# Determine the Value of all six trigonometric functions of theta when given?

I need help. This is for my midterm review, if anyone is willing to help, I'd appreciate it.

a) sin theta= 3/5 and theta is in quadrant 1.

b) tan theta= -2 and theta is in quadrant II

c) sec theta = 2 radical 3/3 and theta is in quadrant II

d) csc theta= -2/3 and theta is in quadrant III.

I need to find all six trigonometric functions for each one ( sin, cos, tan, csc, sec, cot)

## With each function given, you have two sides of a right triangle. Using the pyth theorm, you can find the other side, and thus, the other functions. We will be happy to check your work.

For example, c) If secTheta = 2 sqrt (1/3), then the sides of the triangle are

opposite figure out

adjacent sqrt 3

hypo= 2

and the opposite will be

4=3+opposite^2 or opposite is +-1, Now in quadrant two, that means sec is negative, so it should have been -2sqrt(1/3), and the opposite (vertical) is 1. From that you get the six functions.

## I will do c. You try the rest.

Oh no I won't --> sign of sec is negative in quad 2. I think typo

I will do d.

d) csc T = 1/sin T = -2/3

Hey! what is going on here?

the absolute value of sin T may not be greater than 1. It can not be -3/2

Either we are looking at typos or someone is playing games with you.

I will do b I guess

b) draw triangle in quad 2

-1 along -x axis

+2 up at x = -2

now hypotenuse = sqrt (-1^2 + 2^2) = sqrt 5

so

sin = 2/sqrt 5

cos = -1/sqrt 5 = -(1/5)sqrt 5

tan = -2/1

csc = 1/sin = (1/2)sqrt 5

sec = 1/cos = -sqrt 5

cot = 1/tan = -1/2

## prove the identities (theta + phi), cos theta = -1/3 in Quadrant III, sin theta = 1/4 in quadrant II

## I can help you with that! To find the values of the six trigonometric functions (sin, cos, tan, csc, sec, cot) of an angle, we can use the definitions and relationships between these functions.

a) Given sin(theta) = 3/5 in quadrant 1:

To find the other trigonometric functions, we can use the Pythagorean identity sin^2(theta) + cos^2(theta) = 1.

Since theta is in quadrant 1, we know that sin(theta) is positive. Using sin^2(theta) = (3/5)^2 = 9/25, we can solve for cos^2(theta): 1 - 9/25 = 16/25.

Therefore, cos^2(theta) = 16/25. Taking the square root, we get cos(theta) = ±4/5. But since theta is in quadrant 1, we take the positive value: cos(theta) = 4/5.

Now, we can find the remaining trigonometric functions:

tan(theta) = sin(theta) / cos(theta) = (3/5) / (4/5) = 3/4

csc(theta) = 1 / sin(theta) = 1 / (3/5) = 5/3

sec(theta) = 1 / cos(theta) = 1 / (4/5) = 5/4

cot(theta) = 1 / tan(theta) = 1 / (3/4) = 4/3

So, the values of the six trigonometric functions for theta in this case are:

sin(theta) = 3/5

cos(theta) = 4/5

tan(theta) = 3/4

csc(theta) = 5/3

sec(theta) = 5/4

cot(theta) = 4/3

b) Given tan(theta) = -2 in quadrant II:

We can start by finding sin(theta) using the tangent identity tan(theta) = sin(theta) / cos(theta).

Since tan(theta) is negative in quadrant II and sin(theta) is positive, we have: sin(theta) = 2, cos(theta) = -1 (taking the negative value since cos(theta) is negative in quadrant II).

Now, we can find the remaining trigonometric functions:

csc(theta) = 1 / sin(theta) = 1 / 2 = 1/2

sec(theta) = 1 / cos(theta) = 1 / (-1) = -1

cot(theta) = 1 / tan(theta) = 1 / (-2) = -1/2

So, the values of the six trigonometric functions for theta in this case are:

sin(theta) = 2

cos(theta) = -1

tan(theta) = -2

csc(theta) = 1/2

sec(theta) = -1

cot(theta) = -1/2

c) Given sec(theta) = (2√3)/3 in quadrant II:

To find sin(theta), we can use the reciprocal identity sec(theta) = 1 / cos(theta):

cos(theta) = 3 / (2√3) = √3 / 2

Now, we can find the remaining trigonometric functions:

sin(theta) = √(1 - cos^2(theta)) = √(1 - (3/2)^2) = √((-1/4)) = NaN (since there is no real value)

tan(theta) = sin(theta) / cos(theta) = NaN

csc(theta) = 1 / sin(theta) = 1 / NaN = undefined

sec(theta) = (2√3)/3

cot(theta) = 1 / tan(theta) = 1 / NaN = undefined

In this case, sine, cosine, tangent, cosecant, and cotangent are undefined for the given conditions.

d) Given csc(theta) = -2/3 in quadrant III:

To find sin(theta), we can use the reciprocal identity csc(theta) = 1 / sin(theta):

sin(theta) = 1 / csc(theta) = 1 / (-2/3) = -3/2

Now, we can find the remaining trigonometric functions:

cos(theta) = √(1 - sin^2(theta)) = √(1 - (-3/2)^2) = √((-5/4)) = NaN (since there is no real value)

tan(theta) = sin(theta) / cos(theta) = (-3/2) / NaN = NaN

sec(theta) = 1 / cos(theta) = 1 / NaN = undefined

cot(theta) = 1 / tan(theta) = 1 / NaN = undefined

In this case, cosine, tangent, secant, and cotangent are undefined for the given conditions.

Please note that when the values of sine, cosine, tangent, secant, or cotangent are undefined, it means that the angle does not have a real value for those functions in the given quadrant.