Concept Check Suppose that ―90° \u003c θ \u003c 90° . Find the sign of each function value.sec(―θ)

Recall the definition of the secant function: $$\sec(\theta) = \frac{1}{\cos(\theta)}. To $$find the sign of $$\sec(-\theta)$$, we need to understand the sign of $$\cos(-\theta)$$ first. Use the even-odd property of cosine: $$\cos(-\theta) = \cos(\theta). $$This means the cosine function is even, so its value at $$-\theta is $$the same as at $$\theta. $$Since $$-90^\circ \u003c \theta \u003c 90^\circ$$, $$\theta$$ lies in the first or fourth quadrant. In both these quadrants, $$\cos(\theta) is $$positive. Because $$\cos(\theta) is $$positive in this interval, $$\cos(-\theta) is $$also positive. Therefore, $$\sec(-\theta) = \frac{1}{\cos(-\theta)}$$ will have the same sign as $$\frac{1}{\text{positive}}$$, which is positive. Conclude that $$\sec(-\theta) is $$positive for $$-90^\circ \u003c \theta \u003c 90^\circ.$$

Ch. 1 - Trigonometric Functions

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Lial 12th Edition

Ch. 1 - Trigonometric Functions

문제 101

2장, 문제 101

Concept Check Suppose that ―90° < θ < 90° . Find the sign of each function value.
sec(―θ)

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Recall the definition of the secant function: $\sec(\theta) = \frac{1}{\cos(\theta)}$. To find the sign of $\sec(-\theta)$, we need to understand the sign of $\cos(-\theta)$ first.

Use the even-odd property of cosine: $\cos(-\theta) = \cos(\theta)$. This means the cosine function is even, so its value at $-\theta$ is the same as at $\theta$.

Since $-90^\circ < \theta < 90^\circ$, $\theta$ lies in the first or fourth quadrant. In both these quadrants, $\cos(\theta)$ is positive.

Because $\cos(\theta)$ is positive in this interval, $\cos(-\theta)$ is also positive. Therefore, $\sec(-\theta) = \frac{1}{\cos(-\theta)}$ will have the same sign as $\frac{1}{\text{positive}}$, which is positive.

Conclude that $\sec(-\theta)$ is positive for $-90^\circ < \theta < 90^\circ$.

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주요 개념

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Definition and Domain of the Secant Function

The secant function, sec(θ), is defined as the reciprocal of the cosine function: sec(θ) = 1/cos(θ). It is important to understand that sec(θ) is undefined where cos(θ) = 0. Since θ is between -90° and 90°, cos(θ) is positive in this interval except at the endpoints, so sec(θ) will also be defined and its sign depends on cos(θ).

Even-Odd Properties of Trigonometric Functions

The cosine function is an even function, meaning cos(-θ) = cos(θ). Since sec(θ) = 1/cos(θ), sec(θ) inherits this even property: sec(-θ) = sec(θ). This property helps determine the sign of sec(-θ) by relating it directly to sec(θ) without changing the sign.

Sign of Cosine and Secant in the Interval -90° < θ < 90°

Within the interval -90° < θ < 90°, cosine values are positive because the angle lies in the first and fourth quadrants where cosine is positive. Since sec(θ) = 1/cos(θ), sec(θ) is also positive in this range. Therefore, sec(-θ) will have the same positive sign as sec(θ).

Concept Check Suppose that ―90° < θ < 90° . Find the sign of each function value.sec(―θ)