how to prove (1+sinx+icosx)/(1+sinx-icosx)= cos(p/2-x)+isinx(p/2-x)

Question

plz help

Answer 1

Adnan, use a different kind of Euler's formula's application sin(x)=1/2i(e^ix-e^-ix) and cos(x)=1/2(e^ix+e^-ix)

and you should get (e^-ix+1)/(e^ix+1)=e^-ix

[2/2 + (eix-e-ix)/2i + i((eix+e-ix)/2]/[2/2 + (eix-e-ix)/2i - i((eix+e-ix)/2] = e^i∏/2e^-ix = e^-ix

([2 + i(-eix-e-ix) + i(eix+e-ix)]/2)/([2 + i(eix-e-ix) - i(eix+e-ix)]/2)=e^-ix

 [(2 + 2e^-ix)/2]/[(2 - 2e^ix)/2] = e^-ix

(1 + e^-ix)/(1+e^ix)=e^-ix

(1+e^-jx)/(1-e^jx) = e^j(∏/2-x)  

 

 

 

Euler's formula e^jx=cosx+jsinx

multiply both sides by the denominator

1+e^jx=e^j(∏/2-x)-e^jxe^j(∏/2-x)

        = e^j(∏/2-x)-e^j(x+∏/2-x)

         = ..........  - e^j∏/2        e^j∏/2 = cos∏/2+jsin∏/2=1

 1+e^jx = e^j(∏/2-x)-1

I hope this helps but I am working on it as I write this

e^j(∏/2-x)= cos(∏/2-x) + jsin(∏/2-x)

cos(∏/2)cos(x)+sin(∏/2)sin(x) + j[sin(∏/2)cos(x)-cos(∏/2)sin(x)]

        (1)cos(x)+        (0)sin(x) + j[        (0)cos(x)-       (1)sin(x)]           cos(∏/2)=1, sin(∏/2)=0

<=> cos(x) - jsin(x)

 

 

Answer 2

On the right side, do you mean "+isinx(p/2-x)" or "+isin(p/2-x)"?