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3 years ago

Would appreciate some help with this: Verify (sinq+cosq)^2 - 1 = sin2q

Mathematics

1 answer:

Phantasy [73]3 years ago

7 0

Hello from MrBillDoesMath!

Answer:

See Discussion below

Discussion:

(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2

(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1

1 + 2 sinq*cosq (*)

Setting a = b = q in the trig identity:

sin(a+b) = sina*cosb + cosa*sinb

sin(2q) = (**)

sinq*cosq + cosq*sinq => as both terms are identical

2 sinq*cosq

Combining (*) and (**)

(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)

= 1 + sin(2q)

Hence

(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides

(sinq + cosq)^2 - 1 = sin(2q)

The last statement is what we are trying to prove.

Thank you,

MrB

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3 years ago

2. In your own words, explain HOW to find the standard deviation of this data set. (20 points) 3. What does the standard deviati

Ket [755]

Answer:

$\sigma = 11.28$ --- Standard deviation

Step-by-step explanation:

Given

See attachment for graph

Solving (a): Explain how the standard deviation is calculated.

<u>Start by calculating the mean</u>

To do this, we divide the sum of the products of grade and number of students by the total number of students;

i.e.

$\bar x = \frac{\sum fx}{\sum f}$

So, we have:

$\bar x = \frac{50 * 1 + 57 * 2 + 60 * 4 + 65 * 3 +72 *3 + 75 * 12 + 77 * 10 + 81 * 6 + 83 * 6 + 88 * 9 + 90 * 12 + 92 * 12 +95 * 2 + 99 * 4 + 100 * 5}{1 + 2 + 4 + 3 +3 + 12 + 10 + 6 + 6 + 9 + 12 + 12 +2 + 4 + 5}$

$\bar x = \frac{7531}{91}$

$\bar x = 82.76$

Next, calculate the variance using the following formula:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f}$

i.e subtract the mean from each dataset; take the squares; add up the squares; then divide the sum by the number of dataset

So, we have:

$\sigma^2 = \frac{1 * (50 - 82.76)^2 +...................+ 5 * (100 - 82.76)^2}{91}$

$\sigma^2 = \frac{11580.6816}{91}$

$\sigma^2 = 127.26$

Lastly, take the square root of the variance to get the standard deviation

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{127.26}$

$\sigma = 11.28$ --- approximated

<em>Hence, the standard deviation is approximately 11.28</em>

Considering the calculated mean (i.e. 82.76), the standard deviation (i.e. 11.28) is small and this means that the grade of the students are close to the average grade.

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yKpoI14uk [10]

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