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Question Number 109516 by ajfour last updated on 24/Aug/20

cos (1−i)=a+ib Find a, b.

$$\mathrm{cos}\:\left(\mathrm{1}−{i}\right)={a}+{ib} \\ $$$${Find}\:\:{a},\:{b}. \\ $$

Answered by Dwaipayan Shikari last updated on 24/Aug/20

cos(1−i)=((e^(i(1−i)) +e^(−i(1−i)) )/2)=((e^(1+i) +e^(−(1+i)) )/2)=e(e^i /2)+(1/e).(e^(−i) /2) (1/2)(e(cos1+isin(1))+(1/e)(cos(1)−isin(1))) ((e^2 +1)/(2e))cos(1)+(1/2)isin(1)(e−(1/e)) ((e^2 +1)/(2e))cos(1)+i(((e^2 −1)/(2e)))sin(1) a=((e^2 +1)/(2e))cos(1) b=((e^2 −1)/(2e))sin(1)

$${cos}\left(\mathrm{1}−{i}\right)=\frac{{e}^{{i}\left(\mathrm{1}−{i}\right)} +{e}^{−{i}\left(\mathrm{1}−{i}\right)} }{\mathrm{2}}=\frac{{e}^{\mathrm{1}+{i}} +{e}^{−\left(\mathrm{1}+{i}\right)} }{\mathrm{2}}={e}\frac{{e}^{{i}} }{\mathrm{2}}+\frac{\mathrm{1}}{{e}}.\frac{{e}^{−{i}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({e}\left({cos}\mathrm{1}+{isin}\left(\mathrm{1}\right)\right)+\frac{\mathrm{1}}{{e}}\left({cos}\left(\mathrm{1}\right)−{isin}\left(\mathrm{1}\right)\right)\right) \\ $$$$\frac{{e}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{e}}{cos}\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}{isin}\left(\mathrm{1}\right)\left({e}−\frac{\mathrm{1}}{{e}}\right) \\ $$$$\frac{{e}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{e}}{cos}\left(\mathrm{1}\right)+{i}\left(\frac{{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{e}}\right){sin}\left(\mathrm{1}\right) \\ $$$${a}=\frac{{e}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{e}}{cos}\left(\mathrm{1}\right) \\ $$$${b}=\frac{{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{e}}{sin}\left(\mathrm{1}\right) \\ $$

Commented by ajfour last updated on 24/Aug/20

cos θ=((e^(iθ) +e^(−iθ) )/2) Sir...

$$\mathrm{cos}\:\theta=\frac{{e}^{{i}\theta} +{e}^{−{i}\theta} }{\mathrm{2}}\:\:\:\:\:\:{Sir}... \\ $$

Answered by 1549442205PVT last updated on 24/Aug/20

cos (1−i)=((e^(i(1−i)) +e^(−i(1−i)) )/2) =((e^(1+i) +e^(−(1+i)) )/2)=((e.e^i +e^(−1) .e^(−i) )/2) =(e/2)×(cos(1)+isin(1))+(e^(−1) /2)×(cos(1)−isin(1)) =((e+e^(−1) )/2)cos(1)+i(((e−e^(−1) )/2).sin(1)) =a+bi.Therefore, { ((a=((e+e^(−1) )/2)cos(1))),((b=((e−e^(−1) )/2).sin(1))) :}

$$\mathrm{cos}\:\left(\mathrm{1}−\mathrm{i}\right)=\frac{\mathrm{e}^{\mathrm{i}\left(\mathrm{1}−\mathrm{i}\right)} +\mathrm{e}^{−\mathrm{i}\left(\mathrm{1}−\mathrm{i}\right)} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{1}+\mathrm{i}} +\mathrm{e}^{−\left(\mathrm{1}+\mathrm{i}\right)} }{\mathrm{2}}=\frac{\mathrm{e}.\mathrm{e}^{\mathrm{i}} +\mathrm{e}^{−\mathrm{1}} .\mathrm{e}^{−\mathrm{i}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{e}}{\mathrm{2}}×\left(\mathrm{cos}\left(\mathrm{1}\right)+\mathrm{isin}\left(\mathrm{1}\right)\right)+\frac{\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}×\left(\mathrm{cos}\left(\mathrm{1}\right)−\mathrm{isin}\left(\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{e}+\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}\mathrm{cos}\left(\mathrm{1}\right)+\mathrm{i}\left(\frac{\mathrm{e}−\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}.\mathrm{sin}\left(\mathrm{1}\right)\right) \\ $$$$=\mathrm{a}+\mathrm{bi}.\mathrm{Therefore}, \\ $$$$\begin{cases}{\mathrm{a}=\frac{\mathrm{e}+\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}\mathrm{cos}\left(\mathrm{1}\right)}\\{\mathrm{b}=\frac{\mathrm{e}−\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}.\mathrm{sin}\left(\mathrm{1}\right)}\end{cases} \\ $$

Commented by ajfour last updated on 24/Aug/20

Thank you so much Sir!

$${Thank}\:{you}\:{so}\:{much}\:{Sir}! \\ $$

Commented by 1549442205PVT last updated on 24/Aug/20

You are welcome.

$$\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

Answered by mathmax by abdo last updated on 24/Aug/20

cos(1−i) =ch(i+1) =((e^(i+1) +e^(−i−1) )/2) =(1/2){e (cos1 +isin1)+e^(−1) (cos1−isin(1)) =(1/2){ (e+e^(−1) )cos(1)+i (e−e^(−1) )sin(1)} =a+ib ⇒ a =((e+e^(−1) )/2)cos(1) and b =((e−e^(−1) )/2)sin(1)

$$\mathrm{cos}\left(\mathrm{1}−\mathrm{i}\right)\:=\mathrm{ch}\left(\mathrm{i}+\mathrm{1}\right)\:=\frac{\mathrm{e}^{\mathrm{i}+\mathrm{1}} \:+\mathrm{e}^{−\mathrm{i}−\mathrm{1}} }{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{e}\:\left(\mathrm{cos1}\:+\mathrm{isin1}\right)+\mathrm{e}^{−\mathrm{1}} \left(\mathrm{cos1}−\mathrm{isin}\left(\mathrm{1}\right)\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\left(\mathrm{e}+\mathrm{e}^{−\mathrm{1}} \right)\mathrm{cos}\left(\mathrm{1}\right)+\mathrm{i}\:\left(\mathrm{e}−\mathrm{e}^{−\mathrm{1}} \right)\mathrm{sin}\left(\mathrm{1}\right)\right\}\:=\mathrm{a}+\mathrm{ib}\:\Rightarrow \\ $$$$\mathrm{a}\:=\frac{\mathrm{e}+\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}\mathrm{cos}\left(\mathrm{1}\right)\:\mathrm{and}\:\mathrm{b}\:=\frac{\mathrm{e}−\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}\mathrm{sin}\left(\mathrm{1}\right) \\ $$

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