Mathematics Learning Centre Introduction to Trigonometric Functions Pegg Adamson and Jackie Nicholas c 998 Universit of Sdne
Acknowledgements A significant part of this manuscript has previousl appeared in a version of this booklet published in 986 b Pegg Adamson. In rewriting this booklet, I have relied a great deal on Pegg s ideas and approach for Chapters,, 3, 4, 5 and 7. Chapter 6 appears in a similar form in the booklet, Introduction to Differential Calculus, which was written b Christopher Thomas. In her original booklet, Pegg acknowledged the contributions made b Mar Barnes and Sue Gordon. I would like to etend this list and thank Collin Phillips for his hours of discussion and suggestions. Jackie Nicholas September 998
Contents Introduction. How to use this booklet....... bjectives...... Angles and Angular Measure. Converting from radians to degrees and degrees to radians.... 3. Real numbers as radians...... 4.. Eercise... 5 3 Trigonometric Ratios in a Right Angled Triangle 6 3. Definition of sine, cosine and tangent.... 6 3.. Eercise... 7 3. Some special trigonometric ratios...... 7 4 The Trigonometric Functions 8 4. The cosine function..... 8 4.. Eercise... 9 4. The sine function...... 9 4.. Eercise... 0 4.3 The tangent function.... 0 4.3. Eercise... 4.4 Etending the domain... 4.4. Eercise... 3 5 Graphs of Trigonometric Functions 4 5. Changing the amplitude...... 5 5.. Eercise... 5 5. Changing the period.... 6 5.. Eercise... 7 5.3 Changing the mean level...... 8 5.3. Eercise... 8 5.4 Changing the phase..... 9 5.4. Eercise... 0 i
6 Derivatives of Trigonometric Functions 6. The calculus of trigonometric functions... 6.. Eercise... 7 ABrief Look at Inverse Trigonometric Functions 3 7. Definition of the inverse cosine function... 4 7.. Eercise... 5 8 Solutions to Eercises 6 ii
Mathematics Learning Centre, Universit of Sdne Introduction You have probabl met the trigonometric ratios cosine, sine, and tangent in a right angled triangle, and have used them to calculate the sides and angles of those triangles. In this booklet we review the definition of these trigonometric ratios and etend the concept of cosine, sine and tangent. We define the cosine, sine and tangent as functions of all real numbers. These trigonometric functions are etremel important in science, engineering and mathematics, and some familiarit with them will be assumed in most first ear universit mathematics courses. In Chapter we represent an angle as radian measure and convert degrees to radians and radians to degrees. In Chapter 3 we review the definition of the trigonometric ratios in a right angled triangle. In Chapter 4, we etend these ideas and define cosine, sine and tangent as functions of real numbers. In Chapter 5, we discuss the properties of their graphs. Chapter 6 looks at derivatives of these functions and assumes that ou have studied calculus before. If ou haven t done so, then skip Chapter 6 for now. You ma find the Mathematics Learning Centre booklet: Introduction to Differential Calculus useful if ou need to stud calculus. Chapter 7 gives a brief look at inverse trigonometric functions.. How to use this booklet You will not gain much b just reading this booklet. Mathematics is not a spectator sport! Rather, have pen and paper read and tr to work through the eamples before reading their solutions. Do all the eercises. It is important that ou tr hard to complete the eercises, rather than refer to the solutions as soon as ou are stuck.. bjectives B the time ou have completed this booklet ou should: know what a radian is and know how to convert degrees to radians and radians to degrees; know how cos, sin and tan can be defined as ratios of the sides of a right angled triangle; know how to find the cos, sin and tan of 6, 4 and ; know how cos, sin and tan functions are defined for all real numbers; be able to sketch the graph of certain trigonometric functions; know how to differentiate the cos, sin and tan functions; understand the definition of the inverse function f () =cos ().
