Trigonometry Notes on Unit Circle Trigonometry.

Transcription

1 Trigonometr Notes on Unit Circle Trigonometr. Epanding the Definition of the Trigonometric Functions: As time went b and the concept of angles epanded beond acute and obtuse angles, it was discovered that a more general definition of the trigonometric function was needed. Definitions that allowed for an real number to be used as an angle. First lets remember the definition we had for right triangle, but let s make the hpotenuse have length. Definition of the Trigonometric Functions for Right Triangles w/ Hpotenuse Sine Cosecant opposite hpotenuse sin( θ ) csc( θ ) hpotenuse opposite Cosine Secant hpotenuse cos( θ ) sec( θ ) hpotenuse Tangent Cotangent opposite tan( θ ) cot( θ ) opposite Notice that the triangle fits within a quartercircle as show to the right. Now all we need do to is epand to the full circle and we have the general definitions. Definitions of the Trigonometric Functions Using the Unit Circle: Definition of the Trigonometric Functions Using the Unit Circle Sine Cosecant opposite hpotenuse sin( θ ) csc( θ ) hpotenuse opposite Cosine Secant hpotenuse cos( θ ) sec( θ ) hpotenuse Tangent Cotangent opposite tan( θ ) cot( θ ) opposite or alternatel using some of the identities: Alternate Definition of the Trigonometric Functions Using the Unit Circle Sine Cosecant sin( θ ) the -coordinate on the unit circle csc( θ ) Cosine sin( θ ) cos( θ ) the -coordinate on the unit circle Secant Tangent sec( θ ) sin( θ) tan( θ ) cos( θ ) cos( θ) Cotangent cos( θ) cot( θ ) sin( θ) SCC:Rickman Notes on Unit Circle Trigonometr. Page # of

2 Finall, using the results from previous notes, we get the following unit circle graph showing basic angles. 0, π, ; θ 60 π, ; θ 45 4 π, ; θ 0 6 (-,0 ) (, 0) ( 0,-),-,-,- Evaluating Trigonometric Functions. Eample #: Evaluate all 6 trigonometric functions at θ 90. First, notice that θ 90 takes us to the point (0,). Thus, sin ( 90 ) cos( 90 ) 0 tan ( 90 ) undefined 0 0 csc( 90 ) 0 sec( 90 ) undefined 0 0 cot ( 90 ) 0 SCC:Rickman Notes on Unit Circle Trigonometr. Page # of

3 π Eample #: Evaluate all 6 trigonometric functions at θ. 4 π First, notice that θ takes us to the point 4. Thus, π sin 4 π cos - 4 π tan π csc 4 π sec π - cot - 4 An alternate wa of evaluating trigonometric functions is to use reference angles, θ ' the acute angle measured off the -ais. Eample #: Evaluate all 6 trigonometric functions at θ -40. We first want to find where θ -40 is on the unit circle. Since θ , θ is 60 past a full rotation clockwise. Putting θ in the 4 th quadrant with a reference angle, θ ', of θ ' 60. Thus, the point for θ -40 would be the point for θ ' 60 adjusting the signs of the coordinates for the quadrant. In the 4 th quadrant the -coordinate is positive and the -coordinate is negative. Therefore, the point for θ -40 is,- sin (-40 ) - cos( -40 ). Hence, - tan (-40 ) - csc( - 40 ) sec( -40 ) cot (-40 ) Thus, we can evaluate the trigonometric functions at angles related to basic angles if we know the reference angle, θ ', and the sign of the trigonometric function in that quadrant. Therefore, we need to know the signs of the trigonometric function in each quadrant. Signs of the Trigonometric Functions Quadrant II Positive: Sine, Cosecant Negative: Cosine, Tangent, Secant, Cotangent Quadrant III Positive: Tangent, Cotangent Negative: Sine, Cosine, Cosecant, Secant Quadrant I Positive: ALL Negative: NONE Quadrant IV Positive: Cosine, Secant Negative: Sine, Tangent, Cosecant, Cotangent SCC:Rickman Notes on Unit Circle Trigonometr. Page # of

