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The point given below is on the terminal side of an angle $\theta$. Find the exact value of each of the six trigonometric functions of $\theta$.
$(-9,12)$
\[
\sin \theta=
\]
(Simplify your answer. Type an exact answer, using radicals as needed. Use integers or fractions for any numbers in the expression. Rationalize all denominators.)

Step 1 ：Given the point (-9,12), we can think of -9 as the adjacent side (x) and 12 as the opposite side (y).

Step 2 ：We can find the hypotenuse (r) using the Pythagorean theorem: \(r = \sqrt{x^2 + y^2} = \sqrt{(-9)^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\).

Step 3 ：Now we can find the six trigonometric functions:

Step 4 ：\(\sin(\theta) = \frac{y}{r} = \frac{12}{15} = \frac{4}{5}\)

Step 5 ：\(\cos(\theta) = \frac{x}{r} = \frac{-9}{15} = \frac{-3}{5}\)

Step 6 ：\(\tan(\theta) = \frac{y}{x} = \frac{12}{-9} = \frac{-4}{3}\)

Step 7 ：\(\csc(\theta) = \frac{r}{y} = \frac{15}{12} = \frac{5}{4}\)

Step 8 ：\(\sec(\theta) = \frac{r}{x} = \frac{15}{-9} = \frac{-5}{3}\)

Step 9 ：\(\cot(\theta) = \frac{x}{y} = \frac{-9}{12} = \frac{-3}{4}\)

Step 10 ：So, the exact values of the six trigonometric functions of \(\theta\) are \(\sin(\theta) = \frac{4}{5}\), \(\cos(\theta) = \frac{-3}{5}\), \(\tan(\theta) = \frac{-4}{3}\), \(\csc(\theta) = \frac{5}{4}\), \(\sec(\theta) = \frac{-5}{3}\), and \(\cot(\theta) = \frac{-3}{4}\).

Step 11 ：Final Answer: \(\boxed{\sin(\theta) = \frac{4}{5}, \cos(\theta) = \frac{-3}{5}, \tan(\theta) = \frac{-4}{3}, \csc(\theta) = \frac{5}{4}, \sec(\theta) = \frac{-5}{3}, \cot(\theta) = \frac{-3}{4}}\)