Problem 38 - Contemporary Precalculus

Question

In Exercises \(37-42,\) find \(\sin t,\) cos \(t,\) tan \(t\) when the terminal side of an angle of t radians in standard position passes through the given point. $$(-2,1)$$

Step-by-Step Solution

Answer

Answer: For the given point (-2,1), the values of the trigonometric functions are as follows: 1. \(\sin t = \frac{1}{\sqrt{5}}\) 2. \(\cos t = \frac{-2}{\sqrt{5}}\) 3. \(\tan t = -\frac{1}{2}\)

Step 1: Identify the coordinates of the point

The given point is (-2,1), so the coordinate (-2,1) can be represented as (x, y), where x = -2 and y = 1.

Step 2: Find the distance from the origin (0,0) to the point (-2,1)

To calculate the distance from the origin (0,0) to the point (-2,1), we can use the distance formula: \(\text{distance} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) $$\text{distance} = \sqrt{(-2 - 0)^2 + (1 - 0)^2}$$ $$\text{distance} = \sqrt{(-2)^2 + (1)^2}$$ $$\text{distance} = \sqrt{4 + 1}$$ $$\text{distance} = \sqrt{5}$$ Now, we have found the distance which is the hypotenuse or \(r = \sqrt{5}\).

Step 3: Calculate \(\sin t\) and \(\cos t\)

To find \(\sin t\) and \(\cos t\), we use their definitions in terms of coordinates and the hypotenuse: $$\sin t = \frac{y}{r}$$ $$\cos t = \frac{x}{r}$$ Now, substitute the coordinates and the hypotenuse found in the previous steps: $$\sin t = \frac{1}{\sqrt{5}}$$ $$\cos t = \frac{-2}{\sqrt{5}}$$

Step 4: Calculate \(\tan t\)

To find \(\tan t\), we use the definition of tangent in terms of sine and cosine: $$\tan t = \frac{\sin t}{\cos t}$$ Now, substitute the sine and cosine values found in the previous step: $$\tan t = \frac{\dfrac{1}{\sqrt{5}}}{\dfrac{-2}{\sqrt{5}}}$$ Simplify the fraction by multiplying both numerator and denominator by \(\sqrt{5}\): $$\tan t = \frac{1}{-2}$$

Step 5: Write the final values of the trigonometric functions

Now we have the values of \(\sin t,\) \(\cos t,\) and \(\tan t\) for the given point, so we write them as: $$\sin t = \frac{1}{\sqrt{5}}$$ $$\cos t = \frac{-2}{\sqrt{5}}$$ $$\tan t = -\frac{1}{2}$$

Key Concepts

Sine Function

Understanding the sine function begins with the placement of an angle on the coordinate plane. For an angle formed in standard position, the sine of the angle is found using the vertical coordinate (the \( y \)) and the hypotenuse of the right triangle formed by the angle. The sine function is specifically defined as the ratio of the opposite side to the hypotenuse in a right triangle.
In this exercise, we have a point \((-2, 1)\), which means our \( y \) value is \(1\). The hypotenuse \( r \) was already calculated using the distance formula and found to be \(\sqrt{5}\). Thus, the sine function is calculated as:

• \(\sin t = \frac{y}{r} = \frac{1}{\sqrt{5}}\)

This value is derived from basic trigonometric principles, and even though it gives a fraction with a square root in the denominator, it represents the exact value of sine for the angle \(t\). You might notice \(\sin t\) often appears in various mathematics and physics scenarios, specifically when analyzing oscillatory or wave-like patterns.

Cosine Function

The cosine function works closely with the sine function, as both are foundational components of trigonometry. For the angle in standard position, like \(t\) here, cosine is defined as the ratio of the adjacent side (horizontal coordinate \(x\)) to the hypotenuse \(r\).
In the exercise, our point \((-2, 1)\) provides us with \( x = -2 \). We've computed the hypotenuse \( r = \sqrt{5} \) already. Therefore, the cosine function for this angle is:

• \(\cos t = \frac{x}{r} = \frac{-2}{\sqrt{5}}\)

The result places the cosine value in the negative direction, which is common when dealing with angles in different quadrants of the coordinate plane. Cosine values range from \(-1\) to \(1\), representing how far along or away from the x-axis the point is in the coordinate system. Cosine often appears in contexts involving circles, rotations, or any phenomenon where lateral movement is analyzed.

Tangent Function

The tangent function is naturally derived from the sine and cosine functions. In a right triangle, tangent is often thought of as the ratio of the opposite side to the adjacent side. This translates well into the Cartesian coordinate system as well, represented by \(\tan t = \frac{\sin t}{\cos t}\).
For the exercise point \((-2, 1)\), substituting the values calculated gives:

• \(\tan t = \frac{\frac{1}{\sqrt{5}}}{\frac{-2}{\sqrt{5}}} = -\frac{1}{2} \)

Here, the tangent function simplifies by canceling out the common \(\sqrt{5}\) factors. Its negative result indicates a downward slope for the line through the origin and point \((-2, 1)\). The tangent values can extend infinitely in the positive or negative direction, deeply influencing the behavior of waves and oscillations in physics and engineering scenarios. The tangent function often symbolizes angular inclination or the rate of rise over run in geometry and calculus.