The Calculus of Trigonometric Functions the Calculus of Trigonometric Functions – a Guide for Teachers (Years 11–12)
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1 Supporting Australian Mathematics Project 2 3 4 5 6 12 11 7 10 8 9 A guide for teachers – Years 11 and 12 Calculus: Module 15 The calculus of trigonometric functions The calculus of trigonometric functions – A guide for teachers (Years 11–12) Principal author: Peter Brown, University of NSW Dr Michael Evans, AMSI Associate Professor David Hunt, University of NSW Dr Daniel Mathews, Monash University Editor: Dr Jane Pitkethly, La Trobe University Illustrations and web design: Catherine Tan, Michael Shaw Full bibliographic details are available from Education Services Australia. Published by Education Services Australia PO Box 177 Carlton South Vic 3053 Australia Tel: (03) 9207 9600 Fax: (03) 9910 9800 Email: [email protected] Website: www.esa.edu.au © 2013 Education Services Australia Ltd, except where indicated otherwise. You may copy, distribute and adapt this material free of charge for non-commercial educational purposes, provided you retain all copyright notices and acknowledgements. This publication is funded by the Australian Government Department of Education, Employment and Workplace Relations. Supporting Australian Mathematics Project Australian Mathematical Sciences Institute Building 161 The University of Melbourne VIC 3010 Email: [email protected] Website: www.amsi.org.au Assumed knowledge ..................................... 4 Motivation ........................................... 4 Content ............................................. 5 Review of radian measure ................................. 5 An important limit ..................................... 6 Differentiating trigonometric functions ......................... 8 Applications of the derivatives .............................. 12 Integrating trigonometric functions ........................... 13 Simple harmonic motion .................................. 15 Links forward .......................................... 19 Inverse trigonometric functions .............................. 19 The hyperbolic functions ................................. 21 History and applications ................................... 22 Answers to exercises ..................................... 25 The calculus of trigonometric functions Assumed knowledge The content of the modules: • Trigonometric functions and circular measure • Introduction to differential calculus • Integration. Motivation It is an interesting exercise to sit back and think about how we have developed the topic of trigonometry during the earlier secondary school years. We commenced by looking at ratios of sides in a right-angled triangle. This enabled us to find unknown sides and angles. We extended this to include non-right-angled triangles using the sine and cosine rules. In the module Trigonometric functions and circular measure, we redefined the sine and cosine functions in terms of the coordinates of points on the unit circle. This enabled us to define the sine and cosine of angles greater than 90± and to plot the graphs of the trigonometric functions and discover their periodic nature. The sine and cosine functions are used to model periodic phenomena in nature, such as waves, tides and signals. Indeed, these functions are used to model all sorts of oscillatory motion arising in a range of subjects, including economics and ecology. In this module, we continue this development by applying the ideas and techniques of calculus to the trigonometric functions. For example, if we wish to analyse the motion of a particle modelled by a trigonometric function, we can use calculus to find its velocity and acceleration. The simplicity of the results obtained by doing this is amazing and has wide-ranging impact in physics and electrical engineering, and indeed in any area in which periodic motion is being modelled. A guide for teachers – Years 11 and 12 • {5} Content Review of radian measure We saw in the module Trigonometric functions and circular measure that angles can be naturally defined using arc length. We define 1 radian (written as 1c ) to be the angle subtended in the unit circle by an arc length of one unit. y (0,1) 1 1 x (1,0) O (1,0) � (0,1) Since the circumference of the unit circle is 2¼ and the angle in one revolution is 360±, c we can relate the two units by 2¼ 360± or Æ ¼ 180±. Æ (As usual, we will drop the superscript c when it is clear that the angle under discussion is in radians.) Many