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DaveE

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DaveE

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HallsofIvy

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Even in terms of real numbers, the arguments for the **functions** sin(x) and cos(x) are NOT assumed to be angles!

Do you know that e^{ix}= cos(x)+ i sin(x)?

Then you also know that e^{x+ iy}= e^{x}(cos(y)+ i sin(y))

Of course, then e^{x- iy}= e^{x}(cos(y)- i sin(y))

We can, from those same formulas, derive

cos(x)= (e^{ix}+ e^{-ix})/2 and

sin(x)= (e^{ix}- e^{-ix})/(2i)

While those are derived, originally, with x real, we can easily extend them as**definitions** for functions of complex x.

In particular, if x= 4i+ 3, then

cos(4i+3)= (e^{-4+ 3i}+ e^{4- 3i})/2

= (e^{-4}(cos(3)+ i sin(3))+ e^{4}(cos(3)- i sin(3))

= (e^{-4}+ e^{4})cos(3)/2 + i(e^{
and
sin(4i+ 3)= (e-4i+3- e-4+ 3i)/(2i)
= (e-4(cos(3)+ i sin(3))/2i+ i(e4cos(3)+ i sin(3))}

Do you know that e

Then you also know that e

Of course, then e

We can, from those same formulas, derive

cos(x)= (e

sin(x)= (e

While those are derived, originally, with x real, we can easily extend them as

In particular, if x= 4i+ 3, then

cos(4i+3)= (e

= (e

= (e

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[tex]\sin (x+iy) = \sin x \cos iy + \cos x \sin iy[/tex]

[tex]=\sin x \cosh y + i\cos x\sinh y[/tex]

[tex]=\sin x \cosh y + i\cos x\sinh y[/tex]

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