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3 years ago

Verify the identity. 4 csc 2x = 2 csc2x tan x

Mathematics

1 answer:

vlada-n [284]3 years ago

6 0

Step-by-step explanation:

$4csc(2x) = 2csc^2(x) tan(x)$

We start with Left hand side

We know that csc(x) = 1/ sin(x)

So csc(2x) is replaced by 1/sin(2x)

$4 \frac{1}{sin(2x)}$

Also we use identity

sin(2x) = 2 sin(x) cos(x)

$4 \frac{1}{2sin(x)cos(x)}$

4 divide by 2 is 2

Now we multiply top and bottom by sin(x) because we need tan(x) in our answer

$2\frac{1*sin(x)}{sin(x)cos(x)*sin(x)}$

$2\frac{sin(x)}{sin^2(x)cos(x)}$

$2\frac{1}{sin^2(x)} \frac{sin(x)}{cos(x)}$

We know that sinx/ cosx = tan(x)

Also 1/ sin(x)= csc(x)

so it becomes 2csc^2(x) tan(x) , Right hand side

Hence verified

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The average salary of all residents of a city is thought to be about $39,000. A research team surveys a random sample of 200 res

Rina8888 [55]

Answer:

$z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179$

$p_v =2*P(z>1.179)=2*[1-P(Z$

Step-by-step explanation:

Data given and notation

$\bar X=4000$ represent the average score for the sample

$s=12000$ represent the sample standard deviation

$n=200$ sample size

$\mu_o =39000$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu = 39000$

Alternative hypothesis : $\mu \neq 39000$

Compute the test statistic

The sample size is large enough to assume the distribution for the statisitc normal. The statistic for this case is given by:

$z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{40000-39000}{\frac{12000}{\sqrt{200}}}=1.179$

Give the appropriate conclusion for the test

Since is a two tailed test the p value would be:

$p_v =2*P(z>1.179)=2*[1-P(Z$

Conclusion

If we compare the p value and a significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, then the true population mean for the salary not differs significantly from the value of 39000.

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2 years ago

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AVprozaik [17]

Answer:

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1 year ago

The students in a club are selling flowerpots to raise

BARSIC [14]

Pe is the answer he heft ty high

6 0

3 years ago

The junior class of Eastlake High School has 146 students, and 67 of them are girls. How many boys are there in the class?

Katena32 [7]

Hello!

In order to find how many boys there are in the class simply subtract 67 from 146:

146-67= 79

There are 79 boys in the class

I hope it helps!

6 0

3 years ago

The time taken to deliver a pizza has a uniform probability distribution from 20 minutes to 60 minutes. What is the probability

Whitepunk [10]

Answer:

(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.

(2a) The percentage of results more than 45 is 79.67%.

(2b) The percentage of results less than 85 is 91.77%.

(2c) The percentage of results are between 75 and 90 is 15.58%.

(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.

Step-by-step explanation:

(1)

Let Y = the time taken to deliver a pizza.

The random variable Y follows a Uniform distribution, U (20, 60).

The probability distribution function of a Uniform distribution is:

$f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.$

Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:

$P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70$