commonly occurring angles can be expressed in radians as fractions ¼ 11¼ of ¼. For example, 60± and 330± . Æ 3 Æ 6 We have also seen that the arc length ` of a sector of a circle of radius r , containing an angle θ (in radians), is given by ` r θ, ℓ Æ θ r while the area A of the sector is given by O 1 A r 2θ. Æ 2 {6} • The calculus of trigonometric functions An important limit In order to apply calculus to the trigonometric functions, we will need to evaluate the fundamental limit sinx lim , x 0 x ! which arises when we apply the definition of the derivative of f (x) sinx. Æ It must be stressed that, from here on in this module, we are measuring x in radians. sinx You can see that attempting to substitute x 0 into is fruitless, since we obtain the Æ x indeterminate form ‘zero over zero’. On the other hand, if we try substituting say x 0.1 Æ (in radians), then the calculator gives 0.998..., and if we try x 0.01, we get 0.99998.... Æ sinx We guess that, as x approaches 0, the value of approaches 1. We will now give a nice x geometric proof of this fact. Consider a sector OAB of the unit circle containing an acute angle x, as shown in the following diagram. Drop the perpendicular BD to OA and raise a perpendicular at A to meet OB produced at C. Since OA OB 1, we have BD sinx and AC tanx. Æ Æ Æ Æ B C 1 x O D A 1 The area of the sector OAB is x. The sector OAB clearly contains the triangle OAB and 2 is contained inside the triangle OAC. Hence, comparing their areas, we have 1 1 1 1 Area OAB x Area OAC OA BD x OA AC 4 · 2 · 4 Æ) 2 ¢ · 2 · 2 ¢ 1 1 1 sinx x tanx Æ) 2 · 2 · 2 sinx x tanx. Æ) · · sinx ¼ Now, since tanx and since sinx 0 for 0 x , we can divide both inequalities Æ cosx È Ç Ç 2 by sinx to obtain x 1 1 . · sinx · cosx A guide for teachers – Years 11 and 12 • {7} Since cosx approaches 1 as x approaches 0, we see that x 1 as x 0. sinx ! ! Taking the reciprocal, it follows that sinx 1 as x 0. x ! ! (Here we have used the pinching theorem and the algebra of limits, as discussed in the module Limits and continuity.) Thus we can write sinx lim 1. x 0 x Æ ! Note that this means that, if the angle x is small, then sinx x. This fact is often used by ¼ physicists when analysing such things as a simple pendulum with small angle. Exercise 1 Using the limit above, prove that tanx lim 1. x 0 x Æ ! Other related limits can be found by manipulating this basic limit. Example Find sin2x lim . x 0 3x ! Solution We can write sin2x sin2x 2 . 3x Æ 2x £ 3 If we let u 2x, then as x 0, we have u 0. Hence Æ ! ! sin2x 2 sinu 2 lim lim . x 0 3x Æ 3 u 0 u Æ 3 ! ! {8} • The calculus of trigonometric functions Exercise 2 Find sin3x sin7x lim Å . x 0 5x ! Exercise 3 1 cosx 1 a Show that lim ¡ . x 0 x2 Æ 2 ! b Deduce that cosx 1 1 x2, for small x. ¼ ¡ 2 Differentiating trigonometric functions The derivative of sine Since the graph of y sinx is a smooth curve, we would like to find the gradient of the Æ tangent to the curve at any point on it. Before doing this, we derive a useful trigonometric identity that will assist us. Using the compound-angle formulas, we have sin(A B) sin(A B) sin A cosB cos A sinB (sin A cosB cos A sinB) Å ¡ ¡ Æ Å ¡ ¡ 2 cos A sinB. Æ If we put C A B and D A B, we can add these equations to obtain A 1 (C D) Æ Å Æ ¡ Æ 2 Å and subtract them to obtain B 1 (C D). Substituting these back, we obtain the sine Æ 2 ¡ difference formula: ³C D ´ ³C D ´ sinC sinD 2 cos Å sin ¡ . ¡ Æ 2 2 To find the derivative of sinx, we return to the first principles definition of the derivative of y f (x): Æ d y f (x h) f (x) lim Å ¡ . dx Æ h 0 h ! Substituting y sinx, we have Æ d y sin(x h) sinx lim Å ¡ . dx Æ h 0 h ! A guide for teachers – Years 11 and 12 • {9} Applying the sine difference formula, we have ¡ h ¢ ¡ h ¢ sin(x h) sinx 2 cos x sin lim Å ¡ lim Å 2 2 . h 0 h Æ h 0 h ! ! We can put u h . Then u 0 as h 0. So the limit becomes Æ 2 ! ! d y 2 cos(x u) sinu lim Å dx Æ u 0 2u ! ³ ´ ³ sinu ´ lim cos(x u) lim Æ u 0 Å £ u 0 u ! ! cosx 1 Æ £ cosx. Æ We can thus conclude that d (sinx) cosx. dx Æ This result is both simple and surprising, and students need to commit it to memory. The derivation above involved a number of ingredients and is often difficult for students the first time through. Derivatives of other trigonometric functions Now that the derivative of sine is established, we can use the standard rules of calculus — the chain, product and quotient rules — to proceed. Since cosx sin¡x ¼ ¢, we can apply the chain rule to see that Æ Å 2 d d ³ ¼ ´ (cosx) sin¡x ¢ dx Æ dx Å 2 ¼ cos¡x ¢ Æ Å 2 sinx.