Question

find the value of each expression without using a calculator or table.

1. a. sin 180° b. cos 180° c. sin 270° d. cos 270°
2. a. sin (-90°) b. cos (-90°) c. sin 360° d. cos 360°
3. a. sin (-π) b. cos π c. sin 3π/2 d. cos π/2
4. a. cos 2π b. sin (-π/2) c. sin 3π d. cos (-3π/2)

name each quadrant described.

1. a. sin θ>0 and cos θ<0 b. sin θ<0 and cos θ<0
2. a. sin θ<0 and cos θ>0 b. sin θ>0 and sin (90° + θ)>0

without using a calculator or table, solve each equation for all θ in radians.

1. a. sin θ = 1 b. cos θ = -1 c. sin θ = 0 d. sin θ = 2
2. a. cos θ = 1 b. sin θ = -1 c. cos θ = 0 d. cos θ = -3

without using a calculator or table, state whether each expression is positive, negative, or zero.

1. a. sin 4π b. cos 7π/6 c. sin (-π/4) d. cos 3π/4
2. a. cos 3π b. sin 2π/3 c. sin 11π/6 d. cos (-π/2)
3. a. sin 60° b. cos (-120°) c. cos 300° d. sin (-210°)
4. a. cos 45° b. sin 135° c. cos (-225°) d. sin (-315°)
5. a. sin 7π/4 b. sin (-π/6) c. cos 3π/2 d. cos π/3
6. a. cos (-π/3) b. sin π/6 c. sin 5π/4 d. cos 7π/4
7. a. cos 89° b. cos 91° c. sin 720° d. sin (-270°)
8. a. sin 1° b. sin (-1°) c. cos 90° d. cos 540°

find sin θ and cos θ.

1. 18. 19. 20.

Explanation:

1. a. $\sin180^{\circ}$

Recall unit - circle definition. $\sin\theta$ is the y - coordinate on unit circle. At $180^{\circ}$ (or $\pi$ radians), the point on the unit circle is $(- 1,0)$. So $\sin180^{\circ}=0$.

1. b. $\cos180^{\circ}$

$\cos\theta$ is the x - coordinate on unit circle. At $180^{\circ}$, the point is $(-1,0)$. So $\cos180^{\circ}=-1$.

1. c. $\sin270^{\circ}$

At $270^{\circ}$ (or $\frac{3\pi}{2}$ radians), the point on the unit circle is $(0, - 1)$. So $\sin270^{\circ}=-1$.

1. d. $\cos270^{\circ}$

At $270^{\circ}$, the point on the unit circle is $(0,-1)$. So $\cos270^{\circ}=0$.

2. a. $\sin(-90^{\circ})$

Since $\sin(-\theta)=-\sin\theta$, and $\sin90^{\circ}=1$, then $\sin(-90^{\circ})=-1$.

2. b. $\cos(-90^{\circ})$

Since $\cos(-\theta)=\cos\theta$, and $\cos90^{\circ}=0$, then $\cos(-90^{\circ})=0$.

2. c. $\sin360^{\circ}$

At $360^{\circ}$ (or $2\pi$ radians), the point on the unit circle is $(1,0)$. So $\sin360^{\circ}=0$.

2. d. $\cos360^{\circ}$

At $360^{\circ}$, the point on the unit circle is $(1,0)$. So $\cos360^{\circ}=1$.

3. a. $\sin(-\pi)$

Since $\sin(-\theta)=-\sin\theta$ and $\sin\pi = 0$, then $\sin(-\pi)=0$.

3. b. $\cos\pi$

At $\pi$ radians, the point on the unit circle is $(-1,0)$. So $\cos\pi=-1$.

3. c. $\sin\frac{3\pi}{2}$

At $\frac{3\pi}{2}$ radians, the point on the unit circle is $(0,-1)$. So $\sin\frac{3\pi}{2}=-1$.

3. d. $\cos\frac{\pi}{2}$

At $\frac{\pi}{2}$ radians, the point on the unit circle is $(0,1)$. So $\cos\frac{\pi}{2}=0$.

4. a. $\cos2\pi$

At $2\pi$ radians, the point on the unit circle is $(1,0)$. So $\cos2\pi = 1$.

4. b. $\sin(-\frac{\pi}{2})$

Since $\sin(-\theta)=-\sin\theta$ and $\sin\frac{\pi}{2}=1$, then $\sin(-\frac{\pi}{2})=-1$.

4. c. $\sin3\pi$

$\sin3\pi=\sin(\pi + 2\pi)=\sin\pi=0$.

