( +) ( ) ( -6)( +) = ( +) (4 - -8) ( - 6)( +) = ( +) (4 +3)( -6) ( -6)( +) = ( +)(4 +3) ( - 4) ( - 4) Eample 4: Simplify by rationalizing the numerator = = = = ( ) ( ) CALCULUS SUMMER TUTORIAL JUNE, 08 Page 8
19 b b y = b + y log b + log b y = log b y b b y = b- y ( )y = b y b b = b y XII. Eponents and logarithms b = y Û log b y = æ ö log b - log b y = log b ç è y ø log b y = ylog b or if log b = log b y, then = y log b b = and b log b = log = log 0 and ln = log e log b a = log c a log c b CALCULUS SUMMER TUTORIAL JUNE, 08 Page 9
20 XIII. Natural Logarithms ( ) and y = e (eponential function) are inverse to each other properties of the Natural Log: ( ) = ln( A) + ln( B) Recall that y = ln ln AB Eample: ln( ) + ln( 5) = ln( 0) æ ln A ö ç è B ø = ln A ( ) - ln( B) Eample : ln( 6) - ln( ) = lnç 6 ln( A ) p = pln( A) Eample 3: ln 4 æ è ö ø = ln 3 ( ) ( ) = 4ln( ) and 3ln( ) = ln( ) 3 = ln 8 ( ) =, ln( e) =, ln( ) = 0, y = - f () ln e ( ) Eample 4: Use the properties of natural logs to solve for : 5 = 7 5 = 7 5 ln ln = 7 ( ) ( ) = ( ) ( ) ln 5 ln 7 ln ln ( ) ( ) = ( ) ( ) ( ) ( ) = ln ( ) ln ( ) ln ( 5) ln ( 7) ln 5 ln 7 ln ln ( ln 5 ln 7 ) ln ( ) ln ( ) = CALCULUS SUMMER TUTORIAL JUNE, 08 Page 0
21 XIV. Graphing Trig Functions Domain: (, ) Range: [-,] Period: ODD = + n 3 = + n = n f ( ) = sin( ) Domain: (, ) Range: [-,] Period: EVEN = n = + n = + n f ( ) = cos( ) Domain: + n Range: (, ) Period: ODD No maima or minima = n f ( ) = tan( ) Domain: n Range: (, ) Period: ODD No maima or minima = + n f ( ) = cot( ) CALCULUS SUMMER TUTORIAL JUNE, 08 Page
22 f ( ) = csc( ) Domain: n Range: (, ] [, ) Period: ODD = 3 + n = + n No -intercepts Domain: + n Range: (, ] [, ) Period: EVEN = + n = n No -intercepts f ( ) = sec( ) For f () = Asin(B + C) + K, A = amplitude, period = C B = Phase Shift (positive C/B shift left, negative C/B shift right) and K = vertical shift. p B XV. Transformations of graphs CALCULUS SUMMER TUTORIAL JUNE, 08 Page
23 y = f ( - a) is the graph of y = f () shifted horizontally a units (to the right if a > 0 and to the left if a < 0 ) y = f () + a is the graph of y = f () shifted vertically a units (up if a > 0 and down if a < 0 ) y = af () is the graph of y = f () stretched or shrunk vertically by a factor of a (stretched if a > and shrunk if 0 < a <) y = f (a) is the graph of y = f () stretched or shrunk horizontally by a factor of a (stretched if 0 < a < and shrunk if a >) y = - f () is the graph of y = f () reflected over the -ais y = f (-) is the graph of y = f () reflected over the y-ais Eample: Identify the basic shape and then list all transformations in their proper order ( ) = - ( +) 3-5 f Shape: cubic [from eponent, 3] Horizontal Shift: left [from + ] Vertical Stretch by a factor of [from coefficient, - ] -ais reflection [from negative on coefficient, -] Vertical Shift: 5 down [from - 5 on outside] XVI. Basic graphs CALCULUS SUMMER TUTORIAL JUNE, 08 Page 3
24 y = y = y = y n =, n even y n =, n odd y = a, a y = a, 0 a y = log, a y = log, 0 a a a XVII. Equation Of Circles CALCULUS SUMMER TUTORIAL JUNE, 08 Page 4
