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Chapter 4: Problem 6

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 \(\theta\). $$(5,-5)$$

Short Answer

Expert verified
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 by step solution

01

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: \(r = \sqrt{x^2 + y^2} = \sqrt{(5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\).
02

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.
03

Final Answer

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

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