4. d. $\cos(-\frac{3\pi}{2})$

Since $\cos(-\theta)=\cos\theta$, and $\cos\frac{3\pi}{2}=0$, then $\cos(-\frac{3\pi}{2})=0$.

5. a.

If $\sin\theta>0$ (y - coordinate is positive) and $\cos\theta<0$ (x - coordinate is negative), the angle $\theta$ is in the second quadrant.

5. b.

If $\sin\theta<0$ (y - coordinate is negative) and $\cos\theta<0$ (x - coordinate is negative), the angle $\theta$ is in the third quadrant.

6. a.

If $\sin\theta<0$ (y - coordinate is negative) and $\cos\theta>0$ (x - coordinate is positive), the angle $\theta$ is in the fourth quadrant.

6. b.

Since $\sin(90^{\circ}+\theta)=\cos\theta$, if $\sin\theta>0$ and $\sin(90^{\circ}+\theta)>0$ (i.e., $\cos\theta>0$), the angle $\theta$ is in the first quadrant.

7. a. $\sin\theta = 1$

On the unit - circle, $\sin\theta$ (y - coordinate) is 1 when $\theta=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$.

7. b. $\cos\theta=-1$

On the unit - circle, $\cos\theta$ (x - coordinate) is - 1 when $\theta=\pi + 2k\pi,k\in\mathbb{Z}$.

7. c. $\sin\theta = 0$

On the unit - circle, $\sin\theta$ is 0 when $\theta=k\pi,k\in\mathbb{Z}$.

7. d. $\sin\theta = 2$

Since $-1\leqslant\sin\theta\leqslant1$, there is no solution for $\theta$.

8. a. $\cos\theta = 1$

On the unit - circle, $\cos\theta$ (x - coordinate) is 1 when $\theta = 2k\pi,k\in\mathbb{Z}$.

8. b. $\sin\theta=-1$

On the unit - circle, $\sin\theta$ (y - coordinate) is - 1 when $\theta=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}$.

8. c. $\cos\theta = 0$

On the unit - circle, $\cos\theta$ is 0 when $\theta=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$.

8. d. $\cos\theta=-3$

Since $-1\leqslant\cos\theta\leqslant1$, there is no solution for $\theta$.

9. a. $\sin4\pi$

$\sin4\pi=\sin(2\pi + 2\pi)=\sin2\pi=0$.

9. b. $\cos\frac{7\pi}{6}$

$\frac{7\pi}{6}=\pi+\frac{\pi}{6}$, and $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Here $A=\pi$ and $B = \frac{\pi}{6}$, so $\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}<0$.

9. c. $\sin(-\frac{\pi}{4})$

Since $\sin(-\theta)=-\sin\theta$, and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, then $\sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}<0$.

9. d. $\cos\frac{3\pi}{4}$

$\frac{3\pi}{4}=\pi-\frac{\pi}{4}$, and $\cos(A - B)=\cos A\cos B+\sin A\sin B$. Here $A=\pi$ and $B=\frac{\pi}{4}$, so $\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}<0$.

10. a. $\cos3\pi$

$\cos3\pi=\cos(\pi + 2\pi)=-\cos\pi=-(-1)= - 1<0$.

10. b. $\sin\frac{2\pi}{3}$

$\frac{2\pi}{3}=\pi-\frac{\pi}{3}$, and $\sin(A - B)=\sin A\cos B-\cos A\sin B$. Here $A=\pi$ and $B=\frac{\pi}{3}$, so $\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}>0$.

10. c. $\sin\frac{11\pi}{6}$

$\frac{11\pi}{6}=2\pi-\frac{\pi}{6}$, and $\sin(2\pi-\theta)=-\sin\theta$, so $\sin\frac{11\pi}{6}=-\frac{1}{2}<0$.

10. d. $\cos(-\frac{\pi}{2})$

Since $\cos(-\theta)=\cos\theta$, and $\cos\frac{\pi}{2}=0$, then $\cos(-\frac{\pi}{2})=0$.

11. a. $\sin60^{\circ}=\frac{\sqrt{3}}{2}>0$

11. b. $\cos(-120^{\circ})=\cos120^{\circ}=-\frac{1}{2}<0$

11. c. $\cos300^{\circ}=\cos(360^{\circ}-60^{\circ})=\cos60^{\circ}=\frac{1}{2}>0$

11. d. $\sin(-210^{\circ})=-\sin210^{\circ}=-\sin(180^{\circ}+30^{\circ})=\frac{1}{2}>0$