Mathematics Learning Centre, Universit of Sdne Angles and Angular Measure An angle can be thought of as the amount of rotation required to take one straight line to another line with a common point. Angles are often labelled with Greek letters, for eample θ. Sometimes an arrow is used to indicate the direction of the rotation. If the arrow points in an anticlockwise direction, the angle is positive. If it points clockwise, the angle is negative. B A Angles can be measured in degrees or radians. Measurement in degrees is based on dividing the circumference of the circle into 360 equal parts. You are probabl familiar with this method of measurement. 360 o 80 o. o 90 A complete revolution is 360. A straight angle is 80. A right angle is 90. Fractions of a degree are epressed in minutes ( ) and seconds ( ). There are sit seconds in one minute, and sit minutes in one degree. So an angle of 3 7 can be epressed as 3 + 7 60 =3.8. The radian is a natural unit for measuring angles. We use radian measure in calculus because it makes the derivatives of trigonometric functions simple. You should tr to get used to thinking in radians rather than degrees. To measure an angle in radians, construct a unit circle (radius ) with centre at the verte of the angle. The radian measure of an angle AB is defined to be the length of the circular arc AB around the circumference. B A This definition can be used to find the number of radians corresponding to one complete revolution.
Mathematics Learning Centre, Universit of Sdne 3 In a complete revolution, A moves anticlockwise around the whole circumference of the unit circle, a distance of. So a complete revolution is measured as radians. That is, radians corresponds to 360. A Fractions of a revolution correspond to angles which are fractions of. 4 revolution 90 or radians 3 revolution 0 revolution 60 or radians 6 or radians 3 3. Converting from radians to degrees and degrees to radians Since radians is equal to 360 radians = 80, radian = 80 = 57.3, radians = 80, and similarl = 80 radians, 0.07, = 80 radians. Your calculator has a ke that enters the approimate value of.
Mathematics Learning Centre, Universit of Sdne 4 If ou are going to do calculus, it is important to get used to thinking in terms of radian measure. In particular, think of: 80 as radians, 90 as 60 as 45 as 30 as radians, 3 radians, 4 radians, 6 radians. You should make sure ou are reall familiar with these.. Real numbers as radians An real number can be thought of as a radian measure if we epress the number as a multiple of. B For eample, 5 = ( + 4 )= + corresponds to the arc length of revolutions of the unit circle going 4 anticlockwise from A to B. A Similarl, 7 4.97 = 4 +0.97 corresponds to an arc length of 4.97 revolutions of the unit circle going anticlockwise.
Mathematics Learning Centre, Universit of Sdne 5 We can also think of negative numbers in terms of radians. Remember for negative radians we measure arc length clockwise around the unit circle. For eample, 6.546 = + 0.546 corresponds to the arc length of approimatel.546 revolutions of the unit circle going clockwise from A to B. B A We are, in effect, wrapping the positive real number line anticlockwise around the unit circle and the negative real number line clockwise around the unit circle, starting in each case with 0 at A, (, 0). B doing so we are associating each and ever real number with eactl one point on the unit circle. Real numbers that have a difference of (or a multiple of ) correspond to the same point on the unit circle. Using one of our previous eamples, 5 corresponds to as the differ b a multiple of... Eercise Write the following in both degrees and radians and represent them on a diagram. a. 30 b. c. 0 d. 3 4 e. f. 4 3 g. 70 h. i. Note that we do not indicate the units when we are talking about radians. In the rest of this booklet, we will be using radian measure onl. You ll need to make sure that our calculator is in radian mode.
Mathematics Learning Centre, Universit of Sdne 6 3 Trigonometric Ratios in a Right Angled Triangle If ou have met trigonometr before, ou probabl learned definitions of sin θ, cos θ and tan θ which were epressed as ratios of the sides of a right angled triangle. These definitions are repeated here, just to remind ou, but we shall go on, in the net section, to give a much more useful definition. 3. Definition of sine, cosine and tangent In a right angled triangle, the side opposite to the right angle is called the hpotenuse. If we choose one of the other angles and label it θ, the other sides are often called opposite (the side opposite to θ) and adjacent (the side net to θ). Hpotenuse θ Adjacent pposite For a given θ, there is a whole famil of right angled triangles, that are triangles of different sizes but are the same shape. θ θ θ For each of the triangles above, the ratios of corresponding sides have the same values. The ratio adjacent hpotenuse has the same value for each triangle. This ratio is given a special name, the cosine of θ or cos θ. The ratio opposite hpotenuse has the same value for each triangle. This ratio is the sine of θ or sin θ. The ratio adjacent opposite takes the same value for each triangle. This ratio is called the tangent of θ or tan θ. Summarising, cos θ = sin θ = adjacent hpotenuse, opposite hpotenuse, tan θ = opposite adjacent.
