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In Exercises \(55-60\), find the values of all six trigonometric functions at \(t\) if the given conditions are true. $$\sec t=-13 / 5 \quad \text { and } \quad \tan t<0$$

Short Answer

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Question: Given that \(\sec t = -\frac{13}{5}\) and \(\tan t < 0\), find the values of all six trigonometric functions at \(t\). Answer: The values of the trigonometric functions are: \(\sin t = \frac{12}{13}\), \(\cos t = -\frac{5}{13}\), \(\tan t = -\frac{12}{5}\), \(\csc t = \frac{13}{12}\), \(\sec t = -\frac{13}{5}\), and \(\cot t = -\frac{5}{12}\).

Step by step solution

01

Finding \(\cos t\)

Since we are given the value of \(\sec t = -\frac{13}{5}\), we can find the value of \(\cos t\) as the reciprocal of \(\sec t\): $$\cos t = \frac{1}{\sec t} = \frac{1}{-\frac{13}{5}} = -\frac{5}{13}$$
02

Finding the quadrant

We know that \(\tan t < 0\) and \(\sec t < 0\). Since \(\sec t\) is negative, \(\cos t\) will be negative as well. The only two quadrants where \(\cos\) is negative are the second and third quadrants. In the second quadrant, \(\sin t\) is positive and \(\tan t\) is negative, while in the third quadrant, both \(\sin t\) and \(\tan t\) are negative. Since \(\tan t < 0\), \(t\) must be in the second quadrant.
03

Finding \(\sin t\)

Now that we know that \(t\) is in the second quadrant and we have the value of \(\cos t\), we can use the Pythagorean identity to find the value of \(\sin t\): $$\sin^2 t + \cos^2 t = 1$$ $$\sin^2 t + (-\frac{5}{13})^2 = 1$$ $$\sin^2 t = 1 - (-\frac{5}{13})^2 = 1 - \frac{25}{169} = \frac{144}{169}$$ Since \(\sin t\) is positive in the second quadrant, we have: $$\sin t = \sqrt{\frac{144}{169}} = \frac{12}{13}$$
04

Finding all six trigonometric functions

Now that we have the values of \(\sin t\) and \(\cos t\), we can find the other four trigonometric functions as follows: 1) \(\tan t\): Since we know that \(\sin t = \frac{12}{13}\) and \(\cos t = -\frac{5}{13}\), we can find \(\tan t\) as: $$\tan t = \frac{\sin t}{\cos t} = \frac{12/13}{-5/13} = -\frac{12}{5}$$ 2) \(\csc t\): This is the reciprocal of \(\sin t\), so: $$\csc t = \frac{1}{\sin t} = \frac{1}{\frac{12}{13}} = \frac{13}{12}$$ 3) \(\sec t\): This is given in the problem as: $$\sec t = -\frac{13}{5}$$ 4) \(\cot t\): This is the reciprocal of \(\tan t\), so: $$\cot t = \frac{1}{\tan t} = \frac{1}{-\frac{12}{5}} = -\frac{5}{12}$$ So the six trigonometric function values at \(t\) are: $$\sin t = \frac{12}{13} \hspace{0.5cm} \cos t = -\frac{5}{13} \hspace{0.5cm} \tan t = -\frac{12}{5}$$ $$\csc t = \frac{13}{12} \hspace{0.5cm} \sec t = -\frac{13}{5} \hspace{0.5cm} \cot t = -\frac{5}{12}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reciprocal Identities
Trigonometric functions often pair up as reciprocals of each other. Understanding reciprocal identities is essential to solving problems where some trigonometric functions aren't directly given. For example, if you know the secant (\[\sec t\]) of an angle, you can find its cosine (\[\cos t\]) using:- **\[\sec t = \frac{1}{\cos t}\]**This means \(\cos t\) is the reciprocal of \(\sec t\):- **\[\cos t = \frac{1}{\sec t}\]**Similarly, the sine and its reciprocal, cosecant, are related:- **\[\csc t = \frac{1}{\sin t}\]**And likewise for tangent and cotangent:- **\[\tan t = \frac{1}{\cot t}\]**When given a reciprocal trig function, first find its corresponding trig function by inverting the given value. This is crucial to unravel the puzzle of trigonometric identities in exercises.
Pythagorean Identity
The Pythagorean identity is a key principle that ties together the sine and cosine of an angle in a beautifully simple equation:- **\[\sin^2 t + \cos^2 t = 1\]**This relationship is extremely helpful when you know one function's value and need to find the other. For instance, if \(\cos t\) is known, you can rearrange this identity to find \(\sin t\):\[\sin^2 t = 1 - \cos^2 t\]Substitute the value of \(\cos t\), and solve for \(\sin t\). Remember, in practice, you'll often need to take the square root, so consider the angle's quadrant to decide whether \(\sin t\) or \(\cos t\) should be positive or negative. Quadrant knowledge works hand in hand with the Pythagorean identity to pinpoint the exact values of trig functions.
Angle Quadrants
The coordinate plane is divided into four quadrants, and the sign of each trigonometric function depends on which quadrant the angle sits in. Here's a quick guide: - **Quadrant I:** All trigonometric functions are positive. - **Quadrant II:** Sine is positive, while cosine and tangent are negative. - **Quadrant III:** Tangent is positive, sine and cosine are negative. - **Quadrant IV:** Cosine is positive, while sine and tangent are negative. If you know the secant and tangent of an angle, like in our exercise, and their signs, you can tell which quadrant the angle is in. For example, a negative tangent and cosine suggest Quadrant II. Observing these quadrant rules allows you to confidently assign correct values for trigonometric functions, keeping their signs accurate. Knowing the quadrant is crucial since it directs you on the correct sign of the calculated function value.

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Most popular questions from this chapter

In Exercises \(49-54\), prove the given identity. $$\csc (-t)=-\csc t$$

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