Question

In Exercises \(37-40\) , use the given information to find the values of the six trigonometric functions at the angle \(\theta\) . Give exact answers. The point \(P(-2,2)\) is on the terminal side of \(\theta\)

Short answer

The six trigonometric functions are: \( \sin(\theta) = \frac{\sqrt{2}}{2} \), \( \cos(\theta) = -\frac{\sqrt{2}}{2} \), \( \tan(\theta) = -1 \), \( \csc(\theta) = \sqrt{2} \), \( \sec(\theta) = -\sqrt{2} \), \( \cot(\theta) = -1 \).

Step-by-step solution

Identify the coordinates on the plane

The given point is P(-2,2) on a terminal side of an angle. This means that x = -2 and y = 2.

Calculate the length of the hypotenuse

The length of the hypotenuse, r, is calculated using the Pythagorean theorem, \( r = \sqrt{x^{2} + y^{2}} = \sqrt{(-2)^{2} + (2)^{2}} = \sqrt{4 + 4} = \sqrt{8} \).

Compute the six trigonometric functions

The sine function, \( \sin(\theta) = \frac{y}{r} = \frac{2}{\sqrt{8}} = \frac{\sqrt{2}}{2} \). The cosine function, \( \cos(\theta) = \frac{x}{r} = \frac{-2}{\sqrt{8}} = -\frac{\sqrt{2}}{2} \). The tangent function, \( \tan(\theta) = \frac{y}{x} = \frac{2}{-2} = -1 \). Then, for the reciprocal trigonometric functions: The cosecant function, \( \csc(\theta) = \frac{r}{y} = \frac{\sqrt{8}}{2} = \sqrt{2} \). The secant function, \( \sec(\theta) = \frac{r}{x} = \frac{\sqrt{8}}{-2} = -\sqrt{2} \) and finally, the cotangent function, \( \cot(\theta) = \frac{x}{y} = \frac{-2}{2} = -1 \).

Check the Quadrant

Since both x and y are positive, the angle \(\theta \) lies in the first quadrant. Therefore, all trigonometric function values should match our calculated values.