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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(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
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