Question

Use the given conditions to find the values of all six trigonometric functions. (If an answer is undefined, enter UNDEFINED.)? csc(x) = − 7/6, tan(x) > 0. how do I get :sin(x), cos(x), tan(x), csc(x), sec(x), cot(x)?

Use the given conditions to find the values of all six trigonometric functions. (If an answer is undefined, enter UNDEFINED.)?
csc(x) = − 7/6, tan(x) > 0.
how do I get :sin(x), cos(x), tan(x), csc(x), sec(x), cot(x)?

Answer 1

See answer below

Explanation:

Given: `csc x = -7/6; " "tan x > 0`

Find the correct quadrant:

`tan x > 0` occurs in the 1st and 3rd quadrants. Since the `csc` is negative, this would require the angle `x` to be in the 3rd quadrant, since all of the trigonometric functions are positive in the 1st quadrant.

Since `csc x = -7/6 = 1/(sin x) = (hyp)/(opp)`

This means the hypotenuse `h = 7` and the opposite side `y = -6`

Use Pythagorean Theorem to find the other side of the triangle:

`6^2 + a^2 = 7^2; " "a = sqrt(49 - 36) = sqrt(13)" "` adjacent

In the 3rd quadrant, this value would be negative: `adj = -sqrt(13)`

From these values, we can find all of the other trigonometric functions:

`sin x = 1/(csc x) = -6/7`

`cos x = (adj)/(hyp) = -sqrt(13)/7`

`tan x = (opp)/(adj) = (-6)/(-sqrt(13)) = 6/(sqrt(13)) * (sqrt(13))/(sqrt(13)) = (6 sqrt(13))/13`

`sec x = 1/(cos x) = 7/(-sqrt(13)) = -(7 sqrt(13))/13`

`cot x = 1/(tan x) = (-sqrt(13))/-6 = sqrt(13)/6`

Answer 2

Use the Pythagorean formula to find the ratio for the third side of a triangle and build out your results (shown below):

Explanation:

Understanding that cosecant (csc) is equal to the inverse of cosine, and we know that `cos="adjacent"/"hypotenuse"`, we can figure out the length of the third side of the triangle, and calculate the other terms:

`a^2+b^2=c^2`

`(6)^2+b^2=(7)^2`

`b=sqrt(7^2-6^2)`

`b=sqrt(49-36)`

`b=+-sqrt(13)`

I put the `+-` in because the square root of a number can be either positive or negative, and we don't know which term in the csc solution is the negative value.

We do know that `tan(x)>0` which means that either the hypotenuse is negative and both legs are positive, or that the hypotenuse is positive and both legs are negative.

The first option is impossible, since both legs being in the first quadrant would give a positive tangent and the co-secant would also be positive.

This leaves us with the latter, meaning that the legs of the triangle are in the `color(red)("3rd")` quadrant of the Cartesian unit circle.

Based on the above, we now know:

`"Adjacent"=-6`
`"Opposite"=-sqrt(13)`
`"Hypotenuse"=7`

Now, we assemble our trig values:

`sin="opp"/"hyp"=-sqrt(13)/7`
`cos="adj"/"hyp"=-6/7`
`tan="opp"/"adj"=sqrt(13)/6`
`sec="hyp"/"adj"=-7/sqrt(13)`
`csc="hyp"/"adj"=-7/6`
`cot="hyp"/"adj"=6/sqrt(13)`