Pls help me with my assignment <br> ty

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

2 years ago

Use the theoretical method to determine the probability of the following event. A randomly selected person has a birthday in Nov

MakcuM [25]

Answer:

There is an 8.22% probability that a randomly selected person has a birthday in November.

Step-by-step explanation:

The theoretical method to find the probability is the division of the number of desired outcomes by the number of total outcomes.

A randomly selected person has a birthday in November

There are 365 days in a year, so the number of total outcomes is 365.

There are 30 days in november, so the number of desired outcomes is 30.

So the probability is

$P = \frac{30}{365} = 0.0822$

There is an 8.22% probability that a randomly selected person has a birthday in November.

3 0

3 years ago

6. Describe how we could conduct a simulation to decide whether a simple random sample from this population could produce a samp

BigorU [14]

Step-by-step explanation:

We have proportion p = 50/100 = 0.50

Probability of success = 1-0.50 = 0.50

To conduct simulation, let's assume experiment is to be done 10 times

1. Use binomial parameters n, number = 10, p = 0.50. we then get the necessary size of population To carry out the sampling.

2. We get a random sample and get the size of people who support an against smoking with kids in the car.

3. We repeat this process after drawing sample and also get the number of those in support of ban.

4. After getting n, we find proportion.

8 0

2 years ago

You are buying markers for the students in first grade. Each student will get 5 markers. There are 16 students in one class and

Masja [62]

52 markers that you will need to buy.

6 0

2 years ago

Read 2 more answers

What is the area of the shaded sector?

umka2103 [35]

40pie units squared / esa es

6 0

2 years ago

Pls help me with my assignment ty