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- Thread starter puremath
- Start date

[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.

[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

[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.