The point #(-5,-2)# is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?

To determine the exact values of the six trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent) of the angle whose terminal side contains the point (-5, -2) in standard position, we first need to find the values of the adjacent side, opposite side, and hypotenuse of the right triangle formed by the point (-5, -2) and the origin (0, 0).

Using the distance formula, we find the hypotenuse: [ \text{Hypotenuse} = \sqrt{(-5 - 0)^2 + (-2 - 0)^2} = \sqrt{(-5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29} ]

The angle's adjacent side is the horizontal distance from the point (-5, -2) to the origin: [ \text{Adjacent side} = -5 ]

The angle's opposite side is the vertical distance from the point (-5, -2) to the origin: [ \text{Opposite side} = -2 ]

Now, we can determine the trigonometric functions: [ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{-2}{\sqrt{29}} ] [ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-5}{\sqrt{29}} ] [ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{-2}{-5} = \frac{2}{5} ]

The reciprocal functions can be determined as follows: [ \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{\sqrt{29}}{-2} ] [ \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\sqrt{29}}{-5} ] [ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{5}{2} ]

Therefore, the exact values of the six trigonometric functions of the angle whose terminal side contains the point (-5, -2) in standard position are as calculated above.