Question

In which quadrant is angle x if sin x<0 and cos x>0?

Answer 1

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:

`\begin{gathered} \sin (60^0)=\frac{\sqrt[]{3}}{2}>0 \\ \\ \cos (60^0)=(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}`

Quadrant II: 120 degrees

`\begin{gathered} \sin (120^0)=\frac{\sqrt[]{3}}{2}>0 \\ \\ \cos (120^0)=-(1)/(2)<0 \\ \\ \text{ Quadrant II cannot be correct} \end{gathered}`

Quadrant III: 240 degrees

`\begin{gathered} \sin (240^0)=-\frac{\sqrt[]{3}}{2}<0 \\ \\ \cos (240^0)=-(1)/(2)<0 \\ \\ \text{Quadrant III cannot be correct} \end{gathered}`

Quadrant IV: 300 degrees

`\begin{gathered} \sin (300^0)=-\frac{\sqrt[]{3}}{2}<0 \\ \\ \cos (300^0)=(1)/(2)>0 \\ \\ \text{Quadrant IV MUST be correct} \end{gathered}`

Final Answer

OPTION D (Quadrant IV)