Question

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

Answer 1

i) Expanding by identity, cos(x + iy) = cos(x)*cos(iy) - sin(x)*sin(iy) ----(1) 

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)

Answer 2

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