Question

this is a 2 - page document!! directions: given a point on the terminal side of \\( \theta \\) in standard form, find the exact values of the trigonometric functions of \\( \theta \\). 1. \\( p(-5, 12) \\) 2. \\( p(-4, -10) \\) directions: use the given information to find the exact values of the five remaining trigonometric

Explanation:

Step1: Identify x, y, r for P(-4,-10)

For point $P(x,y)=P(-4,-10)$, calculate $r=\sqrt{x^2+y^2}$:
$$r=\sqrt{(-4)^2+(-10)^2}=\sqrt{16+100}=\sqrt{116}=2\sqrt{29}$$

Step2: Calculate $\cos\theta$

$\cos\theta=\frac{x}{r}$:
$$\cos\theta=\frac{-4}{2\sqrt{29}}=-\frac{2\sqrt{29}}{29}$$

Step3: Calculate $\tan\theta$

$\tan\theta=\frac{y}{x}$:
$$\tan\theta=\frac{-10}{-4}=\frac{5}{2}$$

Step4: Calculate $\csc\theta$

$\csc\theta=\frac{r}{y}$:
$$\csc\theta=\frac{2\sqrt{29}}{-10}=-\frac{\sqrt{29}}{5}$$

Step5: Calculate $\sec\theta$

$\sec\theta=\frac{r}{x}$:
$$\sec\theta=\frac{2\sqrt{29}}{-4}=-\frac{\sqrt{29}}{2}$$

Step6: Calculate $\cot\theta$

$\cot\theta=\frac{x}{y}$:
$$\cot\theta=\frac{-4}{-10}=\frac{2}{5}$$

Answer:

For $P(-5,12)$ (completed for reference):
$\sin\theta=\frac{12}{13}$, $\cos\theta=-\frac{5}{13}$, $\tan\theta=-\frac{12}{5}$, $\csc\theta=\frac{13}{12}$, $\sec\theta=-\frac{13}{5}$, $\cot\theta=-\frac{5}{12}$

For $P(-4,-10)$:
$\sin\theta=-\frac{5\sqrt{29}}{29}$, $\cos\theta=-\frac{2\sqrt{29}}{29}$, $\tan\theta=\frac{5}{2}$, $\csc\theta=-\frac{\sqrt{29}}{5}$, $\sec\theta=-\frac{\sqrt{29}}{2}$, $\cot\theta=\frac{2}{5}$