1 Functions and Inverses

Transcription

1 October, 08 MAT86 Week Justin Ko Functions and Inverses Definition. A function f : D R is a rule that assigns each element in a set D to eactly one element f() in R. The set D is called the domain of f. The set R is called the range of f, and it consists of all possible values of f(). If the domain of a function is not eplicitly specified, then we take D to be the largest set such that the function is well-defined. Eample. Functions can be represented in several ways such as a formula, a graph, or a table of values. For eample, the relation from R [0, ) can be epressed as:. Formula: A mathematical formula f() =. Graph: A graph is the collection of ordered pairs {(, f()) : D} epressed as a curve in the y-plane. y f() = Note: A curve in the y-plane is the graph of a function if any vertical line intersects the curve at most once. This is called the vertical line test. 3. Table: A table of values Definition. A function is one-to-one if it never takes the same value twice; that is, for every, D, f( ) = f( ) = =. () Note: In the y-plane, a function f is one-to-one if any horizontal line intersects the graph of f at most once. This is called the horizontal line test. Definition 3. Let f() and g() be two one-to-one functions. The functions f() and g() are inverses if (f g)() = for all in the domain of g and (g f)() = for all in the domain of f. The function g is called the inverse function of f and is usually denoted by g() = f (). The inverse f satisfies several properties:. The domain of f is the range of f and the range of f is the domain of f. In the y-plane, the graph of f is obtained by reflecting the graph of f about the line y = 3. A point (, y) is on the graph of f if and only if (y, ) is a point on the graph of f. Eample. The functions f() = from (0, ) (0, ) is a one-to-one function and has an inverse given by f () = g() =. To verify this, we can check that (f g)() = ( ) = for all (0, ) and (g f)() = = for all (0, ). More generally, the set R in the notation f : D R refers to the co-domain of f. The co-domain must contain the possible values of f(), but may also contain some impossible values. For eample, one may write: let f : R R be the function f() = even though the range of f is only [0, ). For simplicity, we always take the set R to refer to the range of a function in this course, so it is not important to know what a co-domain is. Page of

2 October, 08 MAT86 Week Justin Ko. Eample Problems.. Check if a function is one-to-one Strategy: We need to check Definition directly.. To show a function is one-to-one, we need to check condition () in Definition holds by assuming f( ) = f( ) and showing that we must have =.. Alternatively, to show a function is one-to-one, it suffices to show that the function is strictly increasing or decreasing on its domain (i.e. f( ) < f( ) whenever < or f( ) > f( ) whenever < ). This essentially shows the contrapositive of (), i.e. if then f( ) f( ). This technique will be useful later on in this course once we build the tools to check this condition. 3. To show that a function is not one-to-one, it suffices to find points such that f( ) = f( ). Problem. ( ) Is the function f() = +5 6 one to one? Solution. To show the function is one-to-one, notice for 6 and 6, we have f( ) = = = f( ) ( + 5)( 6) = ( + 5)( 6) so our function is one-to-one = = =, Problem. ( ) Is the function f() = one to one? Solution. From Table, we know that is not one-to-one function. To show this directly, notice that f() = = = = f( ), so our function is not one-to-one. Problem 3. ( ) Is the function f() = e one to one? Solution 3. We will show the function is one-to-one by proving that is strictly increasing. Taking the derivative of f(), we have f () = e + e = e ( + ) > 0 for all R. In particular, we have f() is strictly increasing on R and therefore also one-to-one. Problem 4. ( ) True of False: If f() is an even function (i.e. f( ) = f()), then f() is not one-to-one. Page of

