0 Review of Precalculus Topics

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1 0 Review of Precalculus Topics 0. The Basics Set Notation.. a A means that a is an element in the set A.. a/ A means that a is not an element in the set A.. For sets A and B we write A = B to mean that both sets have eactly the same elements.. For sets A and B we write A B to mean to mean that the set A is a subset of the set B, thatis,everyelementofa is also an element of B. 5. We write ; for the empty the set, that is, the set with no members. Important Sets in Mathematics. N = { : is a natural number} = {,,,,...}.. Z = { : is an integer} = {...,,,, 0,,,,...}.. Q = { : is a rational number}; andso, Q.. R = { : is a real number}; andso, R. Remark. Anumberr is rational (that is, r Q) ifandonlyifr = a b where b 6= 0. for some integers a, b Table of Intervals For real numbers a and b, welistallofthepossibleintervals: Notation Set description (a, b) { R : a<<b} [a, b] { R : a apple apple b} [a, b) { R : a apple <b} (a, b] { R : a<apple b} (a, ) { R : a<} [a, ) { R : a apple } (,b) { R : <b} (,b] { R : apple b} (, ) R

2 Properties of Inequality For all a, b, c, d R the following hold:. a<bor a = b or a>b (Trichotomy). (a <band b<c) implies a<c (Transitivity). a<bimplies a + c<b+ c (Adding on both sides). (a <band c<d) implies a + c<b+ d. (Additivity) 5. (a < band c > 0) implies ac < bc (Multiplying by a positive) 6. (a < band c < 0) implies ac > bc (Multiplying by a negative) Problem. Solve the following inequalities and illustrate the solution set on the real line.. Solve the inequality ( + )+ for.. Solve the inequality 5 apple +6< for.. Let c<0. Solve c 0 > c 0 for.. Let c<. Solve c 5. Solve <6 for. >c+for. 6. Solve > 0 for. Problem. Suppose and y are real numbers such that >and y <. Show that y>5. Theorem. Suppose a and b are real numbers. If 0 <a<b,then a > b. Proof. Given in class. Theorem. Suppose a and b are real numbers. If 0 <a<b,thena <b. Proof. Suppose a and b are real numbers. We shall prove that if 0 <a<b,thena <b.to do this, assume 0 <a<b.wemustprovethata <b.since0 <a<b,weconcludethat a < b and that a, b are positive numbers. Multiplying both sides of the inequality a < b by the positive a gives the inequality (i) a < ab, andmultiplyingbothsidesoftheinequality a<bby the positive b gives the inequality (ii) ab < b. Thus, (i) and (ii) give a < ab < b. Thus, a <b. Therefore, if 0 <a<b,thena <b. Corollary. Suppose a and b are real numbers. If 0 <a<b,then p a< p b. Proof. Suppose that 0 <a<b. We will prove that p a< p b.suppose,foracontradiction, that p b apple p a.if p b = p a,thenb =( p b) =( p a) = a. Contradiction. If p b< p a,then b =( p b) < ( p a) = a by the above theorem, and so b<a. Contradiction. 5

3 Absolute Value. Given a real number, theabsolutevalueof is denoted by and is defined by, if 0 =, if <0. Properties of Absolute Value. For all a, b,, c R, where c>0, the following hold:. <cif and only if c<<c. >cif and only if < c or c<. c = c. ab = a b 5. a b = a b 6. a + b apple a + b (triangle inequality) 7. a b apple a b. Problem. Suppose is a real number. Show that if <, then < 6. [Hint: y <ameans that a<y<a, that is, a<yand y<a.] Problem. Solving inequalities.. Solve 5 < for. Illustratethesolutionsetontherealline.. Solve + for. Illustratethesolutionsetontherealline. [Hint: y a means that y apple a or y a.] Proofs of Algebraic Equations Proof Strategy: To prove an algebraic equation ' =, start with one side of the equality and write a string of equalities ending with the other side of the equality. Remark: It is not a correct proof to assume the equation ' = is true and then work on both sides of this equation to obtain an identity. Problem. Prove the following theorem. Theorem. Suppose a and b are real numbers. Then (a + b) Proof. We start with (a + b) (a + b) a = a +ab + b a = ab + b = b +ab = b(b +a). Eponents and Radicals a = b(b +a). a and derive the right hand side of the desired equality: Eponents. Given a real number and a natural number n the real number n and is defined by n = (n many multiples of ). Here is called the base of n and n is called the eponent or power. 6

4 Negative Eponents. Given a real number and a natural number n the real number n and is defined so that n = n and n = n. Important Properties of Eponents. For all, y R and all m, n Z the following hold:. 0 =. m n = m+n. ( m ) n = mn Eponent Form of Roots. number n is defined so that. ( y) m = m y m 5. y m = m y m when y 6= 0 6. m n = m n when 6= 0 Given a real number and a natural number n, thereal n = n p for all when n odd and for all 0 when n is even. Moreover, for any real number and any even natural number n, wehavethat np n = and so, in particular, p =. Rational Eponents. Given a real number m n and is defined so that m n = n m 0 and a rational number m n and m n =( m ) n. the real number Problem. Evaluate the following: 6, 9, 8, p 9, p ( 9). Quadratic Equations Aquadraticepressionhastheforma +b+c where a 6= 0.Sometimesquadraticepression can be factored. Eample. Factor the following quadratic epressions: Eample. Find the roots of following quadratic equations by factoring:. 0 = 0. +=0. 6 = 0. Sometimes finding the roots of a quadratic equations by factoring is difficult. In these cases we can use the quadratic formula. Quadratic Formula. The roots of a quadratic equation a + b + c =0where a 6= 0are = b ± p b ac. a Eample. Using the quadratic formula, find the roots of equations : () + = 0 and () +5 =. 7

5 Completing the Square To complete the square of the quadratic perform the following steps: a + b + c (0.) Step : Factor out of (0.) the non-zero leading coefficient a, obtaining a + b + c = a + b a + c. (0.) a Step : Add and subtract the value a + b + c = a Step : The term + b b a + a + b a + b a b within the parentheses of (0.) as follows a + b b a + + c a a is a perfect square and so,! b. (0.) a = + b. (0.) a Substituting equation (0.) into equation (0.), we obtain our completion of the square a + b + c = a + b + c a a! b = a + b + c a a b. (0.5) a Problem. Complete the square of the quadratic +. Solution. We follow the above steps as follows: += +! = + +! = + =! + 7 = factor out the leading coefficient add and subtract perfect square by arithmetic. = Therefore, += is the requested completion of the square

6 Rational Epressions Eample. Perform the following operations:. Reduce +7 and add Multiply + Undefined Operations and divide 6. Solve =5. a is undefined for a 6= 0(division by 0 is undefined). is undefined.. p a is undefined for a<0 (the square root of a negative number is undefined). Rationalizing Numerators and Denominators Suppose that we have a ratio with a sum or difference involving a radical in the denominator (numerator), and we need to re-epress this ratio so that the radicals are removed from the denominator (numerator). We now demonstrate the techniques that are used in calculus to re-epress such ratios. Problem. Rationalize the denominator in the ratio p. Solution. To rationalize the ratio p p p,wemultiplytopandbottomoftheratiobythe conjugate of, which is +. p = p +p + p ( + p ) = p p ( )( + ) = ( + p ) ( ) multiply top and bottom by the conjugate by algebra =(+ p ) by algebra. Therefore, p =(+p ). Problem. Rationalize the numerator in the ratio p 7+ p + p. Solution. We multiply top and bottom of the ratio by the conjugate of p 7+ p p p, which is 7, as follows: p p 7+ + p = ( p p 7+ ) ( + p ( p 7 ) ( p 7 p ) p = ) 7 ( + p )( p 7 p ) = 5 ( + p )( p 7 p ). Therefore, p 7+ p + p = 5 (+ p )( p 7 p ). 9

7 0. Functions Definition. We write f : X! Y to mean that f is a function from the set X to the set Y,thatis,foreveryelementof X there is a unique element f() in Y. We say that the element X gets mapped to the element f() by the function f. X is called the domain of the function f and Y is called the co-domain of the function f. Remarks.. Given a function f and the equation f() =y we call the independent variable and we call y the dependent variable.. The epression f() is read as f of.. Warning: The epression f() is not read as f times. Eample. Let f be the function defined by f() =. Whatisthedomainoff?. Evaluate f(0).. Evaluate f().. Evaluate f( ).. Evaluate and simplify f( + h). 5. Evaluate and simplify f(+h) f() h. Eample. Let f be the function defined by f() =. What is f()? What is f()? What is the domain of f? Definition. Given a function f : X! Y the range of f, denotedbyrange(f), istheset range(f) ={f() : X}. Eample. Let f : R! R be f() =.Whatisf()? Whatistherangeoff? Eample. Let f be the function defined by f() =.Whatistherangeoff? Eample. Let C be the cost function defined by C() =. SupposethatC() gives the cost (in dollars) required to buy many books.. What is the cost of 6 books?. How many books can one buy with 08 dollars? 0

