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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

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

Grade:

1 Answers

Ramesh V
70 Points
12 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

 

    

   

       

 

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