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Chapter 9: Problem 1

An angle \(\theta\) is such that \(\sin \theta<0\) and \(\cos \theta<0 .\) In which quadrant does \(\theta\) lie?

Short Answer

Expert verified
Answer: Quadrant III

Step by step solution

01

Quadrant I

In the first quadrant, all coordinates are positive, which means both \(\sin \theta\) and \(\cos \theta\) are positive. So, \(\theta\) cannot be in Quadrant I.
02

Quadrant II

In the second quadrant, the x-coordinate is negative while the y-coordinate is positive. Since sine represents the y-coordinate, we have \(\sin \theta > 0\). Since cosine represents the x-coordinate, we have \(\cos \theta < 0 \). This means θ cannot be in Quadrant II.
03

Quadrant III

In the third quadrant, both the x and y coordinates are negative. This implies that both the sine and cosine functions are negative, meaning \(\sin \theta < 0\) and \(\cos \theta < 0\). Therefore, the angle \(\theta\) must lie in Quadrant III.
04

Quadrant IV

In the fourth quadrant, the x-coordinate is positive while the y-coordinate is negative. Since sine represents the y-coordinate, we have \(\sin \theta < 0\). Since cosine represents the x-coordinate, we have \(\cos \theta > 0 \). This means θ cannot be in Quadrant IV.
05

Conclusion

Based on the given conditions and the description of the sine and cosine functions in each quadrant, the angle \(\theta\) lies in Quadrant III.

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Most popular questions from this chapter

If \(\tan \phi<0\) and \(\sin \phi>0\), state the quadrant in which \(\phi\) lies.

Solve $$ \tan \left(\frac{2 x}{3}\right)=0.7 \quad 0 \leq x \leq 2 \pi $$

(a) Sketch \(y=\cos \left(x-20^{\circ}\right)\), \(0^{\circ} \leq x \leq 360^{\circ}\) (b) On the same axes, sketch \(y=\sin x\). (c) Use your graphs to obtain approximate solutions of $$ \sin x=\cos \left(x-20^{\circ}\right) $$

Solve (a) \(\sin \theta=0.3510,0^{\circ} \leq \theta \leq 360^{\circ}\) (b) \(\sin \theta=0.4161,0 \leq \theta \leq 2 \pi\) (c) \(\cos t=-0.3778,0 \leq t \leq 2 \pi\) (d) \(\cos x=0.7654,0^{\circ} \leq x \leq 360^{\circ}\) (e) \(\tan y=1.7136,0^{\circ} \leq y \leq 360^{\circ}\) (f) \(\tan y=-0.3006,0^{\circ} \leq y \leq 360^{\circ}\)

Show \(\cos \left(180^{\circ}+\theta\right)=-\cos \theta\)
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