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2 years ago

Review the steps of the proof of the identity

Mathematics

1 answer:

astraxan [27]2 years ago

4 0

Answer:

step 2

and then step 3 : the error neutralized the error of step 2

Step-by-step explanation:

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

and because

sin(-b) = -sin(b)

cos(-b) = cos(b)

we have

sin(a - b) = sin(a)cos(-b) + cos(a)sin(-b) =

= sin(a)cos(b) - cos(a)sin(b)

3pi/2 = 270° or -90°.

sin(3pi/2) = sin(270) = sin(-90) = -1

that means, it is the full radius straight down from the center of the circle.

cos(3pi/2) = cos(270) = cos(-90) = cos(90) = 0

so, step 1 is correct :

sin(A - 3pi/2) = sin(A)cos(3pi/2) - cos(A)sin(3pi/2) =

= sin(A)×0 - cos(A)×(-1) (correct step 2)

but as we see, the provided step 2 is incorrect.

it should have been the indicated

sin(A)×0 - cos(A)×(-1)

and NOT the provided

sin(A)×0 + cos(A)×(-1)

step 3 based on the erroneous step 2 should then have been

sin(A)×0 - (1)cos(A)

but instead another error with the sign was made that neutralized the error of step 2 and we got after this second mistake by pure chance the overall correct step 3

sin(A)×0 + (1)cos(A)

so, again, the first error was made in step 2.

but technically, there was also a consecutive error made in step 3 to bring everything back to the correct approach.

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Consecutive means that they have a difference of one, so the expression will be n-1.

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3 years ago

Maggie spent $50 buying 11 magazines.Some of the magazines cost $4 and the rest cost $5.How many $5 magazines did she buy?

antiseptic1488 [7]

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Step-by-step explanation:

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2 years ago

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4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

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3 years ago

An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insu

ankoles [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insure at least one car. (ii) 70% of the customers insure more than one car. (iii) 20% of the customers insure a sports car. (iv) Of those customers who insure more than one car, 15% insure a sports car. Calculate the probability that a randomly selected customer insures exactly one car, and that car is not a sports car?

Answer:

P( X' ∩ Y' ) = 0.205

Step-by-step explanation:

Let X is the event that the customer insures more than one car.

Let X' is the event that the customer insures exactly one car.

Let Y is the event that customer insures a sport car.

Let Y' is the event that customer insures not a sport car.

From the given information we have

70% of customers insure more than one car.

P(X) = 0.70

20% of customers insure a sports car.

P(Y) = 0.20

Of those customers who insure more than one car, 15% insure a sports car.

P(Y | X) = 0.15

We want to find out the probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

P( X' ∩ Y' ) = ?

Which can be found by

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

From the rules of probability we know that,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y ) (Additive Law)

First, we have to find out P( X ∩ Y )

From the rules of probability we know that,

P( X ∩ Y ) = P(Y | X) × P(X) (Multiplicative law)

P( X ∩ Y ) = 0.15 × 0.70

P( X ∩ Y ) = 0.105

So,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )

P( X ∪ Y ) = 0.70 + 0.20 - 0.105

P( X ∪ Y ) = 0.795

Finally,

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

P( X' ∩ Y' ) = 1 - 0.795

P( X' ∩ Y' ) = 0.205

Therefore, there is 0.205 probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

6 0

3 years ago