Mathematics Learning Centre, Universit of Sdne 7 The values of these ratios can be found using a calculator. Remember, we are working in radians so our calculator must be in radian mode. 3.. Eercise Use our calculator to evaluate the following. Where appropriate, compare our answers with the eact values for the special trigonometric ratios given in the net section. a. sin b. tan c. cos d. tan 6 3 4 e. sin.5 f. tan 3 g. cos 6 h. sin 3 3. Some special trigonometric ratios You will need to be familiar with the trigonometric ratios of, and. 6 3 4 The ratios of and are found with the aid of an equilateral triangle ABC with sides of 6 3 length. BAC is bisected b AD, and ADC is a right angle. Pthagoras theorem tells us that the length of AD = 3. A ACD = 3. DAC = 6. 3 C D B cos 3 sin 3 tan 3 =, 3 =, = 3. cos 6 sin 6 tan 6 = =, 3, = 3. The ratios of are found with the aid of an isosceles 4 right angled triangle XYZ with the two equal sides of length. X Pthagoras theorem tells us that the hpotenuse of the triangle has length. cos 4 =, /4 sin 4 =, Z /4 Y tan 4 =.
Mathematics Learning Centre, Universit of Sdne 8 4 The Trigonometric Functions The definitions in the previous section appl to θ between 0 and, since the angles in a right angle triangle can never be greater than. The definitions given below are useful in calculus, as the etend sin θ, cos θ and tan θ without restrictions on the value of θ. 4. The cosine function Let s begin with a definition of cos θ. Consider a circle of radius, with centre at the origin of the (, ) plane. Let A be the point on the circumference of the circle with coordinates (, 0). A is a radius of the circle with length. Let P be a point on the circumference of the circle with coordinates (a, b). We can represent the angle between A and P, θ, bthe arc length along the unit circle from A to P. This is the radian representation of θ. θ Q P(a,b) A The cosine of θ is defined to be the coordinate of P. Let s, for the moment, consider values of θ between 0and. The cosine of θ is written cos θ, sointhe diagram above, cos θ = a. Notice that as θ increases from 0 to, cos θ decreases from to 0. Forvalues of θ between 0 and, this definition agrees with the definition of cos θ as the ratio adjacent hpotenuse of the sides of a right angled triangle. Draw PQ perpendicular A. In PQ, the hpotenuse P has length, while Q has length a. The ratio adjacent hpotenuse = a = cos θ. The definition of cos θ using the unit circle makes sense for all values of θ. For now, we will consider values of θ between 0 and. The coordinate of P gives the value of cos θ. When θ =,Pison the ais, and it s coordinate is zero. As θ increases beond,pmoves around the circle into the second quadrant and therefore it s coordinate will be negative. When θ =, the coordinate is. P P P θ A θ θ P θ cos θ is positive cos =0 cos θ negative cos =
Mathematics Learning Centre, Universit of Sdne 9 As θ increases further, P moves around into the third quadrant and its coordinate increases from to0. Finall as θ increases from 3 to the coordinate of P increases from 0 to. P P cos θ is negative cos 3 P =0 cos θ positive cos = P 4.. Eercise. Use the cosine (cos) ke on our calculator to complete this table. (Make sure our calculator is in radian mode.) θ 0 6 4 3 5 3 3 4 cos θ θ 5 7 6 6 5 4 4 3 3 5 3 7 4 cos θ. Using this table plot the graph of = cos θ for values of θ ranging from 0 to. 4. The sine function The sine of θ is defined using the same unit circle diagram that we used to define the cosine. P(a,b) The sine of θ is defined to be the coordinate of P. θ Q A The sine of θ is written as sin θ, sointhe diagram above, sin θ = b. Forvalues of θ between 0and, this definition agrees with the definition of sin θ as the ratio opposite hpotenuse of sides of a right angled triangle. In the right angled triangle QP, the hpotenuse P has length while PQ has length b. The ratio opposite hpotenuse = b = sin θ. This definition of sin θ using the unit circle etends to all values of θ. consider values of θ between 0 and. Here, we will