3 October, 08 MAT86 Week Justin Ko Solution 4. This is true. Since f( ) = f() for all D f, if we take D f such that 0 then we have f() = f( ) but so our function is not one-to-one. Problem 5. ( ) True of False: If f() is an odd function (i.e. one-to-one. f( ) = f()), then f() is Solution 5. This is false. For eample, consider f() = sin(). It is well known that sin() is an odd function, but f(0) = sin(0) = 0 = sin(π) = f(π), so f() is not one-to-one. Problem 6. ( ) True or False: If g : A B and f : B C are one-to-one then f g : A C is one-to-one. Solution 6. This is true. We can show Definition directly. Suppose that (f g)( ) = (f g)( ), we need to show that =. Notice that both f and g satisfy () since they are one-to-one, so as required. (f g)( ) = (f g)( ) g( ) = g( ) f satisfies ().. Finding the Inverse of a Function = g satisfies () Strategy: Solve = f(y) for y. The resulting equation is y = f (). Problem. ( ) Find the formula for the inverse of the function f() = Solution. It is easy to check that f() is one-to-one. To find the formula for f, it suffices to solve = f(y) = y + 5 y 6 in terms of y. Notice that = y + 5 y 6 y 6 = y + 5 ( )y = f () = y = Problem. ( ) Find the formula for the inverse of the function f() = + +. Solution. The function f() is not one-to-one since f(0) = = f( ), so the inverse does not eist. Problem 3. ( ) Find a formula for the inverse of the function f() =. Be sure to specify the domain of f. Solution 3. It is easy to check that f() is one-to-one on its domain [, ). To find the formula for f, it suffices to solve = f(y) = y in terms of y. Notice that = y = y f () = y =. Although the function f () makes sense for all R, the domain of this inverse is the range of f(), which is [0, ). Page 3 of

4 October, 08 MAT86 Week Justin Ko Eponential Functions An eponential function with base a > 0 is denoted by a and its inverse is denoted by log a (). The most common base for an eponential function is the mathematical constant e = Definition 4. We call function ep() = e the eponential function, and its inverse ln() = log e () the natural logarithm (some books use log() to refer to the natural logarithm). We summarize several properties of the eponential function and its logarithm. The following properties hold more generally for a > 0 instead of e (just replace the e with a). We will prove some change of base formulas in the eercises that epress eponential functions and logarithms with base a in terms of ep() and ln(). Since we can freely convert into base e, it suffices to just work with base e in most applications. Eponential ep( + y) = ep() ep(y) Logarithm ln( y) = ln() + ln(y) ep() y = ep(y ) ln( y ) = y ln() ep( ) = ep() ln() = ln( ) ep(ln()) = for > 0 ln(ep()) = for R ep(0) =, ep() = e ln() = 0, ln(e) = y f() = e y f() = ln Table : Properties of Eponential Functions Note: Some eponential epressions are defined for a < 0. For eample, ( 8) = 64 and ( 8) /3 =. However, ( 8) is not defined on R (for eample, ( 8) / does not have a real solution).. Eample Problems:.. Solving and Simplifying Epressions Involving log() and e : Problem. ( ) Solve 3 3 = 9 for. Solution. Using the properties of eponents, we have 3 3 = 9 33 = (3 ) ( ) 3 3 = 3 ( ) 3 = ( ) = 7. Problem. ( ) Simplify the epression e ln( 3 ). Page 4 of

5 October, 08 MAT86 Week Justin Ko Solution. Using the properties of eponents, we have ( ( e ln( 3 ) = ep ln = ep(ln(3)) = 3. 3)) Problem 3. ( ) Simplify the epression e ln()+( 7) ln(). Solution 3. Using the properties of eponents, we have e ln()+( 7) ln() = e 3 ln() 7 ln() = e ln(3 )+ln( 7) = e ln(3 7) = e ln((3 7)) = 3 7. Problem 4. ( ) Solve the following epression for : log 54 ( ) + log 54 ( + ) =. Solution 4. Simplifying the left hand side, we combine the logarithms to conclude log 54 ( ) + log 54 ( + ) = log 54 (( ) ( + )) = log 54 ( ). Therefore, raising both sides of our original equation with base 54 implies log 54 ( ) + log 54 ( + ) = log 54 ( ) = = = 0. Notice that 56 = ( 8)( + 7) which means the equation 56 = 0 has solutions = 8 and = 7. However, since the domain of log 54 () is > 0, we have = 7 is not in the domain of either log 54 ( ) or log 54 ( + ), so our only solution is = 8. Problem 5. ( ) Solve the following epression for : 3 + = 54. Solution 5. Taking the natural logarithm of both sides, we have Since 54 = 7 = 3 3, we have Simplifying this equation, we have 3 + = 54 ( ) ln() + ( + ) ln(3) = ln(54) ( ) ln() + ( + ) ln(3) = ln(54) = ln( 3 3 ) = ln() + 3 ln(3). ln() + ln(3) = ln() + ln(3) = ln() + ln(3) ln() + ln(3) =. Page 5 of