25 The general equation for a circle that has center (h, k) and radius r is: ( h) + (y k) = r. For eample, the circle pictured above has this equation: ( 5) + (y 4) = 9. For many circles, the h and/or k are negative; remember that a double negative is often written as a +. Putting Equations in Standard Form Eample: Find the standard form, center, and radius of the following circles: + y + 6 8y = 0 ( + 6) + (y 8y) = Combine like terms. ( ) + (y 8y + 6) = Complete the squares. ( + 3) + (y 4) = 36 Rewrite in standard. Center: (-3, 4) and Radius: 6 XVIII. Lines I. Special products and factoring Sum/difference of two cubes: ( ) ( ) ( ) ( ) a + b = a + b a ab + b 3 3 a b = a b a + ab + b 3 3 XIX. Geometry formulas CALCULUS SUMMER TUTORIAL JUNE, 08 Page 5
26 The following formulas are needed for a topic in AP calculus called related rate. Area: Triangle: bh Trapezoid: ( ) h b + b Circle: r Surface area of Sphere: 4 r Lateral area of cylinder: rh Volume: Cone: 3 rh Sphere: r Cylinder: rh Prism: Bh where B is the area of the base Pyramid: Bh where B is the area of the base.. 3 XX. Trig. Equations and Special Values You are epected to know the special values for trigonometric functions. 80 Use p radians Use p radians to get rid of radians and convert to degrees. to get rid of degrees and convert to radians. 80 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. Eample: sin90 = cos p = 0 You should study the following trig identities and memorize them before school starts: Reciprocal identities sin = csc csc = sin cos = sec sec = cos tan = cot cot = tan Tangent Identities sin tan = cos cos cot = sin CALCULUS SUMMER TUTORIAL JUNE, 08 Page 6
27 Pythagorean Identities sin + cos = tan = sec + cot + = csc Double angle Identities sin = sin cos cos = cos sin = cos = sin.. Reduction Identities sin( ) = sin cos( ) = cos tan( ) = tan We use these special values and identities to solve equations involving trig functions. Solve each of the equations for 0 < p. Isolate the variable, sketch a reference triangle, find all the Solutions within the given domain, 0 < p. Remember to double the domain when solving for a double angle. Use trig identities, if needed, to rewrite the trig functions. Eample : Find all Solutions to sin + sin = sin + sin = Original Problem sin + sin = 0 Get one side equal to 0. ( sin )(sin + ) = 0 Factor ( sin ) = 0 and (sin + ) = 0 Set each factor equal to 0 sin = and sin = Get the trig function by itself = p 6 + pk = 5p 6 + pk and 3 = + k Solve for. A - All trig functions positive in Quad I S - sin & csc functions positive in Quad II T tan & cot functions positive in Quad III C cos & sec function positive in Quad IV CALCULUS SUMMER TUTORIAL JUNE, 08 Page 7
28 CALCULUS SUMMER TUTORIAL JUNE, 08 Page 8
29 XXI. Inverse Trigonometric Functions Inverse Trig Functions can be written in either of these ways: arcsin ( ) sin - ( ) Inverse trig functions are defined only in the quadrants as indicated below due to their restricted domains. cos - tan - sin - cos - tan - sin - Eample : Epress the value of y in radians. y = arctan - 3 Draw a reference triangle. 3 - This means the reference angle is 30 or Answer: y = p 6 Eample : Find the value without a calculator. æ cos arctan 5 ö ç è 6ø p 6. So, y = p 6 so that it falls in the interval from -p < y < p Draw the reference triangle in the correct quadrant first. Find the missing side using Pythagorean Theorem. Find the ratio of the cosine of the reference triangle. 6 5 cosq = 6 6 CALCULUS SUMMER TUTORIAL JUNE, 08 Page 9
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