Mathematics Learning Centre, Universit of Sdne 0 As P moves anticlockwise around the circle from A to B, θ increases from 0 to. When PisatA,sinθ =0,and when P is at B, sin θ =. Soasθ increases from 0 to, sin θ increases from 0 to. The largest value of sin θ is. As θ increases beond, sin θ decreases and equals zero when θ =. As θ increases beond, sin θ becomes negative. P P P θ A θ θ P θ sin θ is positive sin = sin θ positive sin =0 P P sin θ is negative sin 3 P = sin θ negative sin =0 P 4.. Eercise. Use the sin ke on our calculator to complete this table. Make sure our calculator is in radian mode. θ 0 0. 0.4 0.6 0.8..4.6 sin θ θ.4.8 3. 3.6 4.0 4.6 5.4 6. sin θ. Plot the graph of the = sin θ using the table in the previous eercise. 4.3 The tangent function We can define the tangent of θ, written tan θ, interms of sin θ and cos θ. tan θ = sin θ cos θ. Using this definition we can work out tan θ for values of θ between 0 and. You will
Mathematics Learning Centre, Universit of Sdne be asked to do this in Eercise 3.3. In particular, we know from this definition that tan θ is not defined when cos θ =0. This occurs when θ = or θ = 3. When 0 <θ< this definition agrees with the definition of tan θ as the ratio opposite adjacent of the sides of a right angled triangle. As before, consider the unit circle with points, A and P as shown. Drop a perpendicular from the point P to A which intersects A at Q. As before P has coordinates (a, b) and Q coordinates (a, 0). P(a,b) opposite adjacent = PQ Q (in triangle PQ) θ Q A = b a = sin θ cos θ = tan θ. If ou tr to find tan using our calculator, ou will get an error message. Look at the definition. The tangent of is not defined as cos =0. For values of θ near, tan θ is ver large. Tr putting some values in our calculator. (eg.570796. Tr tan(.57), tan(.5707), tan(.57079).) 4.3. Eercise. Use the tan ke on our calculator to complete this table. Make sure our calculator is in radian mode. θ 0 0. 0.4 0.6 0.8.0..4.50.65.4 tan θ θ.8 3. 3.6 4.0 4.4 4.6 4.65 4.78 5.0 5.6 6.0 6.8 tan θ. Use the table above to graph tan θ.
Mathematics Learning Centre, Universit of Sdne Your graph should look like this for values of θ between 0 and. Notice that there is a vertical asmptote at θ =. This is because tan θ is not defined at θ =. You will find another vertical asmptote at θ = 3. When θ = 0or, tan θ =0. Forθ greater than 0 and less than, tan θ is positive. For values of θ greater than and less than, tan θ is negative. / 4.4 Etending the domain The definitions of sine, cosine and tangent can be etended to all real values of θ in the following wa. 5 = + corresponds to the arc length of 4 revolutions around the unit circle going anticlockwise from A to B. Since B has coordinates (0, ) we can use the previous definitions to get: sin 5 =, cos 5 =0, B A tan 5 is undefined. Similarl, 6.546 = + 0.546, sin( 6) sin( 0.546 ) 0.9, cos( 6) 0.96, tan( 6) 0.30. B A
Mathematics Learning Centre, Universit of Sdne 3 4.4. Eercise Evaluate the following trig functions giving eact answers where ou are able.. sin 5. tan 3 3. cos 5 4. tan 4 5. sin 3 6 3 6 Notice The values of sine and cosine functions repeat after ever interval of length. Since the real numbers, +,, +4, 4 etc differ b a multiple of, the correspond to the same point on the unit circle. So, sin = sin( +) =sin( ) = sin( + 4) = sin( 4) etc. We can see the effect of this in the functions below and will discuss it further in the net chapter. sin θ 0 θ cosθ 0 θ The tangent function repeats after ever interval of length. tanθ 0 θ
Mathematics Learning Centre, Universit of Sdne 4 5 Graphs of Trigonometric Functions In this section we use our knowledge of the graphs = sin and = cos to sketch the graphs of more comple trigonometric functions. sin 0 cos 0 Let s look first at some important features of these two graphs. The shape of each graph is repeated after ever interval of length. This makes sense when we think of the wa we have defined sin and cos using the unit circle. We sa that these functions are periodic with period. The sin and cos functions are the most famous eamples of a class of functions called periodic functions. Functions with the propert that f() =f( + a) for all are called periodic functions. Such a function is said to have period a. This means that the function repeats itself after ever interval of length a. Note that ou can have periodic functions that are not trigonometric functions. eample, the function below is periodic with period. For -4-3 - - 0 3 4