6 October, 08 MAT86 Week Justin Ko.. Change of Base Formulas: Problem. ( ) Let b > 0 and suppose b. Prove the change of base formula log b () = ln ln b. Solution. We solve the following equation for y y = log b () b y = ln(b y ) = ln() y ln(b) = ln() y = ln() ln(b). compose both sides with b compose both sides with ln() Therefore, we have log b () = y = ln() ln(b). Note: We used the fact b, to ensure we did not divide by 0. Problem. ( ) Prove the change of base formula b = e ln(b). Solution. We solve the following equation for y b = e y ln b = y y = ln(b). compose both sides with ln() Therefore, we have b = e y = e ln(b). Page 6 of

7 October, 08 MAT86 Week Justin Ko 3 Absolute Value For R, the absolute value of is a piecewise function defined by { 0 = < 0. It is easy to check that for, y R, we have. Non-negativity: 0. Multiplicativity: y = y 3. Positive Definiteness: = 0 if and only if = 0 4. Triangle Inequality: + y + y 5. Reverse Triangle Inequality: y y. The graph is given below: y y = 3. Eample Problems 3.. Absolute Value Inequalities: Strategy: We can proceed in two ways:. In general, we need to break our region into cases where our absolute values change sign and solve the inequality on each region separately.. Shortcut: If we want to compute f() C or f() C then we can replace the absolute value with ± and solve the two cases corresponding to +f() and f(). Problem. ( ) Find all such that Solution. Rearranging the inequality, we have ± Solving for the case with the positive sign, we have Page 7 of

8 October, 08 MAT86 Week Justin Ko and for the case with the negative sign, we have Therefore, we have 3 7. Note: Notice that 3, so we do not run into the issue of dividing by zero in the first step. Problem. ( ) Find the domain of the function f() = sin ( 3 7). Solution. Since the domain of sin is, the domain of f() are the values of such that 3 7. To solve this inequality, we notice 3 7 ±( 3 7) Solving for the case with the positive sign, we have and for the case with the negative sign, we have ( 3 7) 3 8 ( 3 7) /3. Therefore, we have 6 /3. Problem 3. ( ) Find all such that + 8 < Solution 3. The function + 8 changes sign when = 8, so we consider the regions < 8 and > 8.. > 8: In this case we have + 8 = + 8, so solving the inequality gives + 8 < < >. Since we must have both 8 and >, we have our inequality is satisfied when >.. < 8: In this case we have + 8 = ( + 8) so solving the inequality gives + 8 < < > 3. Since we must have both < 8 and > 3, no satisfies our inequality. 3. = 8: When = 8, we have 0 < , so = 8 does not satisfy our inequality. Combining the cases above, our solutions are >. Page 8 of

9 October, 08 MAT86 Week Justin Ko Problem 4. ( ) Find all such that < + 4. Solution 4. The function changes sign when = and + 4 changes sign when = 4, so we consider the cases. < 4: On this region, we have = + and + 4 = 4 so we have < < 4 8 < 0, which is a false epression, so no in this region satisfies our inequality.. 4 < < : On this region, we have = + and + 4 = + 4 so we have < < + 4 > 0. so we must have > 0 and 4 < < which means 0 < < is a solution to the inequality. 3. > : On this region, we have = and + 4 = + 4 so we have < + 4 < < 4, which is a true epression so > is a solution to the inequality. 4. = 4 or = : When = 4, we have 6 < which is false, so = 4 is not a solution. When =, we have 0 < 4 which is true, so = is a solution. Combining our cases above, our solutions are > 0. Page 9 of

10 October, 08 MAT86 Week Justin Ko 4 Trigonometric Functions The 3 main trigonometric functions discussed in this course are sin(), cos() and tan(). The key values of sin() and cos() can be summarized by the table of values π 0 6 sin() 0 cos() 3 π 4 π π The values for other key values can be etrapolated by remembering the shapes of the graphs y = sin y = cos y = tan y y y The key trigonometric identies are. Pythagorean Identity:. Sum and Difference Formulas: sin(α ± β) = sin(α) cos(β) ± sin(β) cos(α), from these, one can derive the following identities sin θ + cos θ =. 3. Symmetry and Periodicity: (Use the Sum and Difference Formulas) cos(α ± β) = cos(α) cos(β) sin(α) sin(β). sin( θ) = sin(θ), cos( θ) = cos(θ), sin(θ + kπ) = sin(θ), cos(θ + kπ) = cos(θ) 4. Complementary and Supplementary Angles: (Use the Sum and Difference Formulas) ( π ) ( π ) sin θ = cos(θ), sin(π θ) = sin(θ), cos θ = sin(θ), cos(π θ) = cos(θ). 5. Double Angle Formulas: (Use the Sum and Difference Formulas and the Pythagorean Identity) sin(θ) = sin(θ) cos(θ), cos(θ) = cos (θ) sin (θ) = sin (θ) = cos (θ). 6. Half Angle Formulas: (Use the Double Angle Formulas for cos(θ)) sin θ = cos(θ), cos θ = + cos θ. 7. Product to Sum Formulas: (Use the Sum and Difference Formulas) cos(α) cos(β) = (cos(α + β) + cos(α β)), sin(α) sin(β) = (cos(α β) cos(α + β)), sin(α) cos(β) = (sin(α + β) + sin(α β)). To solve some word problems, it is also useful to recall the Cosine Law c = a + b ab cos C where C is the angle opposite side c. Page 0 of

