Question

$$f(x)=\sin x, g(x)=\cos x, h(x)=\tan x, F(x)=\csc x, G(x)=\sec x, \text{ and } H(x)=\cot x$$ (a) Find \(g\left(\frac{7 \pi}{6}\right) .\) What point is on the graph of \(g ?\) (b) Find \(F\left(\frac{7 \pi}{6}\right)\). What point is on the graph of \(F ?\) (c) Find \(H\left(-315^{\circ}\right)\). What point is on the graph of \(H ?\)

Short answer

(a) \( -\frac{ \sqrt{ 3 } }{2} \) at point \( \left( \frac{7 \pi }{ 6 }, -\frac{ \sqrt{ 3 } }{ 2 } \right) \). (b) \( -2 \) at point \( \left( \frac{ 7 \pi }{ 6 }, -2 \) \). (c) \( 1 \) at point \( \left( -315 \^{ \circ \ }, 1 \) \).

Step-by-step solution

Understanding the Function g

The function given is \( g(x) = \cos x \). To find \( g\left( \frac{7 \pi}{6} \right) \), substitute \( \frac{7 \pi}{6} \) into the cosine function.

Calculate \( g\left( \frac{7 \pi}{6} \right) \)

Recall that \( \cos\left(\frac{7 \pi}{6}\right) \) is in the third quadrant where cosine is negative. The reference angle is \( \frac{\pi}{6} \). Hence, \( \cos\left(\frac{7 \pi}{6}\right) = -\frac{\sqrt{3}}{2} \).

Determine the Point on the Graph of g

The point is \( \left( \frac{7 \pi}{6}, -\frac{\sqrt{3}}{2} \right) \).

Understanding the Function F

The function is \( F(x) = \csc(x) = \frac{1}{\sin(x)} \). To find \( F\left( \frac{7 \pi}{6} \right) \), substitute \( \frac{7 \pi}{6} \) into the cosecant function.

Calculate \( F\left( \frac{7 \pi}{6} \right) \)

Recall that \( \sin\left(\frac{7 \pi}{6}\right) \) is in the third quadrant where sine is negative. The reference angle is \( \frac{\pi}{6} \). Hence, \( \sin\left(\frac{7 \pi}{6}\right) = -\frac{1}{2} \). Therefore, \( F\left(\frac{7 \pi}{6}\right) = \frac{1}{-\frac{1}{2}} = -2 \).

Determine the Point on the Graph of F

The point is \( \left( \frac{7 \pi}{6}, -2 \right) \).

Understanding the Function H

The function is \( H(x) = \cot(x) = \frac{\cos(x)}{\sin(x)} \). To find \( H(-315\^{\circ}) \), convert \( -315\^{\circ} \) to radians and locate the corresponding angle.

Convert Angle to Radians and Simplify

\( -315\^{\circ} \) is equivalent to \( -\frac{315\pi}{180} = -\frac{7\pi}{4} \). Since the cotangent function is periodic with period \( \pi \), \( -\frac{7\pi}{4} = \frac{\pi}{4} \) in the unit circle.

Calculate \( H(-315\^{\circ}) \)

Recall that \( \cot\left(\frac{\pi}{4}\right) = 1 \). Therefore, \( H\left(-315\^{\circ}\right) = 1 \).

Determine the Point on the Graph of H

The point is \( \left( -315\^{\circ}, 1 \right) \).

