Question

find the exact value of each of the six trigonometric functions of θ, if (-4, -5) is a point on the terminal side of angle θ.
sin θ =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
cos θ =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
tan θ =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
csc θ =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
sec θ =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
cot θ =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Find the radius $r$

Given the point $(x,y)=(-4,-5)$, use the formula $r = \sqrt{x^{2}+y^{2}}$. So $r=\sqrt{(-4)^{2}+(-5)^{2}}=\sqrt{16 + 25}=\sqrt{41}$.

Step2: Calculate $\sin\theta$

By the definition $\sin\theta=\frac{y}{r}$, substituting $y=-5$ and $r = \sqrt{41}$, we get $\sin\theta=\frac{-5}{\sqrt{41}}=-\frac{5\sqrt{41}}{41}$.

Step3: Calculate $\cos\theta$

Using the definition $\cos\theta=\frac{x}{r}$, with $x = - 4$ and $r=\sqrt{41}$, we have $\cos\theta=\frac{-4}{\sqrt{41}}=-\frac{4\sqrt{41}}{41}$.

Step4: Calculate $\tan\theta$

Since $\tan\theta=\frac{y}{x}$, substituting $x=-4$ and $y = - 5$, we obtain $\tan\theta=\frac{-5}{-4}=\frac{5}{4}$.

Step5: Calculate $\csc\theta$

As $\csc\theta=\frac{r}{y}$, with $r=\sqrt{41}$ and $y=-5$, we get $\csc\theta=-\frac{\sqrt{41}}{5}$.

Step6: Calculate $\sec\theta$

Using $\sec\theta=\frac{r}{x}$, and $r=\sqrt{41}$, $x=-4$, we have $\sec\theta=-\frac{\sqrt{41}}{4}$.

Step7: Calculate $\cot\theta$

Since $\cot\theta=\frac{x}{y}$, substituting $x=-4$ and $y=-5$, we obtain $\cot\theta=\frac{4}{5}$.

Answer:

$\sin\theta=-\frac{5\sqrt{41}}{41}$
$\cos\theta=-\frac{4\sqrt{41}}{41}$
$\tan\theta=\frac{5}{4}$
$\csc\theta=-\frac{\sqrt{41}}{5}$
$\sec\theta=-\frac{\sqrt{41}}{4}$
$\cot\theta=\frac{4}{5}$