# PracticeTrig Functions of Theta 1

#### puremath

##### New member
Let P(x, y) denote the point where the terminal side of angle theta (in standard position) meets the unit circle. Use the given information to evaluate the six trigonometric functions of theta.

A. x = -8/15 and pi < theta < (3pi)/2

B. y = 2/9 and 0 < theta < pi/2

#### MarkFL

##### La Villa Strangiato
Math Helper
The unit circle has the equation:

[MATH]x^2+y^2=1[/MATH]
A. We see point $$P$$ is in the third quadrant, which means $$y$$ must be negative, hence:

[MATH]y=-\sqrt{1-\left(-\frac{8}{15}\right)^2}=-\frac{\sqrt{161}}{15}[/MATH]
So, we can use:

[MATH]\sin(\theta)=y[/MATH]
[MATH]\cos(\theta)=x[/MATH]
[MATH]\tan(\theta)=\frac{y}{x}[/MATH]
[MATH]\csc(\theta)=\frac{1}{y}[/MATH]
[MATH]\sec(\theta)=\frac{1}{x}[/MATH]
[MATH]\cot(\theta)=\frac{x}{y}[/MATH]
See if you can complete part A, and then part B.

#### puremath

##### New member
The unit circle has the equation:

[MATH]x^2+y^2=1[/MATH]
A. We see point $$P$$ is in the third quadrant, which means $$y$$ must be negative, hence:

[MATH]y=-\sqrt{1-\left(-\frac{8}{15}\right)^2}=-\frac{\sqrt{161}}{15}[/MATH]
So, we can use:

[MATH]\sin(\theta)=y[/MATH]
[MATH]\cos(\theta)=x[/MATH]
[MATH]\tan(\theta)=\frac{y}{x}[/MATH]
[MATH]\csc(\theta)=\frac{1}{y}[/MATH]
[MATH]\sec(\theta)=\frac{1}{x}[/MATH]
[MATH]\cot(\theta)=\frac{x}{y}[/MATH]
See if you can complete part A, and then part B.

Thanks

#### puremath

##### New member
The unit circle has the equation:

[MATH]x^2+y^2=1[/MATH]
A. We see point $$P$$ is in the third quadrant, which means $$y$$ must be negative, hence:

[MATH]y=-\sqrt{1-\left(-\frac{8}{15}\right)^2}=-\frac{\sqrt{161}}{15}[/MATH]
So, we can use:

[MATH]\sin(\theta)=y[/MATH]
[MATH]\cos(\theta)=x[/MATH]
[MATH]\tan(\theta)=\frac{y}{x}[/MATH]
[MATH]\csc(\theta)=\frac{1}{y}[/MATH]
[MATH]\sec(\theta)=\frac{1}{x}[/MATH]
[MATH]\cot(\theta)=\frac{x}{y}[/MATH]
See if you can complete part A, and then part B.

I got both right on paper. I'm not going to show my work here. It is basically plug and chug. Thanks.