Mathematics Learning Centre, Universit of Sdne 5 The values of the functions = sin and = cos oscillate between and. We sa that = sin and = cos have amplitude. A general definition for the amplitude of an periodic function is: The amplitude of a periodic function is half the distance between its minimum and maimum values. Also, the functions = sin and = cos oscillate about the -ais. We refer to the -ais as the mean level of these functions, or sa that the have a mean level of 0. We notice that the graphs of sin and cos have the same shape. The graph of sin looks like the graph of cos shifted to the right b units. We sa that the phase difference between the two functions is. ther trigonometric functions can be obtained b modifing the graphs of sin and cos. 5. Changing the amplitude Consider the graph of the function =sin. = sin = sin 0 The graph of =sin has the same period as = sin but has been stretched in the direction b a factor of. That is, for ever value of the value for =sin is twice the value for = sin. So, the amplitude of the function =sin is. Its period is. In general we can sa that the amplitude of the function = a sin is a, since in this case = a sin oscillates between a and a. What happens if a is negative? See the solution to number 3 of the following eercise. 5.. Eercise Sketch the graphs of the following functions.. =3cos. = sin 3. = 3 cos
Mathematics Learning Centre, Universit of Sdne 6 5. Changing the period Let s consider the graph of = cos. To sketch the graph of = cos, first think about some specific points. We will look at the points where the function = cos equals 0 or ±. = cos = cos 0 cos = when =0, socos = when =0, ie =0. cos =0 when =, so cos =0 when =, ie =. 4 cos = when =, so cos = when =, ie =. cos =0 when = 3, so cos =0 when = 3, ie = 3. 4 cos = when =, so cos = when =, ie =. As we see from the graph, the function = cos has a period of. The function still oscillates between the values and, so its amplitude is. Now, let s consider the function = cos. Again we will sketch the graph b looking at the points where = cos equals 0 or ±. = cos = cos / 0 3 4 cos = when =0, socos = when =0, ie =0. cos =0 when =, so cos =0 when =, ie =. cos = when =, so cos = when =, ie =. cos =0 when = 3, so cos =0 when = 3, ie =3. cos = when =, so cos = when =, ie =4.
Mathematics Learning Centre, Universit of Sdne 7 In this case our modified function = cos has period 4. It s amplitude is. What happens if we take the function = cos ω where ω>0? = cos ω /ω /ω 0 /ω /ω cos = when =0, socos ω = when ω =0, ie =0. cos =0 when =, so cos ω =0 when ω =, ie = ω. cos = when =, so cos ω = when ω =, ie = ω. cos =0 when = 3, so cos ω =0 when ω = 3, ie = 3 ω. cos = when =, so cos ω = when ω =, ie = ω. In general, if we take the function = cos ω where ω>0, the period of the function is. ω What happens if we have a function like = cos( )? See the solution to number 3 of the following eercise. 5.. Eercise Sketch the graphs of the following functions. function. Give the amplitude and period of each. = cos. =sin 4 3. = cos( ) 4. = sin( ) 5. =3sin 6. = sin 7. Find the equation of a sin or cos function which has amplitude 4 and period.
Mathematics Learning Centre, Universit of Sdne 8 5.3 Changing the mean level We saw above that the functions = sin and = cos both oscillate about the -ais which is sometimes refered to as the mean level for = sin and = cos. We can change the mean level of the function b adding or subtracting a constant. For eample, adding the constant to = cos gives us = cos +and has the effect of shifting the whole graph up b units. So, the mean level of = cos +is. 0 Similarl, we can shift the graph of = cos down b two units. In this case, we have = cos, and this function has mean level. 0 - = cos In general, if d>0, the function = cos + d looks like the function = cos shifted up b d units. If d>0, then the function = cos d looks like the function = cos shifted down b d units. If d<0, sa d =, the function = cos + d = cos +( ) can be writen as = cos soagain looks like the function = cos shifted down b units. 5.3. Eercise Sketch the graphs of the following functions.. = sin +3. =cos 3. Find a cos or sin function which has amplitude, period, and mean level.
