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alexandr402 [8]

3 years ago

14

A 180 degree rotation around the origin

1 answer:

ivanzaharov [21]3 years ago

3 0

Answer:

Rotation of a point through 180°, about the origin when a point M (h, k) is rotated about the origin O through 180° in anticlockwise or clockwise direction, it takes the new position M' (-h, -k)

Step-by-step explanation:

IDK if this helped you out

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Use De Moivre's theorem to write the complex number in trigonometric form: (cos(3pi/5) + i sin (3pi/5))^3

docker41 [41]

By De Moivre's theorem,

$\left(\cos\dfrac{3\pi}5+i\sin\dfrac{3\pi}5\right)^3=\boxed{\cos\dfrac{9\pi}5+i\sin\dfrac{9\pi}5}$

We can stop here ...

# # #

... but we can also express these trig ratios in terms of square roots. Let $x=\dfrac\pi5$ and let $c=\cos x$ . Then recall that

$\cos5x=c^5-10c^3\sin^2x+5c\sin^4x$

$\cos5x=c^5-10c^3(1-c^2)+5c(1-2c^2+c^4)$

$\cos5x=16c^5-20c^3+5c$

On the left, $5x=\pi$ so that

$16c^5-20c^3+5c+1=(c+1)(4c^2-2c-1)^2=0$

Since $c=\cos\dfrac\pi5\neq1$ , we're left with

$4c^2-2c-1=0\implies c=\dfrac{1+\sqrt5}4$

because we know to expect $\cos x>0$ . Then from the Pythagorean identity, and knowing to expect $\sin x>0$ , we get

$\sin x=\sqrt{1-c^2}=\sqrt{\dfrac58-\dfrac{\sqrt5}8}$

Both $\cos$ and $\sin$ are $2\pi$ -periodic, so that

$\cos\dfrac{9\pi}5=\cos\left(\dfrac{9\pi}5-2\pi\right)=\cos\left(-\dfrac\pi5\right)=\cos\dfrac\pi5$

and

$\sin\dfrac{9\pi}5=\sin\left(-\dfrac\pi5\right)=-\sin\dfrac\pi5$

so that the answer we left in trigonometric form above is equal to

$\cos\dfrac\pi5-i\sin\dfrac\pi5=\boxed{\dfrac{1+\sqrt5}4+i\sqrt{\dfrac58+\dfrac{\sqrt5}8}}$

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