8 Graphs of Functions Eample. Graph the function f() =.Whatisthedomainandrangeofthisfunction? Solution. We make a table of values and the graph of the function f() = is f() = Figure : Graph of the function f() =

9 Eample. Graph the function f() = Solution. We make a table of values.whatisthedomainandrangeofthisfunction? f() = and using this table one obtains the following graph of the function f() = : Figure : Graph of the function f() =

10 Eample. Graph the function f() =.Whatisthedomainandrangeofthisfunction? Solution. We make a table of values f() = and using this table one obtains the following graph of the function f() = : Figure : Graph of the function f() =

11 Eample. Graph the function f() = p.whatisthedomainandrangeofthisfunction? Solution. We make a table of values f() = p and using this table one obtains the following graph of the function f() = p : Figure : Graph of the function f() = p Definition 0... Let f be a function. A real number is called a zero (or a root) ofthe function f, iff() =0. Eample. Find all the zeros of each function:. f() =. f() = +

12 Graphing Quadratic Functions. Problem. How to sketch the graph of a quadratic function f() =a + b + c. Solution. By completing the square, we obtain f() =a +b+c = a + b a + c b. The verte is the highest or the lowest point on the graph of a quadratic function. Find the coordinate of the verte by computing v = b a.. Evaluate the function at v,thatis,evaluatey v = f( v )=c b a.. The point ( v,y v ) is the lowest point on the graph if a>0. the highest point on the graph if a<0.. Plot some points on both sides of the verte. 5. Connect the points with a sketch of the graph. Eample.. Sketch the graph of the quadratic function f() = Sketch the graph of the quadratic function f() = Increasing and Decreasing Functions Definition 0... Let f : I! R be a function where I is an interval. a. We say that f is increasing on I when the following holds: For all, <,thenf( ) <f( ). We say that f is decreasing on I when the following holds: For all, <,thenf( ) >f( ). I if I if Even and Odd Functions Definition 0... Let f : I! R be a function where I is an interval. We say that f is even when for all I, f( ) =f(). We say that f is odd when for all I, f( ) = f(). Piecewise Defined Functions Eample. Sketch a graph of the following piecewise defined function: ( +, if apple ; f() =, if >. 5

13 0. A Classification of Functions Linear Functions Eample. Graph the straight line given by the equation y = +.Notethatthepoints (0, ), (, ), (, 5) are on the line because these points satisfy the equation. Horizontal Lines. If c is any real number then, the equation has a horizontal line as its graph. y = c Eample. Graph the straight line given by the equation y =. Vertical Lines. If c is any real number then, the equation has a vertical line as its graph. = c Eample. Graph the straight line given by the equation =. Slope of a Straight Line. If (,y ) and (,y ) are any two distinct points on a line, then the slope m of line is given by m = y y = y. Eample. Graph the straight line given by the equation y = +.Notethatthepoints (0, ), (, ), (, 5) are on this line. Using these points, find the slope of this line. Eample. Graph the straight line given by the equation y = +.Notethatthepoints (0, ), (, ), (, ) are on this line. Using these points, find the slope of this line. Sign of the Slope.. If m>0, thentheline rises fromlefttoright.. If m<0, thentheline falls fromlefttoright. Parallel and Perpendicular Lines. Suppose two lines have slopes m and m.. The two lines are parallel if and only if m = m.. The two lines are perpendicular if and only if m m = ; thatism = m. 6

