Lesson 4-3 Basic Basic Trigonometric Identities Identities Vocabular identit BIG IDEA If ou know cos, ou can easil fi nd cos( ), cos(90º - ), cos(180º - ), and cos(180º + ) without a calculator, and similarl for sin and tan. An identit is an equation that is true for all values of the variables for which the epressions on each side are defined. There are five theorems in this lesson; all are identities. The Pthagorean Identit The first identit we derive in this lesson comes directl from the equation 2 + 2 = 1 for the unit circle. Because, for ever, the point P = (cos, sin ) is on the unit circle, the distance from P to (0, 0) must be 1. Using the Distance Formula, (cos - 0) 2 + (sin 0) 2 = 1. Squaring both sides of the equation gives (cos ) 2 + (sin ) 2 = 1. This argument proves a theorem called the Pthagorean Identit. Pthagorean Identit Theorem For ever, cos 2 + sin 2 = 1. An abbreviated version of (cos ) 2 is cos 2, the square of the cosine of. Similarl, (sin ) 2 is written sin 2 and (tan ) 2 is written tan Notice that we do not write cos 2 for (cos ) QY1 The name of the above identit comes from the Pthagorean Theorem because in the first quadrant, as shown at the right, cos and sin are the sides of a right triangle with hpotenuse 1. Among other things, the Pthagorean Identit enables ou to obtain either cos or sin if ou know the other. P = (cos, sin ) 1 = cos = sin P = (, ) = (cos, sin ) Mental Math l P W O E Q N S True or False a. POE and POW are complementar. b. POE and PON are supplementar. c. m POE = m QOW d. m POW = π m POE QY1 Which two epressions are equal? A tan 2 B tan 2 C (tan ) 2 Basic Trigonometric Identities 23
Chapter 4 Eample 1 If cos = 3_, fi nd sin. Solution Substitute into the Pthagorean Identit. ( 3_ ) 2 + sin 2 = 1 9_ 2 + sin2 = 1 sin 2 = 16 _ 2 sin = ± 4_ Thus, sin = 4_ or sin = 4_. Check Refer to the unit circle. The vertical line = 3_ intersects the unit circle in two points. One is in the fi rst quadrant, in which case the -coordinate (sin ) is 4_. The other is in the fourth quadrant, where sin is 4_. 3 4, 3 4, - = 3 (1, 0) QY2 The Smmetr Identities Man other properties of sines and cosines follow from their definitions and the smmetr of the unit circle. Recall that a circle is smmetric to an line through its center. This means that the reflection image of an point over one of these lines also lies on the circle. QY2 If sin = 0.6, what is cos? Activit 1 MATERIALS DGS or graph paper, compass, and protractor Step 1 Draw a unit circle on a coordinate grid. Plot the point A = (1, 0). Pick a value of between 0º and 90º. Let a point P in the fi rst quadrant be the image of A under the rotation R. Find the values of cos and sin from the coordinates of P. A sample is shown at the right. Step 2 Refl ect P over the -ais. Call its image Q. Notice that Q is the image of (1, 0) under a rotation of magnitude. Consequentl, Q = (cos( ), sin( )). a. What are the values of cos( ) and sin( ) for our point Q? b. How are cos and cos( ) related? What about sin and sin( )? Step 3 Rotate our point P 180º around the circle. Call its image H. Notice that H is the image of (1, 0) under a rotation of magnitude (180º + ). Consequentl, H = (cos(180º + ), sin(180º + )). a. What are the values of cos(180º + ) and sin(180º + ) for our point H? b. How are cos and cos(180º + ) related? How are sin and sin(180º + ) related? 236 Trigonometric Functions