11 October, 08 MAT86 Week Justin Ko 4. Eample Problems 4.. General Trigonometry Problems Problem. ( ) Find all such that ( cos + π ) = 0. Solution. From the graph of cos(), we know cos() = 0 = π + kπ for k Z. Therefore, the solutions of our equation are such that + π = π + kπ = kπ for k Z. Problem. ( ) Derive the Sum and Difference formulas. Solution. Recall Euler s Identity, If we take θ = α + β, then e iθ = cos(θ) + i sin(θ). e i(α+β) = cos(α + β) + i sin(α + β) and using the fact ep(a + b) = ep(a) ep(b), we also have e i(α+β) = e iα e iβ = (cos(α) + i sin(α))(cos(β) + i sin(β)) ( ) ( ) = cos(α) cos(β) sin(α) sin(β) + i sin(α) cos(β) + sin(β) cos(α). Therefore, our two equations above implies ( ) ( ) cos(α+β)+i sin(α+β) = e i(α+β) = cos(α) cos(β) sin(α) sin(β) +i sin(α) cos(β) + sin(β) cos(α). Equating the real and imaginary parts, we have cos(α + β) = cos(α) cos(β) sin(α) sin(β) and sin(α + β) = sin(α) cos(β) + sin(β) cos(α). To derive the formulas for α β, we use the fact sin( ) = sin() and cos( ) = cos() and use our formulas above to conclude and cos(α β) = cos(α + ( β)) = cos(α) cos( β) sin(α) sin( β) = cos(α) cos(β) + sin(α) sin(β) sin(α β) = sin(α + ( β)) = sin(α) cos( β) + sin( β) cos(α) = sin(α) cos(β) sin(β) cos(α). 4.. Inverse Trig Problems Problem. ( ) Find the such that. sin(sin ()) =. sin (sin()) = 3. tan (tan()) = 4. cot (cot()) = 5. csc(csc ()) = 6. cos(cos ()) =. Page of

12 October, 08 MAT86 Week Justin Ko Solution. Recall that the domain of a composition f g is { D g : g() D f } where D g and D f are the respective domains of g and f. Even though the trigonometric functions sin(), cos(),... are defined on large domains, the functions are not one-to-one, so its inverses are not well defined on the larger domain. We always restrict the domain of the trigonometric functions to the range of the inverse functions so that the inverses behave nicely. The answers to these problems can be read directly off the domains and ranges of the inverse trigonometric functions in Table. Let D f and R f be the domains and ranges of f.. The inverse identity sin(sin ()) = holds for all D sin () = [, ].. The inverse identity sin (sin()) = holds for all D sin() = R sin () = [ π, π ]. 3. The inverse identity tan (tan()) = holds for all D tan() = R tan () = ( π, π ). 4. The inverse identity cot (cot()) = holds for all D cot() = (0, π). 5. The inverse identity csc(csc ()) = holds for all D csc () = (, ] [, ). 6. The inverse identity cos(cos ()) = holds for all D cos () = [, ]. Problem. ( ) Rewrite the epression tan(cos ()) without using trigonometric functions. What is the domain of this function? Solution. We can solve this problem either geometrically or algebraically. Geometric Solution: We first find the domain of our function. We have D cos () = [, ] and D tan() = { k+ π}, so our domain consists of points in D cos () such that cos () π 0. Therefore, the domain of our function is [, ] \ {0}. Case > 0: We first consider the case such that > 0 on our domain. On this region, we have θ = cos () [0, π ] (the first quadrant). The triangle corresponding to cos(θ) = in the first quadrant is given by θ From this triangle, we see tan(cos ()) = tan(θ) = for (0, ]. Case < 0: We now consider the case such that < 0 on our domain. On this region, we have θ = cos () [ π, π] (the second quadrant). The triangle corresponding to cos(θ) = in the second quadrant is given by Page of