14 Equation of a Straight Line.. Point-Slope Form: y y = m( ) where (,y ) is a point on the line and m is its slope.. Slope-Intercept Form: y = m + b where (0,b) is a point on the line and m is its slope. In this form b is called the y intercept. Eample. Graph the straight line given by the equation y = +.Notethatthepoint (, ) is on the line. Using this point, find the Point-Slope Form equation of this line. Eercise: The point (, 5) is a point on a line with slope. Find both the point-slope and slope-intercept forms of the equation of this line. Eample. Consider the line +y =. FindtheSlope-InterceptFormequationofthis line. What is the slope and y intercept of this line? Definition 0... Afunctionf is called a linear function if it has the form f() =m + b where m is the slope of the function and b is its y intercept. Polynomial Functions Definition 0... Afunctionf is called a polynomial function if it has the form f() =a n n + a n n + a + a 0 where n is a natural number. The constants a n,a n polynomial function.,...,a 0 are called the coefficients of the Definition 0... Afunctionf is called a quadratic function if it has the form f() =a + a + a 0. Definition 0... Afunctionf is called a cubic function if it has the form f() =a + a + a + a 0. Definition Afunctionf is called a rational function if it has the form f() = p() q() where p() and q() are polynomial functions. 7

15 Polynomial Long Division Polynomial long division is the following algorithm for dividing a polynomial by another polynomial of the same or lower degree:. Divide the highest degree term of the divisor into the highest degree term of the dividend. Write the result above the division line. Go to step.. Multiply the result, obtained in step, by the divisor and subtract the product from the dividend, obtaining a new dividend. Go to step.. If the highest degree term of the divisor is less than or equal to the highest degree term of the new dividend, then go to step ; else, go to step.. Stop. The new dividend is the remainder, because it has lower degree than the divisor. Problem. Find the remainder on division of:. + +7by by. Solution. We perform the requested polynomial long division.. We divide + +7by +and obtain divisor! quotient dividend dividend dividend remainder and thus we have the remainder 0. It thus follows that + +7=( +7)( +) and so, = is a root of the polynomial We divide by and obtain and thus we have the remainder +. It thus follows that =( +)( )

16 0. New Functions from Old Functions Translations of Functions Given the graph G of a function y = f() and a positive constant c>0, then. the graph of y = f( + c) is obtained by shifting G by c units to the left. the graph of y = f( c) is obtained by shifting G by c units to the right. the graph of y = f()+c is obtained by shifting G by c units up. the graph of y = f() c is obtained by shifting G by c units down. When translating the input variable ( that is, f(+a) ), the graph shifts along the ais in the opposite direction of the sign of a Figure 5: f() =cos() Figure 6: f() =cos( +) Figure 7: f() =cos( ) 9

17 When translating the output variable ( that is, f() +a ), the graph shifts along the y ais in the same direction of the sign of a Figure 8: f() =sin() Figure 9: f() = sin() Figure 0: f() = sin() 0

18 0.. Algebra of Functions Definition 0... Let f : D! R and g : D! R be functions. We can define new functions as follows: () The sum (f + g): D! R is defined by (f + g)() =f()+g() for all D. () The product (fg): D! R is defined by (fg)() =f()g() for all D. () For each k R, themultiple (kf): D! R is defined by (kf)() =kf() for all D. f () If g() 6= 0for all D, thenthequotient : D! R is defined by is defined by g f () = f() for all D. g g() 0.. Composition of Functions Definition. Given two functions f and g, oneformsthecomposition function f defining (f g)() =f(g()). Eample. Let g : R! R and f : R! R be the functions defined by f() = g() =. Findformulasfor(f g)() and (g f)(). Isf g = g f? + g by and Solution. Let R. Weevaluatethefunction(f g)() as follows: (f g)() =f(g()) = f( ) = ( ) +. Thus, (f g)() =.Weevaluate(g f)() to obtain ( ) + (g f)() =g(f()) = g + = + = +. Hence, (g f)() =. Since (f g)(0) = + f g 6= g f. and (g f)(0) = 0, weconcludethat Eample. Let (f g)() = p +.Findf and g. Nowfind(g f)(). Solution. Let (f g)() = p +. Since (f g)() =f(g()), wefirstidentifythe inside function in the formula p +.Weseethatg() = and f() = p + is the outside function. We now evaluate (g f)() as follows: (g f)() =g(f()) = g p + = p + p +.

