Problem 71 - Precalculus Mathematics for Calculus

Question

Find the values of the trigonometric functions of \(t\) from the given information. \(\sin t=-\frac{1}{4}, \quad \sec t < 0\)

Step-by-Step Solution

Answer

In the third quadrant, \( \cos t = -\frac{\sqrt{15}}{4} \), \( \sec t = -\frac{4\sqrt{15}}{15} \), \( \csc t = -4 \), \( \tan t = \frac{\sqrt{15}}{15} \), and \( \cot t = \sqrt{15} \).

Step 1: Determine the Quadrant

The given conditions are \( \sin t = -\frac{1}{4} \) and \( \sec t < 0 \). Since sine is negative, \(t\) must be in either the third or fourth quadrant. Because \( \sec t = \frac{1}{\cos t} \) is negative, \( \cos t \) must be negative as well, placing \(t\) in the third quadrant.

Step 2: Use Pythagorean Identity to Find Cosine

We use the identity \( \sin^2 t + \cos^2 t = 1 \). Substituting \( \sin t = -\frac{1}{4} \) into the equation, we have:\[\left(-\frac{1}{4}\right)^2 + \cos^2 t = 1\]\[ \frac{1}{16} + \cos^2 t = 1\]\[\cos^2 t = 1 - \frac{1}{16}\]\[\cos^2 t = \frac{15}{16}\]Since \( \cos t \) is negative in the third quadrant, \( \cos t = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}\).

Step 3: Calculate Secant

Secant is defined as the reciprocal of cosine:\[\sec t = \frac{1}{\cos t} = \frac{1}{-\frac{\sqrt{15}}{4}} = -\frac{4}{\sqrt{15}}\]Rationalizing the denominator gives:\[\sec t = -\frac{4\sqrt{15}}{15}\]

Step 4: Calculate Cosecant

Cosecant is the reciprocal of sine:\[\csc t = \frac{1}{\sin t} = \frac{1}{-\frac{1}{4}} = -4\]

Step 5: Calculate Tangent

Using \( \tan t = \frac{\sin t}{\cos t} \), we substitute the known values:\[\tan t = \frac{-\frac{1}{4}}{-\frac{\sqrt{15}}{4}} = \frac{1}{\sqrt{15}}\]Rationalizing gives:\[\tan t = \frac{\sqrt{15}}{15}\]

Step 6: Calculate Cotangent

Cotangent is the reciprocal of tangent:\[\cot t = \frac{1}{\tan t} = \frac{1}{\frac{\sqrt{15}}{15}} = \frac{15}{\sqrt{15}} = \sqrt{15}\]

Key Concepts

Sine

Sine is a fundamental trigonometric function which represents the y-coordinate of a unit circle. If you visualize a right triangle, the sine of an angle is the ratio of the length of the opposite side to the hypotenuse. In mathematical terms, this is expressed as: \[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\]In this specific problem, we are given that \( \sin t = -\frac{1}{4} \). This value tells us that the angle \( t \) is related to a right triangle where the opposite side is 1 and the hypotenuse is 4. However, the negative sign indicates that the angle \( t \) is either in the third or fourth quadrant of the unit circle. This is because sine values are negative in these quadrants, which are located below the x-axis. Understanding where sine is positive or negative can be important for determining properties of angles.

Cosine

Cosine is another fundamental trigonometric function, representing the x-coordinate on the unit circle. Similar to sine, in a right triangle, cosine of an angle is the ratio of the adjacent side to the hypotenuse, or: \[ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\]In this exercise, we found that \( \cos t \) had to be negative, which reveals that angle \( t \) is located in the third quadrant. Since \( \sec t \) was given as less than zero, this means \( \cos t \) couldn’t be positive. Using the Pythagorean identity, the exact value of cosine was calculated as \( \cos t = -\frac{\sqrt{15}}{4} \). The sign of cosine helps determine which quadrant the angle lies in, with positive values in the first and fourth quadrants and negative values in the second and third quadrants.

Pythagorean identity

The Pythagorean identity is a fundamental identity connecting sine and cosine. It is named after the famous Pythagorean theorem and is expressed as: \[ \sin^2 t + \cos^2 t = 1\]This identity holds true for all angles \( t \) and is a crucial part of solving trigonometric problems. In our problem, the identity was pivotal in finding \( \cos t \). We substituted \( \sin t = -\frac{1}{4} \) into the identity, yielding: \[\left(-\frac{1}{4}\right)^2 + \cos^2 t = 1\]This equation was simplified to find \( \cos^2 t = \frac{15}{16} \), from which the negative cosine value was derived, owing to the third quadrant location. The Pythagorean identity helps connect the dots between the different trigonometric functions.

Trigonometric identities

Trigonometric identities are equations that are true for all values of the involved variables. These identities simplify complex relationships between the angles and sides of triangles. In this problem, we saw several trigonometric identities in action:

• Secant and Cosine: \( \sec t = \frac{1}{\cos t} \)
• Cosecant and Sine: \( \csc t = \frac{1}{\sin t} \)
• Tangent and Cotangent: \( \tan t = \frac{\sin t}{\cos t} \) and \( \cot t = \frac{1}{\tan t} \)

These identities were used in steps to derive the remaining trigonometric values like secant, tangent, and cotangent. Having a good grasp of these identities aids in solving trigonometric exercises, as they allow conversion and manipulation of function expressions into more useful forms.