Ramesh V
Last Activity: 15 Years ago
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(ei(A+iB) +e-i(A+iB))
here A=1 and B=-1
so 1/2(ei(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