Mathematics Learning Centre, Universit of Sdne 9 5.4 Changing the phase Consider the function = sin( 4 ). = sin = sin ( - /4 ) 0 /4 To sketch the graph of = sin( )weagain use the points at which = sin is 0 or 4 ±. sin =0 when =0, sosin( 4 )=0 when 4 =0, ie =0+ 4 = 4. sin = when =, so sin( 4 )= when 4 =, ie = + 4 = 3 4. sin =0 when =, so sin( 4 )=0 when 4 =, ie = + 4 = 5 4. sin = when = 3, so sin( 4 )= when 4 = 3, ie = 3 + 4 = 7 4. sin =0 when =, so sin( 4 )=0 when 4 =, ie = + 4 = 9 4. The graph of = sin( )looks like the graph of = sin shifted 4 4 We sa that there has been a phase shift to the right b. 4 units to the right. Now consider the function = sin( + 4 ). = sin /4 = sin ( + /4 ) 0 sin =0 when =0, sosin( + )=0 4 when + =0, 4 ie =0 =. 4 4 sin = when =, so sin( + )= 4 when + =, 4 ie = =. 4 4 sin =0 when =, so sin( + )=0 4 when + =, 4 ie = = 3. 4 4 sin = when = 3, so sin( + )= when + = 3, 4 4 ie = 3 = 5. 4 4 sin =0 when =, so sin( + )=0 4 when + =, 4 ie = = 7. 4 4
Mathematics Learning Centre, Universit of Sdne 0 The graph of = sin( + 4 )looks like the graph of = sin shifted 4 units to the left. More generall, the graph of = sin( c) where c>0 can be drawn b shifting the graph of = sin to the right b c units. = sin = sin ( - c) for c>0 0 c Similarl, the graph of = sin( + c) where c>0 can be drawn b shifting the graph of = sin to the left b c units. = sin c = sin ( + c) for c>0 0 What happens to = sin( c) ifc<0? Consider, for eample, what happens when c =.Inthis case we can write = +. So, = sin( )=sin( + ) and 4 4 4 4 4 we have a shift to the left b units as before. 4 5.4. Eercise Sketch the graphs of the following functions.. = cos( ). = sin ( + ) (Hint: For this one ou ll need to think about the period as well.) 3. =3cos( + ) 4. = 3 cos( + )+(Hint: Use the previous eercise.) 5. Find a sin or cos function that has amplitude, period, and for which f(0) =.
Mathematics Learning Centre, Universit of Sdne 6 Derivatives of Trigonometric Functions This Chapter assumes ou have a knowledge of differential calculus. studied differential calculus before, go on to the net chapter. If ou have not 6. The calculus of trigonometric functions When differentiating all trigonometric functions there are two things that we need to remember. d sin d = cos d cos d = sin. f course all the rules of differentiation appl to the trigonometric functions. Thus we can use the product, quotient and chain rules to differentiate combinations of trigonometric functions. For eample, tan = sin,sowecan use the quotient rule to calculate the derivative. cos f() = tan = sin cos f () = cos.(cos ) sin.( sin ) (cos ) = cos + sin cos = cos (since cos + sin =) = sec. Note also that so it is also true that Eample Differentiate f() =sin. Solution cos + sin cos = cos cos + sin cos =+tan d d tan = sec =+tan. f() =sin is just another wa of writing f() =(sin ). This is a composite function, with the outside function being ( ) and the inside function being sin. Bthe chain rule,
Mathematics Learning Centre, Universit of Sdne f () =(sin) cos =sin cos. Alternativel, setting u = sin we get f(u) =u and df () d df (u) = du du d =u du d =sin cos. Eample Differentiate g(z) =cos(3z +z + ). Solution Again we should recognise this as a composite function, with the outside function being cos( ) and the inside function being 3z +z +.Bthe chain rule g (z) = sin(3z +z +) (6z +)= (6z +)sin(3z +z +). Eample Differentiate f(t) = et sin t. Solution B the quotient rule f (t) = et sin t e t cos t sin t = et (sin t cos t) sin. t Eample Use the quotient rule or the composite function rule to find the derivatives of cot, sec, and cosec. Solution These functions are defined as follows: cot = cos sin sec = cos csc = sin. B the quotient rule d cot = sin cos d sin Using the composite function rule d sec d(cos ) = d d d csc d(sin ) = d d 6.. Eercise = sin. = (cos ) ( sin ) = sin cos. = (sin ) cos = cos sin. Differentiate the following:. cos 3. sin(4 +5) 3. sin 3 4. sin cos 5. sin 6. cos( +) 7. sin 8. sin 9. tan( ) 0. sin
