Question

find the quadrant containing 0 if the given conditions are true.
csc 0 less than 0 & sec 0 less than 0

Answer 1

What this means is that the csc returns a negative value and the secant returns a negative value. \[csc \Theta = \frac {1}{sin \Theta}\] and \[sec \Theta = \frac {1}{cos \Theta}\]So if csc and sec are negative, what does that tell you about the sign of sin x and cos x? What quadrant is this condition true?

Answer 2

Please explain the answer

Answer 3

The way i posted it up is whats showing on my test paper.

Answer 4

I'm assuming the point 0 is actually O (not zero). In what quadrants are sin(x) negative? Cos(x)?

Answer 5

the zero is written the way u did

Answer 6

So what is the answer to the question? Does that mean its true?

Answer 7

The answer isn't true or false. Before you get the answer, I want you to understand the theory behind the question :). When is sin(x) negative? (Hint: sin(x) is positive in two quadrants, and negative in two quadrants)

Answer 8

would that be quadrant 2

Answer 9

Sin(x) is negative in TWO quadrants, and positive in TWO quadrants. Also remember that \[sin(x) = \frac {y}{H}\]where H is the hypotenuse (Never negative), and y is y-coordinate. Therefore, in which quadrants is y, and therefore sin(x) negative?

Answer 10

So that will be on the iv quadrant then

Answer 11

Remember, TWO quadrants.

Answer 12

confused so that will be quadrant 2

Answer 13

Think about it this way: If you take a line segment starting on the positive x axis and rotate it counterclockwise, when will the y-coordinate for any point on that line segment become negative?

Answer 14

IV

Answer 15

CONFUSED STILL PLEASE PROCEED WITH TEACHING ME

Answer 16

The y-coordinate will be negative in quadrant III and quadrant IV. Another way to think about this is if you were to pick any point in each quadrant, what would the sign of each coordinate be?

Answer 17

Since in quadrants III and IV the y coordinate will be negative, the value of sin(x) is also negative.

Answer 18

SO ITS QUADRANT 3

Answer 19

The answer to the question is quadrant 3, since that is the only quadrant in which both the cos(x) and the sin(x) are negative.

Answer 20

\[cos(x) = \frac {x}{H}\]

Answer 21

THANKS I UNDERSTOOD IT.

Answer 22

CAN U HELP WITH ANOTHER QUESTIN

Answer 23

You're welcome :)

Answer 24

Sure

Answer 25

OK HERE IT IS:
Y=1/3 TAN X
I NEED TO FIND PERIOD OF EQUATION

Answer 26

For any sinusoidal equation (sin(kx), cos(kx), etc.) the period of the function is \[T = \frac {2 \pi}{k}\]where T is the period and k is the coefficient in the sinusoidal argument.

Answer 27

1/3=2PI/X
IS THAT HOW IM SUPPOSE TO RE-WRITE THE EQUATION

Answer 28

VERY CONFUSED PLEASE HELP ME

Answer 29

IM TAKING A TEST RIGHT NOW

Answer 30

In general, a sinusoidal function has the form \[Y = A sin (k(x - b)) + C\]where A is the amplitude, or how high the sine function reaches, k defines the period, b is the phase shift, and c is the vertical shift.

Answer 31

OK PLEASE CONTINUE

Answer 32

The period is \[T = \frac {2\pi}{k}\]What is "k" in your equation?

Answer 33

1/3

Answer 34

No, 1/3 is your amplitude, A. Remember, k is the coefficient in front of your x variable. So what is your value of k?

Answer 35

SORRY MATH I DON'T UNDERSTAND BUT WHEN EXPLAINED STEP BY STEP I GET IT

Answer 36

SO K IS THE SHIFT

Answer 37

AMPLITUDE, SHIFT & PERIOD IS WHAT MAKES UP MY EQUATION

Answer 38

K defines how long the period is.

Answer 39

SO HOW DO I WORK OUT THE PROBLEM

Answer 40

Your equation is \[y = \frac {1}{3} tan(x)\]What is the coefficient of x? (Hint: it's not 1/3)

Answer 41

2PI

Answer 42

2pi is your period!!!

Answer 43

Since the coefficient in front of x is excluded, that means it's 1, and therefore \[T = \frac {2\pi}{k} = \frac {2\pi}{1} = 2\pi\]

Answer 44

THANK YOU SO THE ANSWER WAS 2PI

Answer 45

Yes

Answer 46

WOW YOUR A MATH WIZARD

Answer 47

CAN U HELP ME MORE

Answer 48

CAN U HELP ME WITH 2 MORE PROBLEMS

Answer 49

We'll see, depending on how long it takes.

Answer 50

OK I HAVE AN UPSIDE DOWN TRIANGLE, ONE SIDE IS 21DEGREES AND THE INSIDE AT THE TIP AREA IS 60 DEGREES. I NEED TO FIND EXACT VALUE OF X & Y

Answer 51

Would you be able to draw a diagram using the draw button to help me visualize it?

Answer 52

YES

Answer 53

Thanks, it makes it easier for me to answer and explain.