13 October, 08 MAT86 Week Justin Ko θ Notice that < 0, so this triangle is indeed in the second quadrant. From this triangle, we see tan(cos ()) = tan(θ) = for [, 0). Algebraic Solution: We first find the domain of our function. We have D cos () = [, ] and D tan() = { k+ π}, so our domain consists of points in D cos () such that cos () π 0. Therefore, the domain of our function is [, ] \ {0}. Case > 0: We first consider the case such that 0 on our domain. On this region, we have θ = cos () [0, π ] so trigonometric functions are positive. We now solve the identity algebraically. We want to write tan(θ) in terms of cos(θ). Using the Pythagorean identity, sin (θ) + cos (θ) = tan (θ) + = cos (θ) tan (θ) = cos (θ) cos (θ) cos (θ) tan(θ) =. cos(θ) Since tan(θ) 0 and cos(θ) > 0, we didn t have to worry about absolute values when taking the squareroots of both sides or dividing by zero. Therefore, if we set θ = cos (), we have tan(cos cos (cos ()) = ()) cos(cos = for (0, ]. ()) Case < 0: We now consider the case such that < 0 on our domain. We can easily check that our function tan(cos ()) is odd. To see this, notice that cos () π is odd, and therefore tan((cos () π ) + π ) is a composition of odd functions and therefore odd. Etending our solution for > 0 to make it odd, we have tan(cos ()) = ( ) = for [, 0). Problem 6. ( ) Rewrite the epression tan(csc ()) without using trigonometric functions. What is the domain of this function? Solution 6. We can solve this problem either geometrically or algebraically. Geometric Solution: We first find the domain of our function. We have D csc () = (, ] [, ) and D tan() = { k+ π}, so our domain consists of points in D csc () such that csc () ± π ±. Therefore, the domain of our function is (, ) (, ). Case > 0: We first consider the case such that > 0 on our domain. On this region, we have θ = csc () [0, π ] (the first quadrant). The triangle corresponding to csc(θ) = in the first quadrant is given by Page 3 of

14 October, 08 MAT86 Week Justin Ko θ From this triangle, we see tan(csc ()) = tan(θ) = for (, ). Case < 0: We first consider the case such that > 0 on our domain. On this region, we have θ = csc () [ π, 0] (the fourth quadrant). The triangle corresponding to csc(θ) = in the fourth quadrant is given by θ Notice that < 0, so the side opposite the right angle is positive. From this triangle, we see tan(csc ()) = tan(θ) = for (, ). Algebraic Solution: We first find the domain of our function. We have D csc () = (, ] [, ) and D tan() = { k+ π}, so our domain consists of points in D csc () such that csc () ± π ±. Therefore, the domain of our function is (, ) (, ). Case > 0: We first consider the case such that > 0 on our domain. On this region, we have θ = csc () [0, π ] so trig functions are all positive. We now solve the identity algebraically. We want to write tan(θ) in terms of csc(θ). Using the Pythagorean identity, sin (θ) + cos (θ) = + tan (θ) = csc(θ) tan (θ) = csc (θ) tan(θ) = csc (θ). Since sin(θ) > 0 and cos(θ) > 0 on this domain, we didn t have to worry about absolute values when taking the squareroots of both sides or dividing by zero. Therefore, if we set θ = csc (), we have tan(csc ()) = for (, ). Case < 0: We now consider the case such that < 0 on our domain. We can easily check that our function tan(csc ()) is odd. To see this, notice that sec () π is odd, and therefore Page 4 of

15 October, 08 MAT86 Week Justin Ko tan((sec () π ) + π ) is a composition of odd functions and therefore odd. Etending our solution to make it odd, we have tan(csc ()) = ( ) = for (, ). Remark: We used the following fact in Solutions 5 and Solution 6. Suppose we know f() for > 0. odd or even manner We can use following formulas to etend our functions in an. Odd Etension: For < 0, the odd etension of f is given by f( ).. Even Etension: For < 0, the even etension of f is given by f( ). Page 5 of

