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2 years ago

Verify the following trigonometric identity: (csc x + cot x)^2 = cos x +1/ 1- cos x

Mathematics

1 answer:

Airida [17]2 years ago

3 0

9514 1404 393

Explanation:

As written, the equation is not an identity. Perhaps you want to show ...

(csc(x) +cot(x))² = (cos(x) +1)/(1 -cos(x))

We will transform the left-side expression to the form of the right-side expression.

$(\csc(x)+\cot(x))^2=\left(\dfrac{1}{\sin(x)}+\dfrac{\cos(x)}{\sin(x)}\right)^2=\dfrac{(1+\cos(x))^2}{\sin(x)^2}\\\\=\dfrac{(1+\cos(x))^2}{1-\cos(x)^2}=\dfrac{1+\cos(x)}{1-\cos(x)}\cdot\dfrac{1+\cos(x)}{1+\cos(x)}=\boxed{\dfrac{\cos(x)+1}{1-\cos(x)}}$

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Two consecutive integers have a sum of 73. Find the integers.

patriot [66]

Here's how to do it using algebra:

Let the smaller integer be 'x' . Then the larger integer is (x+1).

Their sum is (x) + (x+1).

That's (2x + 1) .

The problem says their sum is 73, so we can write . . . 2x + 1 = 73 .

Subtract 1 from each side . . . 2x = 72

Divide each side by 2 . . . x = 36

The smaller integer is 36 .

The larger integer is (x+1) . That would be 37.

The two consecutive integers are 36 and 37 .

_______________________________________

Here's how to do it using your brain:

-- If they were both the same number, then their sum would be double it, and each integer would be half of 73 .

-- That's 36.5 .

-- But 36.5 is not even an integer, so neither integer can be 36.5. Is there something we can do to a pair of 36.5s to make them whole numbers (integers), make them different, and make them consecutive, but not let their sum change ?

-- Well, if we slice off a piece from one of them and glue it onto the other one, then we'll wind up with two numbers that are different, and their sum won't change.

-- What happens if we take the .5 off of one of them and glue it onto the other one ? Then one of the 36.5s shrinks to become 36, and the other one grows to become 37 .

-- Well whaddaya know ! Now we have two numbers that are consecutive integers, and they add up to 73 . Great move !

4 0

2 years ago

Read 2 more answers

Cassidy has saved $8,000 this year in an account that earns 9% interest annually. Based on the rule of 72, it will take about ye

den301095 [7]

Answer:

The number of years in which saving gets double is 8 years .

Step-by-step explanation:

Given as :

The principal amount saved into the account = p = $8,000

The rate of interest applied = r = 9%

The Amount gets double in n years = $A

Or, $A = 2 × p = $8,000 × 2 = $16,000

Let the number of years in which saving gets double = n years

Now, From Compound Interest method

Amount = Principal × $(1+\dfrac{\textrm rate}{100})^{\textrm time}$

Or, 2 × p = p × $(1+\dfrac{\textrm r}{100})^{\textrm n}$

Or, $16,000 = $8,000 × $(1+\dfrac{\textrm 9}{100})^{\textrm n}$

Or, $\dfrac{16,000}{8,000}$ = $(1.09)^{n}$

Or, 2 = $(1.09)^{n}$

Now, Taking Log both side

$Log_{10}$ 2 = $Log_{10}$ $(1.09)^{n}$

Or, 0.3010 = n × $Log_{10}$ 1.09

Or, 0.3010 = n × 0.0374

∴ n = $\dfrac{0.3010}{0.0374}$

I.e n = 8.04 ≈ 8

So, The number of years = n = 8

Hence, The number of years in which saving gets double is 8 years . Answer

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2 years ago

190258.50,152698.00, 122753.00, 220523.00, 231951.00 what is the average of these five amounts?

White raven [17]

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2 years ago

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Why do all transformation preserve distance

yuradex [85]

Well, a distance-preserving transformation is called a rigid motion, and the name suggests that it moves the points of the plane around in a rigid fashion.

A transformation is distance-preserving if the distance between the images of any two points and the distance between the two original points are equal.

If that's confusing, I get it; basically if you transform something, the points from the transformation are image points. Take the distance from a pair of image points, and take the distance from a pair of original points, and they should be the same for a rigid motion.

I keep emphasizing this b/c not all transformations preserve distance; a dilation can grow or shrink things. But if you didn't go over dilations, don't say nothin XD

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2 years ago

Which of the sets shown includes the elements of set Z that are both odd numbers and multiples of 7?

vesna_86 [32]

Answer:

B. {-21.-7. 7. 21)

Step-by-step explanation:

The set made up of odd numbers and multiples of 7 is given as;

{-21.-7. 7. 21)

Multiples of 7 are usually those numbers where 7 is factor.

Odd numbers are numbers that cannot be divide by 2.

So, the most fitting answer is B.

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2 years ago