Question

If log sin(x + iy) = a + ib. prove that
1) 2e2a = cosh 2y
cosh 2y -- cos 2x
11) tan b = cotx
tan hy​

Answer 1

Answer:

sorry I didn't know that answer

Answer 2

ta$n^{-1}$ [cot x tan hy]

Explanation:

we know that,

sin (A +B) = sin Acos B + cos A sinB

Given function = log sin (x + iy)

log [sin x cos iy + sin iy cos x

and,

sin (x + iy) = sin x cos(iy) + cos x sin iy

we know that,

cosh x = cos (ix) and

sinh x = -i sin (ix),

tan hx = -itan (ix)

⇒sin (x + iy) = sin x cosh x + cos x (i)(sinh x)

= sin x cosh x + i (cos x sinh x)

∴ log [sin (x + iy)] = log [sin x cosh y + icos x sinh y}

cos (ix) = cosh x;

sin (ix) = i sinh x

log (sin (x + iy)) = 1/2 log [sin²x cos²hy + cos²x sinhy] + i tan”! [ cot x tan hy]

Imaginary part is

ta$n^{-1}$ [cot x tan hy]