Step 4 Use a calculator to fi nd cos and sin for our value of in Step 1. Then fi nd cos( ) and sin( ), and also cos(180º + ) and sin(180º + ). Eplain an differences between the values displaed b the calculator and what ou found in Steps 2 and 3. Save our work for Activit 2. Activit 1 is based on the following ideas: When a point P on the unit circle is reflected over either ais, or when it is rotated through a half turn, either the coordinates of the three images are equal to the coordinates of P or the are opposites of the coodinates of P. The magnitudes of the rotations that map (1, 0) onto these points are (for P at the right), (for Q), 180º + (for H), and 180º - (for J ). So the sines and cosines of these magnitudes are either equal or opposites. J H P - A = (1, 0) Q Sines and Cosines of Opposites Rotations of magnitude and go in opposite directions. The two rotation images are reflection images of each other over the -ais. Thus the have the same first coordinates (cosines) but opposite second coordinates (sines). It follows that the ratios of the -coordinates to the -coordinate are opposites. This argument proves the following theorem. R (1,0) = (cos, sin ) = P R - (1,0) = (cos, -sin ) = Q - (1, 0) Opposites Theorem For all, cos( ) = cos, sin( ) = sin, and tan( ) = tan. Sines and Cosines of + 180º or + π Adding 180º or π to the argument of a trigonometric function is equivalent to rotating halfwa around the unit circle. Half-Turn Theorem For all, cos(180º + ) = cos = cos(π + ) sin(180º + ) = sin = sin(π + ) and tan(180º + ) = tan = tan(π + ). Proof Let A = (1, 0) and let P = R (A) = R (1, 0) = (cos, sin ). Now let Q be the image of P under R 180º. Because R 180º maps (a, b) to ( a, b), Q has coordinates ( cos, sin ). But Q is also the image of A under a rotation of magnitude 180º +. So Q also has coordinates (cos(180º + ), sin(180º + )). Equating the two ordered pairs for Q proves the fi rst two parts of the theorem. The third part follows b dividing the second equation b the fi rst. 180 (-cos, -sin ) = Q O P = (cos, sin ) A = (1, 0) Basic Trigonometric Identities 237
Chapter 4 Sines and Cosines of Supplements Recall that if an angle has measure, then its supplement has measure 180º -, that is, π -. Activit 1 shows that the values of the trigonometric functions of and 180º - are related, as stated in the following theorem. Supplements Theorem For all, sin(180º - ) = sin = sin(π - ) cos(180º - ) = cos = cos(π - ) and tan(180º - ) = tan = tan(π - ). Proof Let P = (cos, sin ). Let Q be the refl ection image of P over the -ais, as in the diagram at the right. Because the refl ection image of (, ) over the -ais is (, ), Q = ( cos, sin ). Recall from geometr that refl ections preserve angle measure, so m QOB = m POA =. Also, since!aoq and!qob are a linear pair, m!aoq = 180º -. So, b the defi nitions of cosine and sine, Q = (cos(180º - ), sin(180º - )). Thus, (cos(180º - ), sin(180º - )) = ( cos, sin ). The -coordinates are equal, so cos(180º - ) = cos. Likewise, the -coordinates are equal, so sin(180º - ) = sin. Dividing the latter of these equations b the former gives the third part of the Supplements Theorem, tan(180º - ) = tan. QY3 (cos (180 - ), sin (180 - )) = (-cos, sin ) = Q (-1, 0) = B Eample 2 Given that sin 10º 0.1736, fi nd a value of other than 10º and between 0º and 360º for which sin = 0.1736. Solution Think: sin 10º is the second 1 coordinate of the image of (1, 0) under R 10º. What other rotation will give the same second coordinate? It is the rotation that gives the refl ection image of the point P in the diagram at the right. That rotation has magnitude 180º 10º, or 170º. So sin 170º = sin 10º = 0.1736, and = 170º. P = (-cos 10, sin 10 ) -1 0-1 O 180 - QY3 P = (cos, sin ) A = (1, 0) Suppose sin = 0.496 and cos = 0.868. Without using a calculator, fi nd a. sin(π ). b. cos(180º ). P = (cos 10, sin 10 ) 1 238 Trigonometric Functions
If the requirement that 0º < < 360º in Eample 2 is relaed, there are other answers. Because ou can add or subtract 360º to the magnitude of an rotation and get the same rotation, sin 10º = sin 170º = sin 30º = sin( 190º). Also, in radians, sin ( 18) π_ = sin ( _ 17π 18 ) = sin ( _ 3π 18 ) = sin ( _ 19π 18 ). QY4 Sines and Cosines of Complements QY4 Given that cos 10º 0.9848, fi nd a value of other than 10º for which cos = 0.9848. Activit 2 MATERIALS DGS or graph paper Step 1 Begin with the graph from Step 3 of Activit 1. Hide points H and Q. Draw the line =. Again pick a value of between 0ºand 90º and let P = R (1, 0). Find cos and sin for our value of. Step 2 Refl