cos(1-i) = a+ib (a,b are reals), then which is correct

a) a=1\2(e - 1\e)cos1, b=1\2(e+ 1\e)sin1

b) a=1\2(e + 1\e)cos1, b=1\2(e - 1\e)sin1

c) a=1\2(e + 1\e)cos1, b=1\2(e+ 1\e)sin1

d) a=1\2(e - 1\e)cos1, b=1\2(e-1\e)sin1

we know that : e^(ix)=cos x + i sin x

so e^(ix)=cos x + i sin x

     e^-(ix)=cos x - i sin x  where x is: A+iB

so we can write:   cos(A+iB)=1/2(e^i(A+iB) +e^-i(A+iB))

here A=1 and B=-1

so 1/2(e^i(1-i1) +e ^-i(1-i1))=a+ib

     1/2(eⁱ⁺¹ +e ^-i-1)=a+ib

     1/2(eⁱ.e +e^-1e ^-i  )=a+ib where ( eⁱ= cos 1+i sin 1)

     e/2( cos 1+i sin1)+1/2e(cos 1-i sin1) = a+ ib

    1/2 (ecos 1+1/e.cos1)+i/2(e.sin1-1/e.sin1) = a+ ib

 equating real and imaginery parts, we'll get

a=1/2 cos 1(e+e^-1)

b=1/2 sin 1(e-e^-1)

the answer is B