1.)P(-15/17,-8/17) is found on the unit crcle. find sin0 and cos0.

2.)solve: x=Arc tan(-sq root3).

3.)find sec0 if sin0=-4/5 and 270 degrees<0<360 degrees.

1. Any point P on the unit circle has coordinates (cosØ, sinØ)

so cosØ = -15/17
sinØ = -8/17

2. x = arctan(-√3)
or
tanx = -√3/1 , we know tan 60°=+√3
x must be in II or IV
x = 120° or 300° or -60° or ....

3. sinØ = -4/5
Ø is in IV
recognize the 3 4 5 triangle
so cosØ = 3/5

1.) To find sinθ and cosθ for the point P(-15/17, -8/17) on the unit circle, we need to use the Pythagorean identity:

sin^2θ + cos^2θ = 1

First, let's find the value of sinθ:

sinθ = y-coordinate of the point P = -8/17

Next, we can use the Pythagorean identity to find the value of cosθ:

cosθ = sqrt(1 - sin^2θ)

cosθ = sqrt(1 - (-8/17)^2)

cosθ = sqrt(1 - 64/289)

cosθ = sqrt(225/289)

cosθ = 15/17

So the values of sinθ and cosθ for the point P are sinθ = -8/17 and cosθ = 15/17.

2.) To solve x = arctan(-√3), we need to find the angle whose tangent is -√3.

By definition, the arctan function returns the angle whose tangent is the given value.

In this case, we have x = arctan(-√3), which means tan(x) = -√3.

To find x, we need to use the inverse tangent function or arctan.

So x = arctan(-√3) implies x = -π/3 or -60 degrees.

Hence, the solution is x = -π/3 or x = -60 degrees.

3.) Given sinθ = -4/5 and 270° < θ < 360°, we can find secθ.

We know that secθ is the reciprocal of cosθ, and we already have the value for sinθ.

Using the Pythagorean identity sin^2θ + cos^2θ = 1, we can find cosθ first:

cosθ = sqrt(1 - sin^2θ)

cosθ = sqrt(1 - (-4/5)^2)

cosθ = sqrt(1 - 16/25)

cosθ = sqrt(9/25)

cosθ = 3/5

Now, we can find secθ as the reciprocal of cosθ:

secθ = 1/cosθ

secθ = 1/(3/5)

secθ = 5/3

Therefore, secθ is equal to 5/3.