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statuscvo [17]
3 years ago
15

Cos2A + cos2A cot2A = cot2A​

Mathematics
1 answer:
Vlad [161]3 years ago
7 0

cos(2<em>A</em>) + cos(2<em>A</em>) cot(2<em>A</em>) = cot(2<em>A</em>)

cos(2<em>A</em>) + cos(2<em>A</em>) cot(2<em>A</em>) - cot(2<em>A</em>) = 0

cot(2<em>A</em>) sin(2<em>A</em>) + cos(2<em>A</em>) cot(2<em>A</em>) - cot(2<em>A</em>) = 0

cot(2<em>A</em>) (sin(2<em>A</em>) + cos(2<em>A</em>) - 1) = 0

cot(2<em>A</em>) = 0   <u>or</u>   sin(2<em>A</em>) + cos(2<em>A</em>) - 1 = 0

cot(2<em>A</em>) = 0   <u>or</u>   sin(2<em>A</em>) + cos(2<em>A</em>) = 1

cot(2<em>A</em>) = 0   <u>or</u>   sin(2<em>A</em>) cos(<em>π</em>/4) + cos(2<em>A</em>) sin(<em>π</em>/4) = 1/√2

cot(2<em>A</em>) = 0   <u>or</u>   sin(2<em>A</em> + <em>π</em>/4) = 1/√2

[2<em>A</em> = cot⁻¹(0) + <em>nπ</em>]   <u>or</u>

[2<em>A</em> + <em>π</em>/4 = sin⁻¹(1/√2) + 2<em>nπ</em>   <u>or</u>   2<em>A</em> + <em>π</em>/4 = <em>π</em> - sin⁻¹(1/√2) + 2<em>nπ</em>]

(where <em>n</em> is any integer)

[2<em>A</em> = <em>π</em>/2 + <em>nπ</em>]   <u>or</u>

[2<em>A</em> + <em>π</em>/4 = <em>π</em>/4 + 2<em>nπ</em>   <u>or</u>   2<em>A</em> + <em>π</em>/4 = 3<em>π</em>/4 + 2<em>nπ</em>]

[<em>A</em> = <em>π</em>/4 + <em>nπ</em>/2]   <u>or</u>

[2<em>A</em> = 2<em>nπ</em>   <u>or</u>   2<em>A</em> = <em>π</em>/2 + 2<em>nπ</em>]

[<em>A</em> = <em>π</em>/4 + <em>nπ</em>/2]   <u>or</u>

[<em>A</em> = <em>nπ</em>   <u>or</u>   <em>A</em> = <em>π</em>/4 + <em>nπ</em>]

<em>A</em> = (1 + 2<em>n</em>) <em>π</em>/4   <u>or</u>   <em>A</em> = <em>nπ</em>   <u>or</u>   <em>A</em> = <em>π</em>/4 + <em>nπ</em>

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What are the vertex , focus and directrix of the parabola with the equation x2+8x+4y+4=0?
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You can rewrite the equation in vertex form to find some of the parameters of interest.
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_____
On a graph, if all you know is the vertex, you can draw a line with slope 1/2 through the vertex. It intersects the parabola at the y-value of the focus. Then the distance from that point of intersection to the axis of symmetry (where the focus lies) is the same as the distance from that point to the directrix.

Every point on the parabola, including the vertex and the point of intersection just described, is the same distance from the focus and the directrix. This point of intersection is on a horizontal line through the focus, so finding the distance is made easy.

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i hope this helps :)

Step-by-step explanation:

6 0
3 years ago
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