Question

Prove that log `[sin(x+iy)/sin(x-iy)]=2tan^-1 (cot x tanhy)`

Answer 1

Consider, log[sin(x+iy)]=log[sin x cos(iy) + cos x sin (iy)]
∴log[sin(x+iy)]=log[sin x cosh y + icos x sinh y] {∵cos(ix)=cosh x; sin (ix)=isinh x;} 

∴` Log [sin(x+iy)]1/2log[sin^2 xcosh^2 y+cos^2 x sinh^2y]+itan^-1|(cosxsinhy)/(sin xcosh y)|`

`{∵ log (x-iy)=1/2 log (x^2+y^2)+itan^-1|y/x|}`

∴` log[sin(x+iy)]1/2log [sin^2x cosh^2 y+cos^2xx sinh^2 y]+|cot x tanh y|` 

Taking Conjugate, `log[sin(x-iy)=1/2log [sin^2 x cosh^2 y +cos^2 x sinh^2 y]-itan^1|cotx tanh y|` 

Now, log `[sin(x+iy)/sin(x-iy)]= log [sin(sin(x-iy)]` 

= `{1/2 log [sin2 x cosh2y + cos2 x sinh2 y]+itan-1|cot x tanh y|}`- `{1/2 log [sin2 x cosh 2 y+ cos 2 x sinh2 y]-itan-1|cot x tanhy|}`

= `2 itan^-1(cot x tanh y)` 

∴ `log [sin(x+iy)/sin(x-iy)]=2itan^-1 (cot x tanh y)`