Mathematics Learning Centre, Universit of Sdne 3 7 A Brief Look at Inverse Trigonometric Functions Before we define the inverse trigonometric functions we need to think about eactl what we mean b a function. A function f from a set of elements A to a set of elements B is a rule that assigns to each element in A eactl one element f() inb. = sin, = cos and = tan are functions in the sense of this definition with A and B being sets of real numbers. Let s look at the function = cos. Asou can see, whatever value we choose for, there is onl ever one accompaning value for. For eample, when =, =0. cos 0 Now lets consider the following question. Suppose we have cos =0.5 and we want to find the value of. cos 0 As ou can see from the diagram above, there are (infinitel) man values of for which = cos =0.5. Indeed, there are infinitel man solutions to the equation cos = a where a. (There are no solutions if a is outside this interval.) If we want an interval for where there is onl one solution to cos = a for a, then we can choose the interval from 0 to. Wecould also choose the interval or man others. It is a mathematical convention to choose 0. cos 0
Mathematics Learning Centre, Universit of Sdne 4 In the interval from 0 to, wecan find a unique solution to the equation cos = a where a is in the interval a. We write this solution as = cos a. Another wa of saing this is that is the number in the interval 0 whose cosine is a. Now that we have found an interval of for which there is onl one solution of the equation cos = a where a, we can define an inverse function for cos. 7. Definition of the inverse cosine function We will describe an inverse function for cos where 0. For, f () =cos () isthe number in the interval 0 to whose cosine is. So, we have: cos ( ) = since cos =, cos ( )= 4 since cos 4 =, cos (0) = since cos =0, cos () = 0 since cos 0 =. Provided we take the function cos where 0, we can define an inverse function f () =cos (). This function is defined for in the interval, and is sketched below. f () / 0 Inverse functions for sin where, and tan where << can be defined in a similar wa. For amore detailed discussion of inverse functions see the Mathematics Learning Centre booklet: Functions.
Mathematics Learning Centre, Universit of Sdne 5 7.. Eercise. The inverse function, f () =sin (), is defined for the function f() =sin where. Complete the following table of values. sin ( ) = since sin =, sin ( ) = since sin 4 =, sin ( ) = since sin 0 = 0, sin ( ) = since sin 3 = 3, sin ( ) = since sin =.. Sketch the function f () =sin () using the values in the previous eercise.
Mathematics Learning Centre, Universit of Sdne 6 8 Solutions to Eercises Eercise.. a. 30 or 6 radians b. radian or 57.3 c. 0 or 3 radians d. 3 4 radians or 35 e. radians or 4.6 f. 4 3 radians or 40 g. 70 or 3 radians h. radians or 57.3 i. radians or 90 Eercise 3.. a. sin 6 =0.5 b. tan =.557 c. cos 3 =0.5 d. tan 4 = e. sin.5 =0.997 f. tan 3 =.73 g. cos 6 =0.866 h. sin 3 =0.866 Eercise 4... θ 0 6 4 cos θ 0.97 0.87 0.7 0.5 0.6 0 0.5 0.7 θ 5 7 6 6 5 4 3 4 3 5 3 5 3 3 3 4 7 4 cos θ 0.87 0.87 0.7 0.5 0 0.5 0.7
Mathematics Learning Centre, Universit of Sdne 7..00 cosθ 0 θ -.00 Eercise 4... θ 0 0. 0.4 0.6 0.8..4.6 sin θ 0 0.0 0.39 0.56 0.7 0.84 0.93 0.99.00 θ.4.8 3. 3.6 4.0 4.6 5.4 6.5 sin θ 0.9 0.68 0.33 0.06 0.44 0.76 0.99 0.77 0.083..00 sin θ 0 θ -.00 Eercise 4.3.. θ 0 0. 0.4 0.6 0.8.0..4 tan θ 0 0.0 0.4 0.68.03.56.57 5.80 θ.50.65.4.8 3. 3.6 4.0 tan θ 4.0.60.9 0.9 0.36 0.06 0.49.6 θ 4.4 4.6 4.65 4.78 5.0 5.6 6.0 6.8 tan θ 3.0 8.86 6.0 4.77 3.38 0.8 0.9 0.00
Mathematics Learning Centre, Universit of Sdne 8. tanθ 0.0 0 θ -0.0 Eercise 4.4.. sin 5 = sin 3 = since 5 and 3 differ b 6 =3.. tan 3 6 = tan 6 = 3. 3. cos 5 = 0.76. 4. tan 4 3 = tan 3 = tan 4 3 = tan 3 = 3. 5. sin 3 3 = sin 5 3 = sin 3 = 3. Eercise 5... =3cos.00 0 - - -.00 The amplitude of this function is 3.
