Problem 40 - Contemporary Precalculus

Question

Assume that $$\cos t=-2 / 5 \quad \text { and } \quad \pi

Step-by-Step Solution

Answer

Answer: -2/5

Step 1: Analyzing the Quadrants

Since π < t < 3π/2, we know that t is in the third quadrant. This means that both the sine and cosine of t are negative in this quadrant. Now we need to find the cosine of -t, which means we need to know in which quadrant -t lies. Since t is in the third quadrant, we can infer that -t will lie in the first quadrant (0 < -t < π/2).

Step 2: Applying Cosine Identity

We will use the identity for the cosine of a negative angle: cos(-t) = cos(t). This identity states that the cosine of a negative angle is equal to the cosine of its positive counterpart.

Step 3: Calculating the Cosine of -t

Now that we have applied the identity, we can find the value of cos(-t). Given cos(t) = -2/5, we can directly substitute this value into the identity equation: $$\cos(-t) = \cos(t)$$ $$\cos(-t) = -\frac{2}{5}$$ Therefore, the value of cos(-t) is -2/5.

Key Concepts

Cosine Function

The cosine function, represented as \( \cos(\theta) \), is a fundamental part of trigonometry. It relates the angle \( \theta \) to the ratio of the adjacent side of a right-angled triangle over its hypotenuse. This function is even, meaning that the cosine of an angle is same as the cosine of its negative counterpart. For example, if \( \cos(t) = -\frac{2}{5} \), then \( \cos(-t) \) will also equal \( -\frac{2}{5} \). This property is incredibly useful when working with angles beyond the basic right angles or when dealing with negative angles.

• An even function means \( \cos(-x) = \cos(x) \), simplifying calculations involving negative angles.
• The cosine value can range from -1 to 1, with each position corresponding to specific points on the unit circle.
• Cosine helps determine the location of an angle on the unit circle which is a circle with radius one centered at the origin.

Quadrants in Trigonometry

In trigonometry, the coordinate plane is divided into four quadrants, each of which influences the sign of the trigonometric functions differently. These quadrants are defined by the angles they contain:

• First Quadrant (\(0 < \theta < \frac{\pi}{2}\)): All trigonometric functions are positive.
• Second Quadrant (\(\frac{\pi}{2} < \theta < \pi\)): Sine is positive, cosine, and tangent are negative.
• Third Quadrant (\(\pi < \theta < \frac{3\pi}{2}\)): Tangent is positive, sine and cosine are negative.
• Fourth Quadrant (\(\frac{3\pi}{2} < \theta < 2\pi\)): Cosine is positive, sine and tangent are negative.

For the given exercise, since \( \pi < t < \frac{3\pi}{2} \), angle \( t \) is situated in the third quadrant, where the sine and cosine functions are negative. However, when we evaluate \( -t \), it falls into the first quadrant, where cosine is positive. However, due to the even nature of the cosine function, its value remains the same as it refers to the principal value.

Negative Angle Identity

Negative angle identities are a set of trigonometric identities that show the relationship between the trigonometric functions of negative angles and their corresponding positive angles. These identities allow for simplification, making trigonometric calculations easier.

• For cosine, the identity states: \( \cos(-\theta) = \cos(\theta) \)
• This property is what makes cosine an even function, unlike sine, which is an odd function where \( \sin(-\theta) = -\sin(\theta) \).
• It aids in calculating the cosine for negative angles without additional complexity, thus saving time.

This identity is particularly useful in trigonometry problems involving symmetry and reflection over the axes. For the exercise presented, using this identity directly gives the value of \( \cos(-t) \) as \( -\frac{2}{5} \), aligning with the basics of the negative angle identity.