Taller Grupal2018.docx
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Taller grupal Forma polar impares 3).-5 π§1 = |π§1| = β(β5)2 + (0)2 β25 = 5 0 π = π‘ππβ1 ( ) + 180Β° 5 0Β° + 180Β° = 180Β° Forma polar π§1 = (5,180Β°) Z=5(cos 180 + isen180)
1+π
5) 1βπ
1+π 1+π 1 + π 2 + 2π 1 β 1 + 2π β = = 1βπ 1+π 1 β π2 1+1 2π =π 2
Determinar el valor principal del argumento β1 β π π‘ππβ1 (
π = πππ π‘ππ
β1 ) = 45Β° β1
β1 = 180Β° + 45Β° β1
π = 225Β°
ConversiΓ³n 17) 3(cos 0.2 + π π ππ 0.2) π₯ = π cos π π¦ = π π ππ π π₯ = 3 cos 0.2 = 3 β 1 = 3 π¦ = 3π ππ0.2 = 3 β 0 = 0 =3
1+π
5) 1βπ
1+π 1+π 1 + π 2 + 2π 1 β 1 + 2π β = = 1βπ 1+π 1 β π2 1+1 2π =π 2
Determinar el valor principal del argumento β1 β π π‘ππβ1 (
π = πππ π‘ππ
β1 ) = 45Β° β1
β1 = 180Β° + 45Β° β1
π = 225Β°
ConversiΓ³n 17) 3(cos 0.2 + π π ππ 0.2) π₯ = π cos π π¦ = π π ππ π π₯ = 3 cos 0.2 = 3 β 1 = 3 π¦ = 3π ππ0.2 = 3 β 0 = 0 =3
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