Mathematics Learning Centre, Universit of Sdne 9. = sin.00 0 - - -.00 The amplitude of this function is. 3. = 3 cos.00 0 - - -.00 The amplitude of this function is 3. Notice that it is just the graph of =3cos (in dots) reflected in the -ais. Eercise 5... = cos.00 0 3 4 -.00 This function has period 4 and amplitude.
Mathematics Learning Centre, Universit of Sdne 30. =sin 4.00 0-5 -4-3 - - 3 4 5 -.00 This function has period 8 and amplitude. 3. = cos( ).00 0 - - -.00 This function has period and amplitude. Notice that this function looks like = cos. The cosine function is an even function, so, cos( ) =cos for all values of. 4. = sin( ).00 0 - - -.00 This function has period and amplitude. Notice that this function is a reflection of = sin in the -ais. The sine function is an odd function, so, sin( ) = sin for all values of.
Mathematics Learning Centre, Universit of Sdne 3 5. =3sin.00-3.00 -.00 -.00.00.00 3.00 -.00 This function has period and amplitude 3. 6. = sin.00-3.00 -.00 -.00.00.00 3.00 -.00 This function has period and amplitude. 7. =4cos (solid line) or =4sin (in dashes). There are man other solutions. 4.00.00-3.00 -.00 -.00.00.00 3.00 -.00-4.00
Mathematics Learning Centre, Universit of Sdne 3 Eercise 5.3.. = sin +3 4.00.00 0 - - This function has amplitude, period and mean level 3.. =cos -6.00-4.00 -.00.00 4.00 6.00 -.00-4.00 This function has amplitude, period and mean level. 3. =cos (solid line) or =sin (in dashes) or man others. -3.00 -.00 -.00.00.00 3.00 -.00-4.00
Mathematics Learning Centre, Universit of Sdne 33 Eercise 5.4.. = cos( ).00 0 - - -.00 The function = cos (in dots) has been shifted to the right b this function looks like = sin. units. Notice that. = sin ( + ).00 0 - - -.00 The period of this function is. The function = sin (in dots) has been shifted to the left b units. 3. =3cos( + ).00 0 - - -.00 This function has period and amplitude 3. The function =3cos (in dots) has been shifted to the left b units.
Mathematics Learning Centre, Universit of Sdne 34 4. = 3 cos( + )+ 4.00.00 0 - - This function has period and amplitude 3. The function 3 cos( + ) (see prevoius eercise) has been reflected in the -ais and shifted up b units. 5. =sin +(in dashes) or =cos (solid line) or man others. 4.00.00 0 - - -.00 Eercise 6.... 3. 4. 5. 6. d (cos 3) = 3 sin 3. d d (sin(4 + 5)) = 4 cos(4 +5). d d d (sin3 )=3sin cos. d d (sin cos ) =sin ( sin )+cos (cos ) =cos sin. d d ( sin ) = cos + sin. d d (cos( + )) = sin( + )() = sin( +).
Mathematics Learning Centre, Universit of Sdne 35 7. 8. 9. 0. d d (sin cos sin )=. d d (sin )=(cos )( )= cos. d d (tan( )) = (sec ( ))( sec )=. d d ( sin )= (cos )( )+(sin )( )= 3 (cos + sin ). Eercise 7... sin ( ) = since sin = sin ( )= 4 since sin 4 = sin (0) = 0 since sin 0 = 0 sin ( 3 )= since sin = 3 3 3 sin () = since sin =. f () =sin () /.00 -.00.00 -.00 /