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statuscvo [17]
2 years ago
15

Cos2A + cos2A cot2A = cot2A​

Mathematics
1 answer:
Vlad [161]2 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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