19 0.5 Trigonometric Functions Radian Measure of Angles In Calculus all angles are measure in radians. Recall that the angle given by 60 degrees is the same as radians. Thus, rad =80deg. The following table gives a few of angles epressed in degree measure and radian measure. Degrees Radians 0 6 Acute Angle Trigonometry For an acute angle in the following right triangle hyp opp adj the si trigonometric functions are define as follows: sin( ) = opp hyp cos( ) = adj hyp tan( ) = opp adj csc( ) = hyp opp sec( ) = hyp adj cot( ) = adj opp General Angle Trigonometry For any angle (obtuse or negative) as in the following diagram y ais y (, y) s A A A r A A A A9 A ais

20 the si trigonometric functions are define as follows: sin( ) = y r cos( ) = r tan( ) = y csc( ) = r y sec( ) = r cot( ) = y Some Values of the Trigonometric Functions 0 6 sin( ) 0 cos( ) p p p p p p 0 0 p p 0 Some Trigonometric Identities A trigonometric identity is an equation involving trigonometric functions which is true for all values of. We now cite a few of the identities that we will be using in calculus: csc( ) = sin( ) sec( ) = cos( ) cot( ) = tan( ) tan( ) = sin( ) cos( ) cot( ) = cos( ) sin( ) sin ( )+cos ( ) = tan ( )+=sec ( ) cot ( )+=csc ( ) sin( ) = sin( ) cos( ) =cos( ) sin( + ) =sin( ) cos( + ) =cos( ) sin( + )=sin( )cos( )+cos( )sin( ) cos( + )=cos( )cos( ) sin( )sin( ) Remark. All of the important trig identities are on page 8.

21 Graphs of three Trigonometric Functions Figure : Graph of the function f() =sin() Figure : Graph of the function f() =cos() Figure : Graph of the function f() =tan() Remark. The graphs of the other trig functions can be found on page 87 in the tet.

22 0.6 Eponential and Logarithmic Functions Eponential Functions Definition An eponential function has the form f() =b where b>0 and b 6=. The real number b is called the base.. Eponential functions are positive; that is, b > 0 for all real numbers.. The function f() =b is increasing if b> and it is decreasing if 0 <b<.. The range of any eponential function f() =b is the set of all positive real numbers. - - Figure : Graph of the function f() = - - Figure 5: Graph of the function f() =( ) Theorem 0.6. (Laws of Eponents). Let b>0 and b 6=. Then. b 0 =. b b y = b +y and b b y = b. b = b and (b ) y = b y. b n = np b when n is a natural number. y 5

23 Logarithmic Functions Logarithmic functions are the inverses of eponential functions. Definition A logarithmic function has the form f() =log b () where b>0 and b 6= and satisfies. b log b () = when >0,. log b (b )= for all real numbers. The domain of log b () is { R : >0}, thesetofallpositiverealnumbers. The range of log b () is R, setofallrealnumbers Figure 6: Graph of the function f() = log () Theorem 0.6. (Laws of Logarithms). Let b>0 and b 6=. Then. log b () = 0 and log b (b) =. log b (y) =log b ()+log b (y) and log b =log y b () log b (y). log b = log y b (y) and log b ( y )=ylog b (). 6

24 Right triangle definition For this definition we assume that! 0 < " < or 0 < " < 90. opposite opposite sin" = hypotenuse adjacent cos" = hypotenuse opposite tan" = adjacent Trig Cheat Sheet Definition of the Trig Functions hypotenuse adjacent hypotenuse csc" = opposite hypotenuse sec" = adjacent adjacent cot" = opposite Unit circle definition For this definition " is any angle. y sin" = = y csc" = y cos" = = sec" = y tan" = cot" = y Facts and Properties Domain The domain is all the values of " that Period can be plugged into the function. The period of a function is the number, T, such that f (" + T) = f ("). So, if # sin", " can be any angle is a fied number and " is any angle we cos", " can be any angle have the following periods. tan", " #! $ n+ " %!, n= 0, ±, ±, & ' csc", " # n!, n= 0, ±, ±, sec", " #! $ n+ " %!, n= 0, ±, ±, & ' cot", " # n!, n= 0, ±, ±, Range The range is all possible values to get out of the function. ( ) sin" ) csc" * and csc" )( ( ) cos" ) sec" * andsec" )( (+ < tan" < + (+ < cot" < + " ( y, ) y y sin (#"), " cos (#"), tan (#"), T csc (#"), sec (#"), cot (#"), T! T = #! T = #! = #! T = #! T = #! = # 7

