Find the principal value, show your full Solutions.Evaluate sec−1(3-j). 3.2 Inverse Trigonometric FunctionsExample 3Evaluate sin -1(1 + 2j)Solution: Let x = sin -1(1+ 2j)sin x = 1+2jeix-e-jx= 1+ 2jelx – e-jx = (1 + 2j)(2j)eix - e-jx = -4 + 2/Multiply both sides by elx(ejx) – 1 = (-4 + 2j)e}*(eix)“ + eix (4 – 2j) – 1 = 0elx - -(4-2j)±y(4-2j)2-4(1)(-1)2(1)-4+2/+v162-161eix =2V162 – 16i = v22.627 L - 22.5°= 4.395 – 1.82ieix =4+2j±(4.395-1.82i)2ejx = 0.1975 + 0,09iand eix = -4.1975 + 1.91ijx = In(0.1975 + 0.091) = In(0.217 e0.4276j)jx = In(0.217) + 0.4276jjx = -1.528 + 0.4276jx =. 4276 + 1.528jThen other value:elx = -4.1975 + 1.91/jx = In(-4.1975 + 1.91j)= In(4.612e2.7146) = In 4.612 + 2.7146jjx = 1.529 + 2.7146jx = 2.7146 – 1. 528jExample 4Evaluate tan-1(3 + 4j)Solution: let x = tan-(y) where y = 3 + 4jtan x = y[(ejx-e-jx)][(eix-e-jx= y2jeix - e-jx = j(eix – e-jx)yMultiply both sides by elx(el*)° – 1 = j(e/*)° + 1)y(el*)°(1 – y) = 1 + jy(eix)? = 1+/y1-jySubstitute y = 3+ 4j and use calculator(e*)²,= -0.7059 + 0.1765eix = (0.728 e/2.897)/2= 0.853 e1.4485jjx = In (0.853 e1.4485)) = In 0.853 + 1.4485jjx = -0.159 + 1.4485jx = 1.4485 + 0.159 jThe other value is:eix = -(0.853 e1.4485j)= 0.10406 – 0.847j= 0.853e-1.693jJx = In (0.853e-1.693/ ) = In 0.853 – 1.693j = -0.159 – 1.693jThen: x = -1.693 + 0.159j

Name: Find the principal value, show your full Solutions.Evaluate sec−1(3-j)
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: Find the principal value, show your full Solutions.Evaluate sec−1(3-j). 3.2 Inverse Trigonometric FunctionsExample 3Evaluate sin -1(1 + 2j)Solution: Let x = sin -1(1+ 2j)sin x = 1+2jeix-e-jx= 1+ 2jelx – e-jx = (1 + 2j)(2j)eix - e-jx = -4 + 2/Multiply