Given: Circumscribed polygon ELPJ

Mathematics

1 answer:

34kurt3 years ago

5 0

$P_{ELPJ} = EK+ES+LU+LK+PV+PU JS+JV \\ \Leftrightarrow P_{ELPJ} = 2EK + 2LU + 2PV + 2JS = 2 \times 2 + 2 \times 4 + 2 \times 1 + 2 \times 2 = 18$

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Please Help!! Use Euler’s formula to write in exponential form.

LekaFEV [45]

Answer:

C, $4e^{i(7\pi/4)}$

Step-by-step explanation:

To remind you, Euler's formula gives a link between trigonometric and exponential functions in a very profound way:

$e^{ix}=\cos{x}+i\sin{x}$

Given the complex number $2\sqrt{2}-2i\sqrt{2}$ , we want to try to get it in the same form as the right side of Euler's formula. As things are, though, we're unable to, and the reason for that has to do with the fact that both the sine and cosine functions are bound between the values 1 and -1, and 2√2 and -2√2 both lie outside that range.

One thing we could try would be to factor out a 2 to reduce both of those terms, giving us the expression $2(\sqrt{2}-i\sqrt{2})$

Still no good. √2 and -√2 are still greater than 1 and less than -1 respectively, so we'll have to reduce them a little more. With some clever thinking, you could factor out another 2, giving us the expression $4\left(\frac{\sqrt{2}}{2} -i\frac{\sqrt{2}}{2}\right)$ , and now we have something to work with.

Looking back at Euler's formula $e^{ix}=\cos{x}+i\sin{x}$ , we can map our expression inside the parentheses to the one on the right side of the formula, giving us $\cos{x}=\frac{\sqrt2}{2}$ and $\sin{x}=-\frac{\sqrt2}{2}$ , or equivalently:

$\cos^{-1}{\frac{\sqrt2}{2} }=\sin^{-1}-\frac{\sqrt2}{2} =x$

At this point, we can look at the unit circle (attached) to see the angle satisfying these two values for sine and cosine is 7π/4, so $x=\frac{7\pi}{4}$ , and we can finally replace our expression in parentheses with its exponential equivalent: