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Darya [45]
3 years ago
14

Write cos (19°) in terms of sine​

Mathematics
1 answer:
seraphim [82]3 years ago
8 0

Answer:

sin(71°)

Step-by-step explanation:

Sine lags behind cosine by 90°

In other word, cos(∅) = sin(90° - ∅)

cos(19°) = sin(90° - 19°)

= sin(71°)

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Solve the system of equations: <br> X+y+z=5<br> -3x-4y+4z=15<br> 2x-y-4z=-8
kolezko [41]

Answer:

Solution By Gauss jordan elimination method

x = 3, y = 2 and z = 4

6 0
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
Math help please? &lt;3
ipn [44]
A linear equation cannot be used.
because 60/1 isn't equal to 70/5 and isn't equal to 80/20 and so on.
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3 years ago
What is the next term ofthe sequence 5z, 8x, 11v,...?
nexus9112 [7]

Answer:

<u>14t</u>

Step-by-step explanation:

stuvwxyz

567891011121314

6 0
3 years ago
The formula for finding the perimeter of equilateral triangle is P=3s. how can you transform the formula so that it could be use
Mandarinka [93]

Your answer is D. Hope this helps -John



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3 years ago
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