16 October, 08 MAT86 Week Justin Ko 5 Hyperbolic Functions The 3 main hyperbolic functions discussed in this course are sinh() = e e, cosh() = e + e, tanh() = sinh() cosh(). The graphs of these functions are given by y = sinh y y = cosh y y = tanh y Just like the trigonometric functions, the hyperbolic functions satisfy a similar set of properties. Analogue of the Pythagorean Identity: cosh () sinh () =.. Sum and Difference Formulas: sinh(±y) = sinh() cosh(y)±sinh() cosh(y), cosh(±y) = cosh() cosh(y)±sinh() sinh(y). 3. Double Angle Formulas: (Use sum and difference formulas and the Pythagorean identity) sinh() = sinh() cosh(), cosh() = cosh ()+sinh () = sinh ()+ = cosh (). 4. Half Angle Formulas: (Use the double angle formula for cosh()) sinh () = cosh(), cosh () = cosh() +. Similarly to the trigonometric functions, we can also define the following lesser used hyperbolic functions csch() = sinh(), sech() = cosh(), coth() = tanh(). 5. Eample Problems Problem. ( ) Derive the formula for cosh (). Solution. We eplicitly compute the inverse of f() = cosh(). Since cosh() is not one to one, we have to restrict its domain to 0 to make our function one to one. We know that the range of Page 6 of

17 October, 08 MAT86 Week Justin Ko cosh() is [, ). Setting f(y) = and solving for y, we have cosh(y) = ey + e y = e y + e y = 0 e y e y + = 0 e y = ± 4 4 multiply both sides by e y using the quadratic formula e y = + since e y must be > 0 for all y = ln( + ). This solution makes sense because, so we have y 0 as needed. Therefore, the formula for the inverse function is f () = ln( + ). Note: If we multiplied both sides by e y, in the computation above, we would have deduced that f () = ln( + ). This is the other possible inverse of cosh () if we choose to restrict its domain to 0. Problem. ( ) Verify the Pythagorean identity cosh () sinh () =. Solution. This is a direct computation. We have ( e cosh () sinh y + e y ) ( e y e y ) e y + + e y e y + e y () = + = 4 =. Page 7 of

18 October, 08 MAT86 Week Justin Ko 6 Appendi: Essential Functions and Their Graphs Below is a non-ehaustive list of the basic functions we will encounter in this class. Elementary Functions Function Domain Range One-to-One n (where n is even) R [0, ) No n (where n is odd) R R Yes [0, ) [0, ) Yes R \ {0} R \ {0} Yes R [0, ) No Eponential Functions Function Domain Range One-to-One a (where a > 0) R (0, ) Yes log a () (where a > 0) (0, ) R Yes Trigonometric Functions Function Domain Range One-to-One sin() R [, ] No cos() R [, ] No tan() = sin() cos() { : k+ π, k Z} R No sin () [, ] [ π, π ] Yes cos () [, ] [0, π] Yes tan () R ( π, π ) Yes csc() = sin() { : kπ, k Z} (, ] [, ) No sec() = cos() { : k+ π, k Z} (, ] [, ) No cot() = tan() { : kπ, k Z} R No csc () (, ] [, ) [ π, 0) (0, π ] Yes sec () (, ] [, ) [0, π ) ( π, π] Yes cot () R (0, π) Yes Hyperbolic Functions Function Domain Range One-to-One sinh() = e e R R Yes cosh() = e +e R [, ) No tanh() = sinh() cosh() R (, ) Yes sinh () = ln( + + ) R R Yes cosh () = ln( + ) [, ) [0, ) Yes tanh () = + ln( ) (, ) R Yes csch() = sinh() R \ {0} R \ {0} Yes sech() = cosh() R (0, ] No coth() = tanh() R \ {0} (, ) (, ) Yes Table : Table of Functions Page 8 of

19 October, 08 MAT86 Week Justin Ko 6. Graphs 6.. Trigonometric Functions The trigonometric functions are not one-to-one, so we first restrict the domain of the trig functions to a region such that the functions is one-to-one. Trigonometric functions with leading sine terms are restricted to a subset of the domain [ π/, π/] and the Trigonometric functions with leading cosine terms are restricted to the domain [0, π]. Note: In the following pictures, the dotted blue graph is the full function. The blue graph is the one-to-one function on the restricted domain. And the red graph is the inverse function. Sine Function: Cosine Function: Tangent Function: Page 9 of

20 October, 08 MAT86 Week Justin Ko Cosecant Function: Secant Function: Cotangent Function: 6.. Eponential Functions Eponential Function with base e: Page 0 of

21 October, 08 MAT86 Week Justin Ko 6..3 Hyperbolic Functions Hyperbolic Sine: Hyperbolic Cosine: Hyperbolic Tangent: Page of