If cos(x+iy)=R(cosA+isinB) then prove that e^2y={sin(x-A) / sin(x+A) } ; where i=iota?

2
by dessimunda

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If cos(x+iy)=R(cosA+isinB) then prove that e^2y={sin(x-A) / sin(x+A) } ; where i=iota?

2
by dessimunda

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ii) Applying hyperbolic function properties, cos(iy) = cosh(y) and

-i*sin(iy) = sinh(y); ==> -sin(iy) = -i*sinh(y)

Substituting these in (1) above,

cos(x + iy) = cos(x)*cosh(y) -i*sin(x)*sinh(y) = R*cos(A) + i*Rsin(A) [As given]

Equating the real and imaginary parts on either side of the above equation,

R*cos(A) = cos(x)*cosh(y) -------- (2) and

R*sin(A) = -sin(x)*sinh(y) ------ (3)

iii) sin(x - A)/sin(x + A) = {sin(x)*cos(A) - cos(x)*sin(A)}/{sin(x)*cos(A) + cos(x)*sin(A)}

= {sin(x)*Rcos(A) - cos(x)*Rsin(A)}/{sin(x)*Rcos(A) + cos(x)*Rsin(A)}

Substituting the values of Rcos(A) & Rsin(A) from (2) & (3) above,

sin(x - A)/sin(x + A) =

{sin(x)cos(x)cosh(y) + sin(x)cos(x)sinh(y)/{sin(x)cos(x)cosh(y) - cos(x)sin(x)sinh(y)}

= {cosh(y) + sinh(y)}/{cosh(y) - sinh(y)}

iv) By hyperbolic functions properties,

cosh(y) = {e^y + e^(-y)}/2 and sinh(y) = {e^y - e^(-y)}/2

Substituting these in the above,

sin(x - A)/sin(x + A) = [{e^y + e^(-y)}/2 + {e^y - e^(-y)}/2]/[{e^y + e^(-y)}/2 - {e^y - e^(-y)}/2]

This simplifies to: 2(e^y)/2(e^-y) = e^y*e^y = e^(2y)

Thus it is proved that, sin(x - A)/sin(x + A) = e^(2y)