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Determine $\sin(2 - 2i)$ and write the answer in the form $a + ib$.

I managed to get it into the form $\dfrac{e^{2i + 2} - e^{-2 -2i}}{2i}$.

The solution has the following calculations:

$\sin(2 - 2i) = \dfrac{e^{i(2 - 2i)} - e^{-i(2 - 2i)}}{2i}$

$ = \dfrac{(e^2 - e^{-2})\cos(2) + i(e^2 + e^{-2})\sin(2)}{2i}$

$= \dfrac{(e^2 + e^{-2})\sin(2)}{2} - i\dfrac{(e^2 - e^{-2})\cos(2)}{2}$

I am struggling to get from my solution into the form of the provided solution.

I can see that the algebraic property $e^{x + iy} = e^xe^{iy} = e^x(\cos(y) + i\sin(y))$ was used, but I still do not understand how to get from my solution to the provided solution.

I would appreciate it if people could please take the time to post a solution that shows the steps that lead from my intermediate solution to the provided solution.

Mark Viola
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The Pointer
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  • Your solution is already in the desired form, with $a= \dfrac{(e^2 + e^{-2})\sin(2)}{2}, b= -\dfrac{(e^2 - e^{-2})\cos(2)}{2}$. – marty cohen May 23 '17 at 20:54

1 Answers1

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$$\sin (2-2i)=\frac { { e^{ i(2-2i) }-e^{ -i(2-2i) } } }{ 2i } { = }\frac { { e }^{ 2i }\cdot { e }^{ 2 }-{ e }^{ -2i }\cdot { e }^{ -2 } }{ 2i } =\\ =\frac { \left( \cos { 2 } +i\sin { 2 } \right) { e }^{ 2 }-\left( \cos { \left( -2 \right) } +i\sin { \left( -2 \right) } \right) { e }^{ -2 } }{ 2i } =\\ =\frac { \cos { 2 } \left( { e }^{ 2 }-{ e }^{ -2 } \right) +i\sin { 2 } \left( { e }^{ 2 }+{ e }^{ -2 } \right) }{ 2i } =\frac { \sin { 2 } \left( { e }^{ 2 }+{ e }^{ -2 } \right) }{ 2 } +\frac { \cos { 2 } \left( { e }^{ 2 }-{ e }^{ -2 } \right) }{ 2i } =\\ =\frac { \sin { 2 } \left( { e }^{ 2 }+{ e }^{ -2 } \right) }{ 2 } -i\frac { \cos { 2 } \left( { e }^{ 2 }-{ e }^{ -2 } \right) }{ 2 } $$

haqnatural
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