Chapter 4.2, Problem 47E

To determine

Find the value of $\sin (-\frac{9\text{π}}{4})$ by using its period.

Answer to Problem 47E

  $-\frac{\sqrt{2}}{2}$

Explanation of Solution

Given: The angle $\sin (-\frac{9\text{π}}{4})$

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: $\sin (-\text{θ})=-\sin \text{θ}$

  $\begin{array}{l}\sin (-\frac{9\text{π}}{4})=-\sin \frac{9\text{π}}{4}\\ -\sin \frac{9\text{π}}{4}=-\sin \text{( 2π}+\frac{\text{π}}{4})=-\sin \frac{\text{π}}{4}=-\frac{\sqrt{2}}{2}\end{array}$

Thus the value of $\sin (-\frac{9\text{π}}{4})=-\frac{\sqrt{2}}{2}$