2 cos sen π π 2 2 3 tg sen cos 2 2 3 π π π 2 2 047 07158 06983 10251 1 2 06983

2 cos sen π π 2 2 3 tg sen cos 2 2 3 π π π 2 2

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2 cos sen π π 2 2 3 tg sen cos 2 2 3 π π π 2 2 047 = − 0,7158 0,6983 1,0251 1 2 = − 0,6983 0,7158 1 = 0,4657 0,8849 2 1 1 2 + − = − ( ) 1,9004 0,4657 = 0,7822 0,623 1,2555 1 2 = − 0,7822 0,623 1 2 = − 0,8016 0,5979 1 1 2 + − = ( ) 0,7459 0,8016 3 SOLUCIONARIO
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126 Completa, sin usar la calculadora, los valores de la tangente de los siguientes ángulos. Utiliza la calculadora para hallar las razones. a) sen 319° 12 ' 52 " e) cosec 200° 16 ' i) tg b) cos 434° 26 ' f ) sec j) cos 3,845 c) tg 7,03 g) tg 183° 13 ' 53 " k) cotg d) sen h) sen 333° 55 " l) cosec 5,24 a) sen 319° 12 ' 52 " = − 0,6532 g) tg 183° 13 ' 53 " = 0,0565 b) cos 434° 26 ' = 0,2684 h) sen 333° 55 " = − 0,4538 c) tg 7,03 = 0,9257 i) tg = 2,4142 d) sen = − 0,9511 j) cos 3,845 = − 0,7626 e) cosec 200° 16 ' = − 2,8869 k) cotg = − 1,7321 f) sec = − 1,4142 l) cosec 5,24 = − 1,1574 Reduce los ángulos al 1. er cuadrante y calcula estas razones. a) sen 131° e) sec 156° 23 ' 6 " i ) cotg 295° 12 ' 45 " b) cos 334° 46 ' f ) tg 238° 24 ' j ) sec 203° 36 ' 54 " c) tg 146° 22 " g) sen 302° 15 " d) cosec 122° 53 ' h) cos 192° 21 ' 32 " a) sen 131° = sen 49° = 0,7547 b) cos 334° 46 ' = cos 25° 14 ' = 0,9046 c) tg 146° 22 " = − tg 33° 59 ' 38 " = − 0,6744 d) cosec 122° 53 ' = cosec 57° 7 ' = 1,1908 e) sec 156° 23 ' 6 " = − sec 23° 36 ' 54 " = − 1,0914 f) tg 238° 24 ' = tg 58° 24 ' = 1,6255 g) sen 302° 15 " = − sen 57° 45 ' = − 0,8480 h) cos 192° 21 ' 32 " = − cos 12° 21 ' 32 " = − 0,9767 i) cotg 295° 12 ' 45 " = − cotg 64° 47 ' 15 " = − 0,4708 j) sec 203° 36 ' 54 " = − sec 23° 36 ' 54 " = − 1,0914 052 5 4 π 11 6 π 8 5 π 11 8 π 8 π 5 11 π 6 5 π 4 11 π 8 051 60° 45° 30° 30° 45° 60° 90° 3 1 3 3 0 3 3 1 3 No definido 120° 135° 150° 180° 210° 225° 240° 270° 3 1 3 3 0 3 3 1 3 No definido 050 Trigonometría
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127 En la circunferencia goniométrica, dibuja y obtén con ayuda de la calculadora. a) Dos ángulos cuyo seno valga 0,36. b) Dos ángulos cuya tangente valga 3,54. a) sen α = 0,36 α 1 = 21° 6 ' 0,706 " α 2 = 158° 53 ' 59,2 " b) tg β = − 3,54 β 1 = 285° 46 ' 27 " β 2 = 105° 46 ' 27 " Halla los siguientes ángulos. a) arc cos 0,4539 d) arc co s ( 0,3996) g) arc sen 0,6862 b) arc sen 0,9284 e) arc tg 2,1618 h) arc sen ( 0,3308) c) arc tg ( 0,5459) f ) arc cos ( 0,2926) a) arc cos 0,4539 = 63° 20,95 " e) arc tg 2,1618 = 65° 10 ' 32,9 " b) arc sen 0,9284 = 68° 11 ' 12,3 " f ) arc cos ( 0,2926) = 107° 49,2 " c) arc tg ( 0,5459) = 331° 22 ' 12 " g) arc sen 0,6862 = 43° 19 ' 48,2 " d) arc cos ( 0,3996) = 113° 33 ' 11 " h) arc sen ( 0,3308) = 340° 40 ' 58 " Conocidas las razones de 30°, 45° y 60°, obtén, sin usar la calculadora, las razones de 120°, 225°, 240° y 300°. sen 120° = sen 60° = sen 240° = − sen 60° = cos 120° = − cos 60° = cos 240° = − cos 60° = tg 120° = − tg 60° = − tg 240° = tg 60° = sen 225° = − sen 45° = sen 300° = − sen 60° = cos 225° = − cos 45° = cos 300° = cos 60° = tg 225° = tg 45° = 1 tg 300° = − tg 60° = − 3 1 2 2 2 3 2 2 2 3 3 1 2 1 2 3 2 3 2 055 054 β 2 β 1 α 2 α 1 053 3 SOLUCIONARIO
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128 Determina el ángulo α del 1. er cuadrante cuyas razones trigonométricas verifican: sen α= sen 249° 31 ' cos α= cos 249° 31 ' tg α= tg 249° 31 ' Halla cuáles son sus razones trigonométricas. sen α = sen 249° 31 ' = 0,9368 α = 69° 31 ' cos α = cos 249° 31 ' = 0,3499 α = 69° 31 ' tg α = tg 249° 31 ' = 2,6770 α = 69° 31 ' ¿Qué ángulo del 3. er cuadrante tiene el mismo coseno que 132° 24 ' 18 " ?
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