4 Eample #4: Evaluate all 6 trigonometric functions at 5π θ -. 6 Since 5π π θ - -π+, it s in the rd π quadrant with a reference angle of θ '. Hence, sin - -sin - csc- -csc cos- -cos tan - tan 6 6 sec- -sec cot - cot 6 6 Thus, there are a number of was to evaluate trigonometric functions. Ultimatel, use whichever method works best for ou, but ou do need to know how to do these WITHOUT a calculator. If ou don t learn how to do these without a calculator, ou won t understand what the trigonometric functions are giving ou, and thus ou won t understand the material in this course. More on Trigonometric Functions of Negative Angles: Look at the graph on the right. Notice that rotating to -θ instead of θ changes onl the sign on the -coordinate. Thus, onl the trigonometric functions that deal with the -coordinate will change sign, while the others won t change. This leads to the following identities. Negative Number Identities for the Trigonometric Functions sin(- θ ) -sin( θ ) csc(- θ ) -csc( θ ) cos(- θ ) cos( θ ) tan(- θ ) - tan( θ ) This actuall gives another wa to deal with negative angles. sec(- θ ) sec( θ ) cot(- θ ) -cot( θ ) Eample #5: Evaluate all 6 trigonometric functions at θ -45. Using the above identities, sin (-45 ) -sin ( 45 ) - csc( - 45 ) -csc( 45 ) cos( - 45 ) cos( 45 ) sec( -45 ) sec( 45 ) tan (-45 ) - tan ( 45 ) - - cot (-45 ) -cot ( 45 ) - - SCC:Rickman Notes on Unit Circle Trigonometr. Page #4 of

5 Pthagorean Identities: Now let s go back to the graph of the unit circle. Since we are dealing with a unit circle, and have the relation of + from the Pthagorean theorem. Working with this, we get: + cos( θ ) + sin ( θ ) For trigonometric functions, it s common to write cos( θ) as cos ( θ) used with trigonometric functions. For almost all other functions, f () f( f( ) ), but this is a convention. So the convention for trigonometric functions is summarized b the following, using sine for the eample, but is the same for the other trigonometric functions. n n sin,if n,, 4,5,... sin The inverse sine function,if n - Thus, we could now write the equation as cos ( θ ) + sin ( θ ). This is the st of the Pthagorean identities. From this one we get: cos ( θ ) + sin ( θ ) cos ( θ ) + sin ( θ ) cos ( θ ) + sin ( θ ) cos ( θ ) + sin ( θ ) cos ( θ) cos ( θ) sin ( θ) sin ( θ) cos ( θ) sin ( θ) cos ( θ) sin ( θ) + + cos ( θ) cos ( θ) cos ( θ) sin ( θ) sin ( θ) sin ( θ) + tan θ sec θ cot θ + csc θ Therefore, we finall get: Eample #6: Given that tan cot ( θ ) tan ( θ) θ and π θ + θ θ tan sec + sec θ sec θ sec ± - 7 sec θ 64 7 sec θ θ π The Pthagorean Identities cos θ + sin θ ( θ ) + ( θ ) + tan θ sec θ cot csc, find the other 5 trigonometric functions. cos( θ ) sec( θ) θ + θ cos sin + sin ( θ ) sin + θ sin ( θ ) sin ( θ ) ± Note that the whenever we got a ±, the decision as to which it is was made based on the restriction π θ rd quadrant. π csc( θ ) sin( θ) , and putting θ in the SCC:Rickman Notes on Unit Circle Trigonometr. Page #5 of

6 Solving Trigonometric Equations Involving Angles in the Interval [0, 60 ): Eample #7: Solve eactl cos( θ ) for 0 θ< 60. First notice that we had this problem in a earlier set of notes but with the restriction that 0 <θ< 90. Thus, since this time we have a larger interval to work with, we could possibl get more solutions. We need to start b identifing what quadrants we are working with. Since cosine is positive in both the st and 4 th quadrants, we need to think about both of them as shown to the right. Because, is a basic result for cosine, we can get that ' 0 θ. Thus, we need the angles in the 0,90 70,60 with a reference angle of θ ' 0. intervals ( ) and The first one would just be 0, since 0 ( 0,90 ). While the second, would be Therefore, θ 0,0. Eample #: Solve eactl csc( θ ) - for 0 θ< π. First manipulate the equation into or more of the basic trigonometric functions. Eample #9: Solve eactl Eample #0: Solve eactl csc( θ ) - - sin ( θ) sin ( θ) - π - sin ( θ),thus, wearein quadrantsor 4 with referenceangleof θ ' 4 5π 7π, θ 4 4 sec ( θ) sec θ 0 for 0 θ< 60. tan θ for 0 θ< π sec ( θ) sec θ 0 sec θ sec θ + 0 sec θ,- cos ( θ ),- θ 60,00,0 tan ( θ ) tan( θ ) ± tan( θ ) ± sin( θ ) ± cos( θ) π 4π π 5π θ,,, SCC:Rickman Notes on Unit Circle Trigonometr. Page #6 of

7 Eample #: Solve sin( θ ) 0.76 for 0 θ< 60. Round answer to the 4 th decimal place. Since, sine is positive in the st and nd quadrants, we need solutions. Inverse sine will give the - θ ' sin 0.76, but for the nd quadrant we reference angle which is the answer in the st quadrant, need to subtract the reference angle from 0. sin( θ ) 0.76 θ - - sin 0.76,0 sin 0.76 Make sure ou are in degree mode. θ 6.767,6. Eample #: Solve cot( θ ) for 0 θ< π. Round answer to the 4 th decimal place. First, change it to tangent. cot( θ ) tan( θ ) Again, we can get the reference angle from θ ' tan. Notice that I didn t use the negative 0.46 from Since the answers will be in quadrants and 4. cot( θ ) tan( θ ) θπ tan,π tan Makesure ou arein radian mode θ.7, Eample #: Solve cos ( θ ) for 0 θ< π. Round answer to the 4 th decimal place. 5 θ 5 - cos( θ ) ± θ ' cos 5 5 cos θ cos, cos, cos, cos 5 π π π θ.07, 5.760,.044, 4.47 θ ' SCC:Rickman Notes on Unit Circle Trigonometr. Page #7 of

8 Applications of Trigonometr: Eample #4: A weight is attached to a string and is being swung around in a circle of radius 4m. Assume the circle is horizontal and that θ 0 is due east. Find both the total distance traveled b the weight and the displacement, the distance between the starting and ending points, of the weight as θ changes from 0 to 40. Remember the formula s rθ when θ is in radians. Thus, the distance traveled is s rθ π π ( 4m) 40 the is to convert degrees to radians π m 5.64m For the displacement, let s first start with the unit circle. The points on the unit circle at θ 0 and θ 40 are given b ( cos( θ),sin( θ )). Therefore, the points are P0 ( cos(0 ),sin(0 )) (,0 ) Pf ( cos(40 ), sin(40 )) ( cos(0 ), sin(0 )) (- cos(60 ), sin(60 )) Note, the change from 40 to 0 comes from subtracting out the full rotations from the 40. Now let s find the distance between these points. 9 d( P 0,Pf ) But, this is the distance for points on a unit circle. We need to epand to points a circle of radius of 4m. Look at the graph on the right. The triangles are similar because the share the same angles. Thus, their sides are in proportion to each other. Therefore, since the sides adjacent to θ on the larger triangle are 4 times larger than the corresponding sides on the smaller triangle, the displacement on the larger circle will be 4 times the displacement on the unit circle. Hence, displacement 4( ) m 6.9 m Thus, while the weight travels a total of 5.64m going around in circles, its displacement is onl 6.9m from its original position. Notice that this etrapolation from the unit circle to an circle centered at the origin leads to a generalization of the definition. Epanded Definition of the Trigonometric Functions Using An Circle Centered at the Origin Sine Cosecant the -coordinate on the unit circle r r sin( θ ) csc( θ ) radius r sin( θ ) Cosine Secant the -coordinate on the unit circle cos( θ ) r r sec( θ ) radius r cos( θ ) Tangent Cotangent sin( θ) tan( θ ) cos( θ) cos( θ) cot( θ ) sin( θ) SCC:Rickman Notes on Unit Circle Trigonometr. Page # of