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find the exact value of each of the six trigonometric functions of θ, i…

Question

find the exact value of each of the six trigonometric functions of θ, if (6, - 3) 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: Calculate the value of $r$

Given the point $(x = 6,y=- 3)$, use the formula $r=\sqrt{x^{2}+y^{2}}$. So, $r=\sqrt{6^{2}+(-3)^{2}}=\sqrt{36 + 9}=\sqrt{45}=3\sqrt{5}$.

Step2: Calculate $\sin\theta$

By the definition $\sin\theta=\frac{y}{r}$, substituting $y = - 3$ and $r=3\sqrt{5}$, we get $\sin\theta=\frac{-3}{3\sqrt{5}}=-\frac{1}{\sqrt{5}}=-\frac{\sqrt{5}}{5}$.

Step3: Calculate $\cos\theta$

Using the definition $\cos\theta=\frac{x}{r}$, substituting $x = 6$ and $r = 3\sqrt{5}$, we have $\cos\theta=\frac{6}{3\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$.

Step4: Calculate $\tan\theta$

By the definition $\tan\theta=\frac{y}{x}$, substituting $x = 6$ and $y=-3$, we obtain $\tan\theta=\frac{-3}{6}=-\frac{1}{2}$.

Step5: Calculate $\csc\theta$

Since $\csc\theta=\frac{1}{\sin\theta}$, substituting $\sin\theta=-\frac{\sqrt{5}}{5}$, we get $\csc\theta=-\sqrt{5}$.

Step6: Calculate $\sec\theta$

As $\sec\theta=\frac{1}{\cos\theta}$, substituting $\cos\theta=\frac{2\sqrt{5}}{5}$, we have $\sec\theta=\frac{\sqrt{5}}{2}$.

Step7: Calculate $\cot\theta$

Since $\cot\theta=\frac{1}{\tan\theta}$, substituting $\tan\theta=-\frac{1}{2}$, we obtain $\cot\theta=-2$.

Answer:

$\sin\theta=-\frac{\sqrt{5}}{5}$
$\cos\theta=\frac{2\sqrt{5}}{5}$
$\tan\theta=-\frac{1}{2}$
$\csc\theta=-\sqrt{5}$
$\sec\theta=\frac{\sqrt{5}}{2}$
$\cot\theta=-2$