ect point P over = and call its image K. From our knowledge of refl ections, what are the coordinates of K? Step 3 In terms of, what is the magnitude of the rotation that maps (1, 0) onto K? (Hint: K is as far from A along the circle as P is from the point (0, 1).) Answer in both degrees and radians. Step 4 Develop an identit that relates the sine and cosine of our answers to Step 3 to the sine and cosine of. If an angle has measure, then its complement has measure 90º - or π_ -. Activit 2 shows that the sines and cosines of and 90º - are 2 related. Complements Theorem For all, sin(90º - ) = cos = sin ( π _ 2 - ) and cos(90º - ) = sin = cos ( π _ 2 - ). These theorems can help etend our knowledge of circular functions. Eample 3 Given that sin 30º = 1_ 2, compute the eact value of each function below. a. cos 60º b. cos 30º c. sin 10º d. cos 210º e. sin( 30º) (continued on net page) Basic Trigonometric Identities 239
Chapter 4 Solution a. Use the Complements Theorem. cos 60º = sin(90º - 60º) = sin 30º. So cos 60º = 1_ b. Use the Pthagorean Identit Theorem. sin 2 30º + cos 2 30º = 1. So cos 2 30º = 1 - ( 2) 1_ 2 = 3_ 4. Thus, cos 30º = ± 3_ 4 = ± _! 3 However, we know cos 30º is positive, so cos 30º = _! 3 c. Use the Supplements Theorem. sin 10º = sin(180º - 10º) = sin 30º. So sin 10º = 1_ d. Use the Half-Turn Theorem. cos 210º = cos(180º + 30º) = cos 30º = _! 3 e. Use the Opposites Theorem. sin( 30º) = sin 30º = 1_ In using these identities, ou should also be able to use the unit circle to do a visual check of our answers or to derive a propert if ou forget one. Questions COVERING THE IDEAS 1. True or False When = 180º, cos 2 + sin 2 = 1. 2. a. If sin = _ 24, what are two possible values of cos? 2 b. Draw a picture to justif our answers to Part a. 3. If tan = 3, what is tan( )? 4. a. True or False cos 14º = cos( 14º) b. Justif our answer to Part a with a unit circle diagram. In and 6, refer to the figure at the right. P = R (1, 0), P = r -ais (P), P = R 180º (P), and P = r -ais (P).. Which coordinates equal cos(180º )? 6. Which coordinates equal sin(180º + )? 7. True or False sin( ) = sin In 8 and 9, sin = 1_ 3. Evaluate without using a calculator. 8. sin( ) 9. sin(180º ) P = (c, d) P = (e, f) P = (a, b) (1, 0) P = (g, h) 10. Using what ou know about sin(180º ) and cos(180º ), eplain wh tan(180º ) = tan. 11. Use a calculator to verif the three parts of the Supplements Theorem when = 146.º. In 12 and 13, suppose cos = _ 13. Evaluate without using a calculator. 12. cos(180º + ) 13. sin(90º ) In 14 and 1, tan = k. Evaluate. 14. tan( ) 1. tan(180º ) 240 Trigonometric Functions
16. Cop the table below, filling in the blank entries and completing the diagrams, to summarize the theorems in this lesson. 180º 180º + 90º 180-180 + 90 - Diagram - cos cos??? sin? sin?? tan tan? tan? APPLYING THE MATHEMATICS In 17 21, the displa below shows inputs and outputs of a CAS in degree mode. What theorem justifies each statement? 17. 18. 19. 20. 21. 22. Prove that sin(π ) = sin π sin is not an identit. In 23 26, from the fact that sin 18º = _ 1 4, find each value. 23. sin 162º 24. sin( 18º) 2. sin _ 11π 26. cos 2π_ 10 REVIEW In 27 29, without using a calculator, give eact values. (Lesson 4-2) 27. sin 90º 28. cos 810º 29. tan(90º + 90º) 30. Convert _ 11 clockwise revolutions to degrees. (Lesson 4-1) 6 31. a. What is the magnitude of the rotation of the minute hand of a clock in 6 minutes? b. What is the measure of the angle between the minute hand and the second hand of a clock at eactl 12:06 A.M.? (Lesson 4-1) 32. Find an equation for the image of the graph of = 2 under the scale change (, ) ( 1_ 2, ). (Lesson 3-) EXPLORATION 33. Use a calculator to investigate whether _ cos2 = 1 + sin is 1 sin an identit. Tr to prove our conclusion, either b providing a countereample or b using definitions and properties. QY ANSWERS 1. A and C 2. cos = ± 1 0.6 2 = ±0.8 3. a. 0.496 b. 0.868 4. Answers var. Samples: 10º, 370º, 30º Basic Trigonometric Identities 241