Question

If cos(1−i) =a+ib where a, b ∈ R and $ i = \sqrt {{\text{ - 1}}} $ then

A. $ {\text{a = }}\dfrac{1}{2}\left( {e - \dfrac{1}{e}} \right){\text{cos1, b = }}\dfrac{1}{2}\left( {e + \dfrac{1}{e}} \right){\text{sin1}} $
B. $ {\text{a = }}\dfrac{1}{2}\left( {e + \dfrac{1}{e}} \right){\text{cos1, b = }}\dfrac{1}{2}\left( {e - \dfrac{1}{e}} \right){\text{sin1}} $
C. $ {\text{a = }}\dfrac{1}{2}\left( {e + \dfrac{1}{e}} \right){\text{cos1, b = }}\dfrac{1}{2}\left( {e + \dfrac{1}{e}} \right){\text{sin1}} $
D. $ {\text{a = }}\dfrac{1}{2}\left( {e - \dfrac{1}{e}} \right){\text{cos1, b = }}\dfrac{1}{2}\left( {e - \dfrac{1}{e}} \right){\text{sin1}} $

Answer 1

Hint: Proceed the solution of this question first by writing value of $ {\text{cos}}\theta $ in complex exponential form then on putting the value of θ according to given in question, further solving and comparing the real and imaginary part we can reach to our answer.

Complete step-by-step answer:
We know that $ {\text{cos}}\theta $ can be written as complex exponential form as
\[ \Rightarrow {\text{cos}}\theta {\text{ = }}\dfrac{{{e^{i\theta }} + {e^{ - i\theta }}}}{2}\] ………..(1) In the question it is given cos(1−i) so here, value of θ will be equal to (1-i)
So on putting the value of θ = (1-i) in (1)
\[ \Rightarrow {\text{cos(1 - i) = }}\dfrac{{{e^{i{\text{(1 - i)}}}} + {e^{ - i{\text{(1 - i)}}}}}}{2}\]​
fFurther
Further solving with the use of $ {{\text{i}}^2} = - 1 $
\[ \Rightarrow {\text{cos(1 - i) = }}\dfrac{{{e^{i + 1}} + {e^{ - i - 1}}}}{2}\]
This can be written as
\[ \Rightarrow {\text{cos(1 - i) = }}\dfrac{{{e^{i + 1}} + {e^{ - (i + 1)}}}}{2}\]
\[ \Rightarrow {\text{ }}\dfrac{{{e^{i + 1}} + {e^{ - (i + 1)}}}}{2}\]
This can be written using exponential simplification
\[ \Rightarrow {\text{ }}\dfrac{{{e^i} \times {e^1} + {e^{ - i}} \times {e^{ - 1}}}}{2}\]
We know that
\[ \Rightarrow {e^{i\theta }} = {\text{cos}}\theta + i\sin \theta {\text{ & }}{e^{ - i\theta }} = {\text{cos}}\theta - i\sin \theta \]
So in the above expression the value of θ is 1 and -1 in 1st and 2nd term respectively.
So replacing \[{e^i} = \cos 1 + i\sin 1 \ and \ { e^{ - i}} = \cos 1 - i\sin 1\]
 $ \Rightarrow \dfrac{{{\text{e(cos1 + isin1) + }}{{\text{e}}^{ - 1}}{\text{(cos1 - isin1)}}}}{2} $
Separate real and imaginary part
 $ \Rightarrow \dfrac{{\left( {{\text{e + }}{{\text{e}}^{ - 1}}} \right){\text{cos1}}}}{2}{\text{ + i}}\dfrac{{\left( {{\text{e - }}{{\text{e}}^{ - 1}}} \right){\text{sin1}}}}{2} = a + ib $
Hence on comparing LHS and RHS
 $ \Rightarrow {\text{a = }}\dfrac{{\left( {{\text{e + }}{{\text{e}}^{ - 1}}} \right){\text{cos1}}}}{2}{\text{ & b = }}\dfrac{{\left( {{\text{e - }}{{\text{e}}^{ - 1}}} \right){\text{sin1}}}}{2} $


Note- In this particular a student should know that by remembering only this equation\[{e^{i\theta }} = {\text{cos}}\theta + i\sin \theta {\text{ }}\], he can develop all result with slight help of trigonometry. As by replacing θ= - θ, we can get \[{e^{ - i\theta }} = {\text{cos}}\theta - i\sin \theta \]
And adding both \[ \Rightarrow {e^{i\theta }} + {e^{ - i\theta }} = {\text{cos}}\theta + i\sin \theta {\text{ + cos}}\theta - i\sin \theta \]
\[ \Rightarrow {e^{i\theta }} + {e^{ - i\theta }} = 2{\text{cos}}\theta \]
Dividing by 2 , we will get the value of \[{\text{cos}}\theta {\text{ = }}\dfrac{{{e^{i\theta }} + {e^{ - i\theta }}}}{2}\]hence no need to remember anything. Similarly by subtracting \[ \Rightarrow {e^{i\theta }} - {e^{ - i\theta }} = {\text{cos}}\theta + i\sin \theta {\text{ - cos}}\theta + i\sin \theta \] we can find the value of \[\sin \theta {\text{ = }}\dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}\].