Section 6.1 Reciprocal, Quotient and Pythagorean Identities

Chapter 6 Section 6.1 Reciprocal, Quotient and Pythagorean Identities Trigonometric Identities Name: ________________________ Date: __________________ Using identities to reduce complexity in expressions and solve equations. Learning Goal 6.1 Identity is a statement that is which the identity Example true for always is defined values for all is an identity. Identities can be proven or verified. a. Verify the function Numerically Graphically 2 9 9 2 9 r 9 3 9 3 81 9 12 6 72 72 a b. Prove the identity ON I x 3 3 9 o 1 22 32 32 g x x 3 x Assignment 3 III 22 o d 0x 9 X 2 g 3 2 3 2 3 p. 296 # 1 12, 14 17, C1, C2 Quiz Next Day! Chapter 6 Section 6.1 Reciprocal, Quotient and Pythagorean Identities Pythagorean Identities Quotient Identities Example a. Verify that using tank f I b. Verify that E cost 2 could be an identity by starting from the original Pythagorean identity. sect Es 113 1 3 43 1 Trigonometric Identities i g É tank denominator Example State any restrictions (non-permissible values) for the identity I gÉ esto RIEPE Ina then simplify. cotaffI g Example Prove I sit Pythagorean identity Assignment seek O novalues 1 0 sing 8 x.at . fye she decomp to sink cosh III p. 296 # 1 12, 14 17, C1, C2 Quiz Next Day! my unit circle Ysino go too Eye sino yr costs coso x sino y sin o tan 0 1 jig o secy 00520 1 I cot O csc O sin sink E sing sq sine 0.2588 É cost't LEXI If It I sing cos HH HE HE 0.2588 A B A B sink E sin 5g E sinks Cos 250 50 230 sin j cos 300 54 Sin I sin ITI Itf II sink É45 E E I I I AB Sina cost cosasint sina.li sink 0052107 tan É E E tant EE E tank tank BE E IE tan ET 17 1 5pA x Éztj2 I I Iz 5 y Iz 15 22 4 2 I I Hane 331 433,13 15300s 3 173 5 33 317 9 345 3 15 3 9 345 345 3 B COSACOSB 42 12 22 1172 1572 1412 42 1772 22 1 2,5 46 42 196 452 15 Fat 1135 2 1424 r2 tank 11 2 12 66 1 sinAsinB 74 83165 333g 846331332 z f 433J EYE S 29 a E sing COSTE TV cosy o I 1 H y I Matchinic Question sinka 2sinAcost A 2A E 21200520 1 260520 20051700 ZI 2sink cos g 40520 2 SEO 2singcost Sin 2 60 2 sin 1207 sink 2 we can evaluate without a calc buttheguestiontold us not to SMA A f 12 13 iii 11232 42 11332 42 25 y 1125 5 sissies 20052 2Sinacosa Intesa 25in cosy E E 55 21 2212 Klatt 222 7 singing s cosiglossiising SEEtootsy sings ayyy ng sink in cos cos nI B 2911 76 42 cosx sina.tl 169 9 yet y 4 Hannett Hana 4 Npr P sit cost 7 singing singgtsint t 1 22 2 1 240 212 172 0 2 ay x I 0 SX NPV sin tf y 5g1 6 16 ly ly IT tan Denominator O N OHRT 10585 R 215,0 EXPRESSIONUNDEFINED tank DME Eff ONE COSTÉEBE a'sinks I of lift 2 cos a tl 2Sinacosx 2 20052 2sinacosx 211 cos 3 zsinxcos sings i 48in't cosa sink 2 sink I I 4sinza I cosalasina 1 asinatillasina Cosa 1 I3 a l sino sino sin20 cosh coszoli sino 0052011 Sino cosa 2sinx 17 tsing lltylll y IIIs p cosa Y 1IM c sites 25nsecosx costa 125in ISI joggling It Ing Nous 1 sing OE A 0520 0 cos to O E 3E f tank I Sina 20052 1 sink O 2 0 I o I 1 252 o 2 0 IT ZIT Otten he 27 In NED O 0 ICOSK 3 0 1 2 Cosa L II 215 22 I K 3 0 COSI 120052 3 0052 1 f Cosa COSI137 cosy 2 I 5 55 3 0 It 0 152733 I 2in NED 2in int 22 25 Costa É It It K xp E3 2 Cosa 42 24 E Xp I É 1 73 43 I I NY 0 Ésinx I sink O 2sinzcosa sink I tanalitzsina 052 0,25 Otan NE27 3 I 3 pm no solution LIN NED sink 1183 sink Isin Sink sink I 1 sink cosx Isinx 0 2 sinselsinx 27 0 t stink 12 0 sing 0 I I 2 091800 sink I If 2 3 24 300 21 300 300 1500 22 1800