Assume that alpha is an angle in the standard position whose terminal side contains the given point. Find the exact values of sin alpha, cos alpha, tan alpha, csc alpha, sec alpha, and cot alpha. (-2,6)
To find the values of the trigonometric functions for an angle in standard position, we can use the following formulas:
\[ \sin(\alpha) = \frac{y}{r}, \cos(\alpha) = \frac{x}{r}, \tan(\alpha) = \frac{y}{x}, \]
\[ \csc(\alpha) = \frac{r}{y}, \sec(\alpha) = \frac{r}{x}, \cot(\alpha) = \frac{x}{y}, \]
where \( x \) and \( y \) are the coordinates of the point and \( r \) is the distance from the origin to the point, which can be found using the Pythagorean theorem: \( r = \sqrt{x^2 + y^2} \).
In this case, the coordinates of the point are \( x = -2 \) and \( y = 6 \). To find the distance \( r \), we calculate:
\[ r = \sqrt{(-2)^2 + 6^2} = \sqrt{40} = 2\sqrt{10}. \]
Now, we can find the values of the trigonometric functions:
\[ \sin(\alpha) = \frac{y}{r} = \frac{6}{2\sqrt{10}} = \frac{3\sqrt{10}}{10}, \]
\[ \cos(\alpha) = \frac{x}{r} = \frac{-2}{2\sqrt{10}} = -\frac{\sqrt{10}}{10}, \]
\[ \tan(\alpha) = \frac{y}{x} = \frac{6}{-2} = -3, \]
\[ \csc(\alpha) = \frac{r}{y} = \frac{2\sqrt{10}}{6} = \frac{\sqrt{10}}{3}, \]
\[ \sec(\alpha) = \frac{r}{x} = \frac{2\sqrt{10}}{-2} = -\sqrt{10}, \]
\[ \cot(\alpha) = \frac{x}{y} = \frac{-2}{6} = -\frac{1}{3}. \]
Therefore, the exact values of the trigonometric functions for the angle with the terminal point (-2,6) are:
\[ \sin(\alpha) = \frac{3\sqrt{10}}{10}, \cos(\alpha) = -\frac{\sqrt{10}}{10}, \tan(\alpha) = -3, \]
\[ \csc(\alpha) = \frac{\sqrt{10}}{3}, \sec(\alpha) = -\sqrt{10}, \cot(\alpha) = -\frac{1}{3}. \]