12. a. $\cos45^{\circ}=\frac{\sqrt{2}}{2}>0$

12. b. $\sin135^{\circ}=\sin(180^{\circ}-45^{\circ})=\frac{\sqrt{2}}{2}>0$

12. c. $\cos(-225^{\circ})=\cos225^{\circ}=-\frac{\sqrt{2}}{2}<0$

12. d. $\sin(-315^{\circ})=-\sin315^{\circ}=-\sin(360^{\circ}-45^{\circ})=\frac{\sqrt{2}}{2}>0$

13. a. $\sin\frac{7\pi}{4}=\sin(2\pi-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}<0$

13. b. $\sin(-\frac{\pi}{6})=-\frac{1}{2}<0$

13. c. $\cos\frac{3\pi}{2}=0$

13. d. $\cos\frac{\pi}{3}=\frac{1}{2}>0$

14. a. $\cos(-\frac{\pi}{3})=\cos\frac{\pi}{3}=\frac{1}{2}>0$

14. b. $\sin\frac{\pi}{6}=\frac{1}{2}>0$

14. c. $\sin\frac{5\pi}{4}=\sin(\pi+\frac{\pi}{4})=-\frac{\sqrt{2}}{2}<0$

14. d. $\cos\frac{7\pi}{4}=\cos(2\pi-\frac{\pi}{4})=\frac{\sqrt{2}}{2}>0$

15. a. $\cos89^{\circ}>0$

15. b. $\cos91^{\circ}<0$

15. c. $\sin720^{\circ}=\sin(2\times360^{\circ})=0$

15. d. $\sin(-270^{\circ})=-\sin270^{\circ}=1>0$

16. a. $\sin1^{\circ}>0$

16. b. $\sin(-1^{\circ})<0$

16. c. $\cos90^{\circ}=0$

16. d. $\cos540^{\circ}=\cos(360^{\circ}+180^{\circ})=-1<0$

17.

For the point $(x = 3,y = 4)$ on the terminal - side of the angle $\theta$, $r=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+4^{2}} = 5$. Then $\sin\theta=\frac{y}{r}=\frac{4}{5}$ and $\cos\theta=\frac{x}{r}=\frac{3}{5}$.

18.

For the point $(x=-3,y = - 1)$ on the terminal - side of the angle $\theta$, $r=\sqrt{x^{2}+y^{2}}=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{9 + 1}=\sqrt{10}$. Then $\sin\theta=\frac{y}{r}=-\frac{1}{\sqrt{10}}=-\frac{\sqrt{10}}{10}$ and $\cos\theta=\frac{x}{r}=-\frac{3}{\sqrt{10}}=-\frac{3\sqrt{10}}{10}$.

19.

For the right - triangle with opposite side $y = 5$ and hypotenuse $r = 13$, by the Pythagorean theorem $x=\sqrt{r^{2}-y^{2}}=\sqrt{13^{2}-5^{2}} = 12$. Then $\sin\theta=\frac{5}{13}$ and $\cos\theta=\frac{12}{13}$.

20.

(No clear information about the point in question 20. Assuming a general right - triangle or unit - circle approach, but without enough data, we can't solve it completely. If we assume a right - triangle with some given side - length relationships, we would use $r=\sqrt{x^{2}+y^{2}}$, $\sin\theta=\frac{y}{r}$, and $\cos\theta=\frac{x}{r}$)

Answer:

1. a. $0$; b. $-1$; c. $-1$; d. $0$
2. a. $-1$; b. $0$; c. $0$; d. $1$
3. a. $0$; b. $-1$; c. $-1$; d. $0$
4. a. $1$; b. $-1$; c. $0$; d. $0$
5. a. Second quadrant; b. Third quadrant
6. a. Fourth quadrant; b. First quadrant
7. a. $\theta=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$; b. $\theta=\pi + 2k\pi,k\in\mathbb{Z}$; c. $\theta=k\pi,k\in\mathbb{Z}$; d. No solution
8. a. $\theta = 2k\pi,k\in\mathbb{Z}$; b. $\theta=\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}$; c. $\theta=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$; d. No solution
9. a. $0$; b. Negative; c. Negative; d. Negative
10. a. Negative; b. Positive; c. Negative; d. $0$
11. a. Positive; b. Negative; c. Positive; d. Positive
12. a. Positive; b. Positive; c. Negative; d. Positive
13. a. Negative; b. Negative; c. $0$; d. Positive
14. a. Positive; b. Positive; c. Negative; d. Positive
15. a. Positive; b. Negative; c. $0$; d. Positive
16. a. Positive; b. Negative; c. $0$; d. Negative
17. $\sin\theta=\frac{4}{5},\cos\theta=\frac{3}{5}$
18. $\sin\theta=-\frac{\sqrt{10}}{10},\cos\theta=-\frac{3\sqrt{10}}{10}$
19. $\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13}$
20. Insufficient data for a complete answer.