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2 years ago

Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A

1 answer:

4 0

Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
= -2cos4A*sin3A + 2cos4A*sinA
= 2cos4A(sinA - sin3A)
= 2cos4A*2cos(2A)sin(-A)
= -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
= 2cos4A*cos3A - 2cos4A*cosA
= 2cos4A(cos3A - cosA)
= 2cos4A*(-2)sin2A*sinA
= -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
= cos2A/sin2A
= cot2A

This verifies the identity.

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2 years ago

What is 3x +99 2(8*3)= 135<br> what is x equal to?

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Step-by-step explanation: I am guessing you meant ...99 + 2(8*3)...

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3 years ago

Find the greatest common factor of 4w and 6c^3

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ

liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

$\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy$

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Mx(x,y) = d/dx 8x²y = 16xy
Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy$

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2$

We conclude that the line integral is 1/2

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2 years ago

What is the equation of the line that passes through 1,4 and 2,1

sashaice [31]

Answer:

y = -3x + 7

Step-by-step explanation:

The equation of a line

y = mx + c

y - intercept point y

m - slope of the line

x - intercept point x

c - intercept point of the line

Step 1: find the slope

m = y2 - y1 / x2 - x1

Given two points

( 1 , 4) ( 2 , 1)

x1 = 1

y1 = 4

x2 = 2

y2 = 1

insert the values

m = 1 - 4 / 2 - 1

m = -3/1

m = -3

y = -3x + c

Step 2: substitute any of the two points given into the equation of a line

y = -3x + c

( 1 ,4)

x = 1

y = 4

4 = -3(1) + c

4 = -3 + c

4 + 3 = c

c = 7

Step 3: sub c into the equation

y = -3x + 7

The equation of the line is

y = -3x + 7

3 0

3 years ago