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3 years ago

Sketch one cycle of the cosine function y= cos 2θ

1 answer:

7 0

Based on function transformations, the graph of y= cos(2θ), is the same as the graph of <span>y= cos(θ), with a component B of 2, in which case, it means the period is the only thing that differs

well the period for this function will then be </span> $\bf \cfrac{2\pi }{B}\iff \cfrac{2\pi }{2}\implies \pi$

so, the graph of is, the same as y= cos(θ), just shrunk by half horizontally

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(50Pts) When using the vertical method to multiply polynomials, your like terms must be lined up in ____________.

vaieri [72.5K]

I think you line them up in columns

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3 years ago

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lesantik [10]

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3 years ago

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

3 years ago

As Darius is skiing down a ski slope, Jorge is on the chairlift on the same slope. The chair lift has stopped. Darius stops dire

Jobisdone [24]

Answer:

0 chances.

2.3 seconds.

Step-by-step explanation:

25=-16t^2+35t+5

0=-16t^2+35t+5

4 0

3 years ago

What is the slope of (-9, 8) and (3,- 12)

earnstyle [38]

Answer:

Your slope would be -5/3

Step-by-step explanation:

To find the slope from two points, you would use the formula y2-y1/x2-x1

In this case, the equation would look like -12-8/3-(-9)

-12 - 8 = -20

3 - (-9) = 12

-20/12 = -5/3

3 0

3 years ago

Read 2 more answers