Trigonometric Review Part 2

Transcription

1 1 Trigonometric Review Prt Identities If, y is point on the unit circle corresponding to the ngle, we know nd y, y is point on the unit circle, then Becuse y 1 This mens tht 1 This is usully denoted s 1 nd is clled Pythgoren s Identity This identity genertes two more identities The identity 1 tn sec cn be found by dividing both sides of Pythgoren s Identity by How cn Pythgoren s Identity generte cot 1 csc? By dividing both sides of Pythgoren s Identity by Now suppose tht The two digrms below show two congruent tringles in the unit circle Ech tringle hs two sides of length 1 tht lie on the sides of n ngle whose mesure is You could think of the tringle on the right s coming from the tringle on the left fter the tringle on the left ws rotted clockwise by rdins (( ), ( )) - ((), ()) (( -), ( -)) - (1, 0) In the first tringle, the length of the third side of the tringle is the distnce between the points (,) nd (, ) In the second tringle, the length of the third side is the distnce between the points (( ),( )) nd ( 1,0)

2 Since these re the sme length, we hve: ( ) ( ) (1 ( )) (( )) Squring both sides nd epnding gives us This cn be rerrnged s 1 ( ) ( ) ( ) 1 ( ) ( ) ( ) Ug the Pythgoren Identity three times (twice on the left nd once on the right) gives us ( ) Subtrcting from both sides nd then dividing by gives us ( ) (1) We strted by ssuming tht, but if, both sides (1) re equl to 1 nd the equlity holds If we hd strted by ssuming tht, then this would hve led to the formul ( ) () Becuse the coe is n even function, ( ) ( ), so this is equivlent to (1) We cn use this formul nd the fct tht coe is even nd e is odd to get ( ) ( ( )) ( ) ( ) Thus the sum nd difference formuls for coe re: ( ) Notice, + on the left mens on the right nd vice-vers

3 3 Looking t the grphs of e nd coe you cn see tht: ( ) nd ( ) We cn use these two identities nd the sum formul for coe to find the sum formul for e: ) ( ) ( ) ( ) ( ) ( Ug this formul nd the fct tht coe is even nd e is odd gives us: ( ) ( ( )) ( ) ( ) Thus the sum nd difference formuls for e re given by: ( ) Notice, + on the left mens + on the right nd on the left mens on the right Hence, we cn sy For the e, the signs mtch The double ngle formul for coe is ( ) nd cn be found by ug the sum formul for coe: ( ) ( )

4 Likewise, you cn show tht the double ngle formul for e is by ug the sum formul for e ( ) Two other trigonometric identities tht re often used re the hlf ngle formuls 1 nd 1 The first one follows becuse Identity 1 Replcewith 1 Then you cn solve 1 for The second one follows becuse 1 Identity Replcewith 1 Then you cn solve 1 for Two other formuls tht come in hndy during clculus or differentil course re Lw of Coes: b c bc c nd the Lw of Sines: b c b To show why these two formuls work, it will be helpful to review some concepts from right tringle trigonometry

5 5 Right Tringle Trigonometry You re probbly fmilir with the formuls Opposite Adjcent Hypotenuse nd Hypotenuse Let s tke look t why they re true We will strt by ssuming tht we hve right tringle nd tht is the mesure of one of the non-right ngles We will plce this tringle on set of es s shown below In the cse shown, r 1, where r is the rdius of the lrger circle nd the length of the hypotenuse of the tringle The unit circle is shown s well s tringle with hypotenuse of mesure 1 tht is similr to, but smller thn the originl tringle The coordintes of the verte on the smller tringle tht lie on the unit circle re (, ) Lbel the verte on the lrger tringle s (, y) Ug the properties of similr tringles 1 (0, 0) (, y) = (r(), r()) r ((), ()) (1, 0) (r, 0) r y r nd y r 1 r 1 Hence, (, y) ( r, r ) Now it is esily observed tht in the lrger right tringle Length of the leg oppositeof r, Length of the hypotenuse r Length of the leg djcent to Length of the hypotenuse r, nd r Length of the leg oppositeof r tn Length of the leg djcent to r If r 1, the formuls come out similrly Of course when r 1, these formuls utomticlly hold

6 6 Now we will emine the lw of coes b c bc We re going to see why it is true on the tringle below First drop n ltitude from the ngle with mesure nd denote the length of this ltitude s h This divides the originl tringle into two right tringles The newly formed right tringle on the left hs horizontl leg with length c (becuse length of segment ) c Since the length of the horizontl side of the originl tringle is b, tht mens tht the length of the horizontl side of the newly formed tringle on the right is b c c h c() b - c() Ug Pythgoren s formul on ech of the smller tringles gives us: c c nd h b c h Substituting the first eqution into the second gives the lw of coes: c c b c c c b bc c c b bc We cn use this sme digrm to generte prt of the Lw of Sines From the digrm bove we cn come up with h c nd h Setting these two equl to ech other gives us c or, c which is prt of the lw of es The rest cn be shown similrly

7 7 In summry, if you know the Pythgoren Identity 1, you cn esily deduce the identities nd 1 tn sec cot 1 csc If you lso know the sum nd difference identities for coe nd e nd ( ) ( ), then you cn deduce the double ngle identities nd ( ) ( ) Combining the double ngle identity for coe with the Pythgoren Identity gives the hlf ngle identities nd 1 1 Finlly, ug right tringle trigonometry nd Pythgoren s formul gives the Lw of Coes nd Lw of Sines nd b c bc b c These 13 identities re the ones typiclly used in clculus course

8 8 1 Only four of the following identities re true Eplin why the true ones re true nd the flse ones re flse For ech identity, ssume the vlues of re restricted so tht you do not hve denomintor of 0 ( 3) tn( 3) () b () c ( ) d e sec 1 tn f 1 g y 1 h ( ) i (t ) (8t ) (10t) j ( k ( ) ) Show tht the following re identities (where they re defined) 1 b c (sec tn ) 1 d ( y)( y) y Hint: Strt with the left side nd pply the sum nd difference formuls for e followed by the Pythgoren identity

9 9 Solutions 1 The true ones re b, e, f, k ( 3) ( 3) tn( 3) Flse, tn( 3) ( 3) b () True becuse ( ) is nd identity () c Flse You cn see this by wht ws true in prt b nd lso, you cnnot simply pull the two outside of the e function ( ) d Flse for the sme reson tht prt c is flse The e of is not e times e sec 1 tn True from the identity sec tn 1 1 f 1 True from the identity g y 1 Flse, if the rguments re not the sme, the left hnd side might not sum to one h ( ) Flse, coe is not n odd function, it is even i (t ) (8t ) (10t) Flse, you cnnot fctor e out of () j ( ) Flse, ( ) nd ( ) ( )( ) k ( ) True, this is n ccepted nottion Show tht the following re identities (where they re defined) ( )( b c 1 d (sec tn ) ( y)( y) (sec tn ) 1 ) ( y)( y) ( y y )( y y ) y y (1 y) y y y y y y (1 )