25 Tangent and Cotangent Identities sin cos tan! =! cot! =! cos! sin! Reciprocal Identities csc! = sin! = sin! csc! sec! = cos! = cos! sec! cot! = tan! = tan! cot! Pythagorean Identities sin! + cos! = tan! + = sec! + cot! = csc! Even/Odd Formulas sin!! =! sin! csc!! =! csc! ( ) ( ) ( ) ( ) ( ) ( ) cos!! = cos! sec!! = sec! tan!! =! tan! cot!! =! cot! Periodic Formulas If n is an integer. sin! + " n = sin! csc! + " n = csc! ( ) ( ) ( n) ( n) ( n) ( n) cos! + " = cos! sec! + " = sec! tan! + " = tan! cot! + " = cot! Double Angle Formulas ( ) ( ) sin! = sin! cos! cos cos sin tan! =!!! (! ) =! cos! =! sin! tan! =! tan! Degrees to Radians Formulas If is an angle in degrees and t is an angle in radians then " 80 = t " t = " and = t " Formulas and Identities Half Angle Formulas sin! =! cos(! ) cos! = + cos(! )! cos(! ) tan! = + cos! ( ) ( ) ( ) Sum and Difference Formulas sin # ± $ = sin# cos $ ± cos# sin $ ( ) ( ) cos # ± $ = cos# cos $! sin# sin $ tan# ± tan $ tan (# ± $ ) =! tan# tan $ Product to Sum Formulas sin# sin $ = # cos( ) cos( ) % #! $! # + $ $ & cos# cos $ = # cos( ) cos( ) % #! $ + # + $ $ & sin# cos $ = # sin (# $ ) sin (# $ ) % + +! $ & cos# sin $ = # sin (# + $ )! sin (#! $ ) $ % & Sum to Product Formulas ' # + $ ( '#! $ ( sin# + sin $ = sin ) * cos) * +, +, ' # + $ ( '#! $ ( sin#! sin $ = cos) * sin ) * +, +, ' # + $ ( '#! $ ( cos# + cos $ = cos) * cos) * +, +, ' # + $ ( '#! $ ( cos#! cos $ =! sin ) * sin ) * +, +, Cofunction Formulas ' " ( '" ( sin )!! * = cos! cos)!!* = sin! +, +, ' " ( '" ( csc)!! * = sec! sec)!!* = csc! +, +, ' " ( '" ( tan )!! * = cot! cot )!!* = tan! +, +, 8

26 Unit Circle! " $ #, % & '! " $ #, % & ' 5" 6! " $ #, % & ' " 50 " 0 5 y ( 0, ) " " 5! " $, % & ' " " 6 0! " $, % & '! " $, % & ' (#,0) " " (,0 )! " $ #,# % & ' 7" 6! " $ #,# % & ' 0 5"! " $ #,# % & ' 5 " 0 70 " ( 0, # ) 00 5" 0 5 7"! " $,# % & ' " 6! " $,# % & '! " $,# % & ' For any ordered pair on the unit circle (, ) y : cos! = and sin! = y Eample! 5 5 cos " "! sin " " $ % = $ % = # & ' & ' 9

27 Definition! y = sin is equivalent to = sin y! y = cos is equivalent to = cos y! y = tan is equivalent to = tan y Domain and Range Function Domain Range! y = sin! " "!!! " y "! y = cos! " " 0 " y "!! y = tan!# < < #!!! < y < Inverse Trig Functions Inverse Properties!! cos cos = cos cos " = " ( ( )) ( ( )) (! ( ) )! ( ( ))! ( )! ( ) sin sin = sin sin " = " ( ) ( ) tan tan = tan tan " = " Alternate Notation! sin = arcsin cos tan!! = arccos = arctan Law of Sines, Cosines and Tangents c $ a # % b Law of Sines sin# sin $ sin% = = a b c Law of Cosines a = b + c! bc cos# b = a + c! ac c = a + b! ab cos $ cos% Mollweide s Formula a+ b cos (#! $ = ) c sin % Law of Tangents a! b tan #! $ = a+ b tan # + $ ( ) ( ) ( $! %) ( $ %) (#!%) (# %) b! c tan = b+ c tan + a! c tan = a+ c tan + 0