1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Natasha2012 [34]
2 years ago
9

Let cos A =

"absmiddle" class="latex-formula"> with A in QI and find sin(2A)
Mathematics
1 answer:
Rzqust [24]2 years ago
3 0
Umm i’m sorry but what is the question here .

This isn’t english
You might be interested in
Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.
elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}

Consider a vector field F=x\^i+y\^j+z\^k

Then the divergence of the vector F is,

div F = \nabla.F = (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)

and the curl of the vector F is,

curl F = \nabla\times F = \^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})

<h3>Calculation:</h3>

The given vector fields are:

F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k

1) Verifying the identity: \nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

Consider L.H.S

⇒ \nabla.(aF1+bF2)

⇒ \nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))

⇒ \nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the dot product between these two vectors,

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(1)

Consider R.H.S

⇒ a\nabla.F1+b\nabla.F2

So,

\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)

⇒ \nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}

\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)

⇒ \nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}

Then,

a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(2)

From (1) and (2),

\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

2) Verifying the identity: \nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Consider L.H.S

⇒ \nabla\times(aF1+bF2)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the cross product,

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y}) ...(3)

Consider R.H.S,

⇒ a\nabla\times F1+b\nabla\times F2

So,

a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))

⇒ \^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})

a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))

⇒ \^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})

Then,

a\nabla\times F1+b\nabla\times F2 =

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})

...(4)

Thus, from (3) and (4),

\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

8 0
1 year ago
Abby surveys every 25th sutend from an alphabetized list of students.l in the school. is this a representative sapmle at Abby's
Roman55 [17]
Yes, because the sample is random. If she had chosen the participants in any way other than randomly, it would not be representative of her entire school.
4 0
3 years ago
Please answer ASAP!!!
Gwar [14]

Answer:

Wheres the question

Step-by-step explanation:

4 0
3 years ago
Cottage cheese is being sold in different size containers. The sizes and prices are given below: 4 - 5 ounce containers for $5.
Karolina [17]

Division is the process of finding the number of times a number is multiplied by itself. The 16-ounce container is the better buy.

<h3>What is Division?</h3>

The division is the process of finding the number of times a number is multiplied by itself.

The prices of the different size containers are given to us,

4-5 ounce containers for $5. 00

1-12 ounce container for $3. 24

1-16 ounce container for $4. 19

Now, find the cost of the cheese per ounce when we buy the minimum cheese and when we but the maximum amount of cheese.

1.   For the first Container, 4-5 ounce containers for $5.00

If we buy 4 ounces of cheese, then the cost of each ounce of cheese will be $1.25 for each ounce, while if we buy the maximum amount of cheese therefore 5 ounces for $5 then the cost of each ounce of cheese will be $1.

2.  For, the second container, 1-12 ounce container for $3.24

If we buy 1 ounce of cheese, then the cost of each ounce of cheese will be $3.24 for each ounce, while if we buy the maximum amount of cheese therefore 5 ounces for $5 then the cost of each ounce of cheese will be $0.27.

3.  For, the third container, 1-16 ounce container for $4.19.

If we buy 1 ounce of cheese, then the cost of each ounce of cheese will be $4.19 for each ounce, while if we buy the maximum amount of cheese therefore 16 ounces for $4.19 then the cost of each ounce of cheese will be $0.26.

Thus, if we buy the maximum amount of cheese from each of the boxes, the third box will be the cheapest.

Hence, The 16-ounce container is the better buy.

Learn more about Division:

brainly.com/question/369266

3 0
2 years ago
Why was an attack against Vicksburg by river just not possible for Union forces?
Vsevolod [243]
<span>Because the town was built above sheer cliffs that were heavily defended by artillery.</span>
8 0
3 years ago
Other questions:
  • Wilma saves $12,000 at the end of every six months for 10 years. Assume 10% compounded semiannually and find the present value.
    8·1 answer
  • PLEASE HELP!! 20 POINTS
    6·1 answer
  • A section of wall is being framed, A model of the framing
    5·1 answer
  • HELP QUICK! which graph represents the solution set for the inequality PICTURE INCLUDED
    11·1 answer
  • 5.5 over x = 1.375 over 11
    11·1 answer
  • ****BRAINLIEST GOES TO THE FIRST CORRECT ANSWER****
    7·1 answer
  • Carmen used a random number generator to simulate a survey of how many children live in the households in her town. There are 1,
    5·1 answer
  • What is the slope of the line
    11·1 answer
  • 1.2+3x6= <br> 2.4(8-3)=<br> 3.7+8(3+2)= <br> 4. (6-5)+(8x10)=
    7·2 answers
  • A delivery company charges a flat rate of $3 for a large envelope plus an additional $0.25 per ounce for every ounce over a poun
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!