Answer :
Answer
OPTION D (Quadrant IV)
SOLUTION
Problem Statement
The question asks us for the quadrant where sin x < 0 and cos x > 0
Method
- A quadrant here represents a set of angles delineated by 90, 180, 270, and 360 degrees.
- Quadrant I has all angles from 0 - 90 degrees.
- Quadrant II has all angles from 91 - 180 degrees.
- Quadrant III has all angles from 181 - 270 degrees.
- Quadrant IV has all angles from 271 - 360 degrees.
- In each Quadrant, the signs of tan (x), cos (x), and sin (x) vary between positive and negative values.
- For us to know which Quadrant sin (x) < 0 and cos (x) > 0, we need to test angles from each of those quadrants.
- For this test, we shall use the following angles:
Quadrant I: 60 degrees.
Quadrant II: 120 degrees
Quadrant III: 240 degrees
Quadrant IV: 300 degrees
Implementation
Note: The criterion we must satisfy is: sin (x) < 0 and cos (x) > 0
Quadrant I: 60 degrees:
[tex]\begin{gathered} \sin (60^0)=\frac{\sqrt[]{3}}{2}>0 \\ \\ \cos (60^0)=\frac{1}{2}>0 \\ \\ \text{ Since both sin}(60^0)\text{ and cos(}60^0)\text{ are greater than zero, Quadrant I cannot be correct} \end{gathered}[/tex]Quadrant II: 120 degrees
[tex]\begin{gathered} \sin (120^0)=\frac{\sqrt[]{3}}{2}>0 \\ \\ \cos (120^0)=-\frac{1}{2}<0 \\ \\ \text{ Quadrant II cannot be correct} \end{gathered}[/tex]Quadrant III: 240 degrees
[tex]\begin{gathered} \sin (240^0)=-\frac{\sqrt[]{3}}{2}<0 \\ \\ \cos (240^0)=-\frac{1}{2}<0 \\ \\ \text{Quadrant III cannot be correct} \end{gathered}[/tex]Quadrant IV: 300 degrees
[tex]\begin{gathered} \sin (300^0)=-\frac{\sqrt[]{3}}{2}<0 \\ \\ \cos (300^0)=\frac{1}{2}>0 \\ \\ \text{Quadrant IV MUST be correct} \end{gathered}[/tex]Final Answer
OPTION D (Quadrant IV)