cos-1-i-a-ib-Find-a-b-

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

\cos (1 - i) = a + i b

F i n d a, b .

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

c o s (1 - i) = \frac{e^{i (1 - i)} + e^{- i (1 - i)}}{2} = \frac{e^{1 + i} + e^{- (1 + i)}}{2} = e \frac{e^{i}}{2} + \frac{1}{e} . \frac{e^{- i}}{2}

\frac{1}{2} (e (c o s 1 + i s i n (1)) + \frac{1}{e} (c o s (1) - i s i n (1)))

\frac{e^{2} + 1}{2 e} c o s (1) + \frac{1}{2} i s i n (1) (e - \frac{1}{e})

\frac{e^{2} + 1}{2 e} c o s (1) + i (\frac{e^{2} - 1}{2 e}) s i n (1)

a = \frac{e^{2} + 1}{2 e} c o s (1)

b = \frac{e^{2} - 1}{2 e} s i n (1)

Commented by ajfour last updated on 24/Aug/20

\cos θ = \frac{e^{i θ} + e^{- i θ}}{2} S i r \dots

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

\cos (1 - i) = \frac{e^{i (1 - i)} + e^{- i (1 - i)}}{2}

= \frac{e^{1 + i} + e^{- (1 + i)}}{2} = \frac{e . e^{i} + e^{- 1} . e^{- i}}{2}

= \frac{e}{2} \times (\cos (1) + isin (1)) + \frac{e^{- 1}}{2} \times (\cos (1) - isin (1))

= \frac{e + e^{- 1}}{2} \cos (1) + i (\frac{e - e^{- 1}}{2} . \sin (1))

= a + bi . Therefore,

{\begin{cases} a = \frac{e + e^{- 1}}{2} \cos (1) \\ b = \frac{e - e^{- 1}}{2} . \sin (1) \end{cases}

Commented by ajfour last updated on 24/Aug/20

T h a n k y o u s o m u c h S i r!

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

You are welcome .

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

\cos (1 - i) = ch (i + 1) = \frac{e^{i + 1} + e^{- i - 1}}{2} = \frac{1}{2} {e (\cos 1 + isin 1) + e^{- 1} (\cos 1 - isin (1))

= \frac{1}{2} {(e + e^{- 1}) \cos (1) + i (e - e^{- 1}) \sin (1)} = a + ib \Rightarrow

a = \frac{e + e^{- 1}}{2} \cos (1) and b = \frac{e - e^{- 1}}{2} \sin (1)