Problem 6 - Precalculus Essentials

Question

In Exercises \(1-8,\) a point on the terminal side of angle \(\theta\) is given. Find the exact value of each of the six trigonometric functions of \(\theta\). $$(5,-5)$$

Step-by-Step Solution

Answer

The six trigonometric functions for point (5,-5) are: \(\sin(\theta) = -\sqrt{2}/2, \cos(\theta) = \sqrt{2}/2, \tan(\theta) = -1, \csc(\theta) = -\sqrt{2}, \sec(\theta) = \sqrt{2}, \cot(\theta) = 1\).

Step 1: Identify given coordinates and calculate hypotenuse

The given point on the terminal side of angle \(\theta\) is (5,-5). Here, the x-coordinate (5) is the adjacent side and the y-coordinate (-5) is the opposite side of the right triangle that represents this point. Calculating the hypotenuse (r) involves using the Pythagorean theorem: \(r = \sqrt{x^2 + y^2} = \sqrt{(5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\).

Step 2: Compute the six trigonometric functions

Now, recall that sine (\(\sin\)) is opposite/hypotenuse, cosine (\(\cos\)) is adjacent/hypotenuse, and tangent (\(\tan\)) is opposite/adjacent. These give: \(\sin(\theta) = -5/5\sqrt{2} = -\sqrt{2}/2, \cos(\theta) = 5/5\sqrt{2} = \sqrt{2}/2, \tan(\theta) = -5/5 = -1\). Cosecant (\(\csc\)), secant (\(\sec\)), and cotangent (\(\cot\)) are the reciprocals of sine, cosine, and tangent respectively, giving: \(\csc(\theta) = -2/\sqrt{2} = -\sqrt{2}, \sec(\theta) = 2/\sqrt{2} = \sqrt{2}, \cot(\theta) = -1/-1 = 1\). The negative sign on the \(\sin\) and \(\csc\) is because the terminal point is in the fourth quadrant, where both these functions are negative.

Step 3: Final Answer

Upon simplifying, the six trigonometric functions for point (5,-5) on angle \(\theta\) are: \(\sin(\theta) = -\sqrt{2}/2, \cos(\theta) = \sqrt{2}/2, \tan(\theta) = -1, \csc(\theta) = -\sqrt{2}, \sec(\theta) = \sqrt{2}, \cot(\theta) = 1\).