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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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If two matrices, A and B, are equal, which of the
alexandr402 [8]

If A and B are equal:

Matrix A must be a diagonal matrix: FALSE.

We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:

A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]

Both matrices must be square: FALSE.

We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works

Both matrices must be the same size: TRUE

If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.

For any value of i, j; aij = bij: TRUE

Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.

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3 years ago
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astraxan [27]
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Suppose the lengths of two sides of a right triangle are represented by 2x and 3 (x + 1), and the longest side is 17 units. Find
densk [106]

Answer:

x=4

Step-by-step explanation:

<u>Step 1</u>:-

given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.

AB = 2x and BC = 3(x+1) and longest side AC= 17

by using Pythagoras theorem

AC^2 = AB^2 + BC^2

<u>step 2:-</u>

The hypotenuse is longest side is AC = 17 units

(17)^2 = 4x^2 +9(X+1)^2

on simplification, we will use formula

(a + b)^2 = a^2 +2ab+b^2

289 = 4x^2 +9(x^2+2x+1)

13x^2 +18x-280 = 0

finding factors  70 X 52 = 3640

13x^2 +70x-52x-280 = 0

13x^2 -52x+ 70x-280 = 0

Taking common , we get

13x(x-4)+70(x-4)=0

x-4=0 and 13x+70=0

x=4 and 13x =-70

x=4 and x=\frac{-70}{13}

we can not choose negative value so x value is 4

Final answer:- x = 4

<u>verification:-</u>

<u></u>AC^2 = AB^2 + BC^2<u></u>

289 = 4(4)^2+9(4+1)^2

289 = 64 +9(25)

289=289

6 0
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MatroZZZ [7]

Given:

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Taking sin on both sides, we get

\sin \sin^{-1}\left(\dfrac{5}{8}\right)=\sin A

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Interchanging the sides, we get

\sin A=\dfrac{5}{8}

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