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

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

A. a = 12(e−1e)cos1, b = 12(e+1e)sin1{\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. a = 12(e+1e)cos1, b = 12(e−1e)sin1{\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. a = 12(e+1e)cos1, b = 12(e+1e)sin1{\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. a = 12(e−1e)cos1, b = 12(e−1e)sin1{\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}}

Explanation

Solution

Hint: Proceed the solution of this question first by writing value of cosθ{\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 cosθ{\text{cos}}\theta can be written as complex exponential form as
⇒cosθ = eiθ+e−iθ2\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)
⇒cos(1 - i) = ei(1 - i)+e−i(1 - i)2\Rightarrow {\text{cos(1 - i) = }}\dfrac{{{e^{i{\text{(1 - i)}}}} + {e^{ - i{\text{(1 - i)}}}}}}{2}​
fFurther
Further solving with the use of i2=−1{{\text{i}}^2} = - 1
⇒cos(1 - i) = ei+1+e−i−12\Rightarrow {\text{cos(1 - i) = }}\dfrac{{{e^{i + 1}} + {e^{ - i - 1}}}}{2}
This can be written as
⇒cos(1 - i) = ei+1+e−(i+1)2\Rightarrow {\text{cos(1 - i) = }}\dfrac{{{e^{i + 1}} + {e^{ - (i + 1)}}}}{2}
⇒ ei+1+e−(i+1)2\Rightarrow {\text{ }}\dfrac{{{e^{i + 1}} + {e^{ - (i + 1)}}}}{2}
This can be written using exponential simplification
⇒ ei×e1+e−i×e−12\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 ei=cos⁡1+isin⁡1 and e−i=cos⁡1−isin⁡1{e^i} = \cos 1 + i\sin 1 \ and \ { e^{ - i}} = \cos 1 - i\sin 1
⇒e(cos1 + isin1) + e−1(cos1 - isin1)2\Rightarrow \dfrac{{{\text{e(cos1 + isin1) + }}{{\text{e}}^{ - 1}}{\text{(cos1 - isin1)}}}}{2}
Separate real and imaginary part
⇒(e + e−1)cos12 + i(e - e−1)sin12=a+ib\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 equationeiθ=cosθ+isin⁡θ {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−iθ=cosθ−isin⁡θ{e^{ - i\theta }} = {\text{cos}}\theta - i\sin \theta
And adding both ⇒eiθ+e−iθ=cosθ+isin⁡θ + cosθ−isin⁡θ\Rightarrow {e^{i\theta }} + {e^{ - i\theta }} = {\text{cos}}\theta + i\sin \theta {\text{ + cos}}\theta - i\sin \theta
⇒eiθ+e−iθ=2cosθ\Rightarrow {e^{i\theta }} + {e^{ - i\theta }} = 2{\text{cos}}\theta
Dividing by 2 , we will get the value of cosθ = eiθ+e−iθ2{\text{cos}}\theta {\text{ = }}\dfrac{{{e^{i\theta }} + {e^{ - i\theta }}}}{2}hence no need to remember anything. Similarly by subtracting ⇒eiθ−e−iθ=cosθ+isin⁡θ - cosθ+isin⁡θ\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⁡θ = eiθ−e−iθ2i\sin \theta {\text{ = }}\dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}.