Chapter 4.2, Problem 49E

(a)

To determine

Find the value of $\sin (-\text{t})\text{if}\sin \text{t}=\frac{1}{3}$ by using the Trigonometric Rule.

Answer to Problem 49E

  $-\frac{1}{3}$

Explanation of Solution

Given: The angle $\sin \text{t}=\frac{1}{3}$

Concept Used:

In Trigonometry we know that: $\text{sin (θ}\text{+}\text{2πn}\text{)}\text{=}\pm \text{sin}\text{θ}\text{;}\text{cos}\text{(θ}\text{+}\text{2πn}\text{)}=\pm \cos \text{θ}$ (positive or negative) depending on the values of n and the quadrant

For any integer $\text{n}$ and real number $\text{t}=\text{θ}$ , the functions that behave in such a repetitive (cyclic) manner are called periodic.

Sine and Cosine functions are periodic and have a period $\text{2π}$

The cosine and secant functions are even

  $\text{cos}(-\text{θ})=\cos \text{θ}\text{;}\sec (-\text{θ})=\sec \text{θ}$

The sine, cosecant, tangent and cotangent functions are odd.

  $\begin{array}{l}\sin (-\text{θ})=-\sin \text{θ}\text{;}\csc (-\text{θ})=-\csc \text{θ}\text{;}\\ \text{tan}(-\text{θ})=-\tan \text{θ}\text{;}\text{cot}(-\text{θ})=-\cot \text{θ}\text{;}\end{array}$

Calculation:

Using the rule: $\sin (-\text{θ})=-\sin \text{θ}\text{;}\csc (-\text{θ})=-\csc \text{θ}\text{;}$

  $\sin (-\text{θ})=-\sin \text{θ}$

Given: $\sin \text{t}=\frac{1}{3}$

  $\begin{array}{l}\sin (-\text{t})=-\sin \text{t}=-\frac{1}{3}\\ \sin (-\text{t})=-\frac{1}{3}\end{array}$

Thus the value of $\sin (-\text{t})=-\frac{1}{3}$

(b)

To determine

Find the value of $\csc (-\text{t})\text{if}\sin \text{t}=\frac{1}{3}$ by using the Trigonometric Rule

Answer to Problem 49E

  $-3$

Explanation of Solution

Given: The angle $\sin \text{t}=\frac{1}{3}$

Concept Used:

In Trigonometry we know that: $\text{sin (θ}\text{+}\text{2πn}\text{)}\text{=}\pm \text{sin}\text{θ}\text{;}\text{cos}\text{(θ}\text{+}\text{2πn}\text{)}=\pm \cos \text{θ}$ (positive or negative) depending on the values of n and the quadrant

For any integer $\text{n}$ and real number $\text{t}=\text{θ}$ , the functions that behave in such a repetitive (cyclic) manner are called periodic.

Sine and Cosine functions are periodic and have a period $\text{2π}$

The cosine and secant functions are even

  $\text{cos}(-\text{θ})=\cos \text{θ}\text{;}\sec (-\text{θ})=\sec \text{θ}$

The sine, cosecant, tangent and cotangent functions are odd.

  $\begin{array}{l}\sin (-\text{θ})=-\sin \text{θ}\text{;}\csc (-\text{θ})=-\csc \text{θ}\text{;}\\ \text{tan}(-\text{θ})=-\tan \text{θ}\text{;}\text{cot}(-\text{θ})=-\cot \text{θ}\text{;}\end{array}$

Calculation:

Using the rule: $\sin (-\text{θ})=-\sin \text{θ}\text{;}\csc (-\text{θ})=-\csc \text{θ}\text{;}$

  $\csc (-\text{t})=-\csc \text{t}$

Given: $\sin \text{t}=\frac{1}{3}$

  $\begin{array}{l}\csc (-\text{t})=-\csc \text{t}=-\frac{1}{\sin \text{t}}=-\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=-3\\ \csc (-\text{t})=-3\end{array}$

Thus the value of $\csc (-\text{t})=-3$