Key concepts

cosine function

The cosine function, denoted as \( \text{cos}(x) \), is one of the fundamental trigonometric functions. It measures the horizontal distance from the point on the unit circle's circumference to the y-axis when an angle x is formed with the positive x-axis.
For example, in the given exercise, we have \( g(x) = \text{cos}(x) \). To find \( g\bigg(\frac{7\text{π}}{6}\bigg) \), we note that the angle \( \frac{7\text{π}}{6} \) places us in the third quadrant, where cosine values are negative.
Thus, \( \text{cos}\bigg(\frac{7\text{π}}{6}\bigg) = -\frac{\text{√3}}{2} \). This is because the reference angle is \( \frac{π}{6} \), and in this quadrant, the cosine is the negative value of cosine at the reference angle.
A point on the graph of \( g(x) \) would then be \( \bigg(\frac{7\text{π}}{6}, -\frac{\text{√3}}{2}\bigg) \).

cosecant function

The cosecant function is the reciprocal of the sine function. It is represented as \( \text{csc}(x) = \frac{1}{\text{sin}(x)} \).
In the problem, we utilize the function \( F(x) = \text{csc}(x) \). To find \( F\bigg(\frac{7\text{π}}{6}\bigg) \), we first determine \( \text{sin}\bigg(\frac{7\text{π}}{6}\bigg) \). The angle \( \frac{7\text{π}}{6} \) falls in the third quadrant, where sine is negative.
The reference angle is \( \frac{π}{6} \), thus \( \text{sin}\bigg(\frac{7\text{π}}{6}\bigg) = -\frac{1}{2} \).
Consequently, \( \text{csc}\bigg(\frac{7\text{π}}{6}\bigg) = \frac{1}{-\frac{1}{2}} = -2 \).
A point on the graph of \( F(x) \) will be \( \bigg(\frac{7\text{π}}{6}, -2\bigg) \).

cotangent function

The cotangent function is the reciprocal of the tangent function, represented as \( \text{cot}(x) = \frac{\text{cos}(x)}{\text{sin}(x)} \).
For the given function \( H(x) = \text{cot}(x) \), we need to find \( H(-315^\text{°}) \). First, we convert \( -315^\text{°} \) to radians:
\( -315^\text{°} = -\frac{315\text{π}}{180} = -\frac{7\text{π}}{4} \).
Because the cotangent is a periodic function with period \( \text{π} \), \( -\frac{7\text{π}}{4} \) is equivalent to \( \frac{\text{π}}{4} \) on the unit circle.
Hence, \( \text{cot}\bigg(\frac{\text{π}}{4}\bigg) = 1 \). Therefore, \( H(-315^\text{°}) = 1 \).
The point on the graph of \( H(x) \) is \( (-315^\text{°}, 1) \).

reference angle

A reference angle is the smallest angle formed by the terminal side of an angle and the x-axis. It is always between \( 0 \) and \( \frac{\text{π}}{2} \).
To determine the value of trigonometric functions for any angle, first find its reference angle.
For instance, the reference angle for \( \frac{7\text{π}}{6} \) is calculated as \( \frac{\text{π}}{6} \). This is because \( \frac{7\text{π}}{6} \) is in the third quadrant, and we subtract \( \text{π} \) from \( \frac{7\text{π}}{6} \) to get it.
Similarly, for \( -315^\text{°} \), which falls in the fourth quadrant when mapped into the unit circle, the reference angle is \( 45^\text{°} \).

trigonometric graph points

Points on the graphs of trigonometric functions represent the values of these functions at specific angles.
For \( g(x) = \text{cos}(x) \), the point at \( \frac{7\text{π}}{6} \) is \( \big(\frac{7\text{π}}{6}, -\frac{\text{√3}}{2}\big) \). This means the cosine value is \( -\frac{\text{√3}}{2} \) when the angle is \( \frac{7\text{π}}{6} \).
For \( F(x) = \text{csc}(x) \), the point at \( \frac{7\text{π}}{6} \) is \( \big(\frac{7\text{π}}{6}, -2\big) \). This shows the cosecant value is \( -2 \) when the angle is \( \frac{7\text{π}}{6} \).
For \( H(x) = \text{cot}(x) \), at \( -315^\text{°} \), the point is \( (-315^\text{°}, 1) \), indicating the cotangent value is \( 1 \) at this angle.
Each point gives insight into the behavior of the function with respect to the angles.