#### (a)

$dfrac{3pi}{4}$ is located in the second quadrant.The related angle is $dfrac{pi}{4}$, and $sin$ is positive in the second quadrant.

#### (b)

$dfrac{5pi}{3}$ is located in the fourth quadrant.The related angle is $dfrac{pi}{3}$, and $cos$ is positive in the fourth quadrant.

#### (c)

$dfrac{4pi}{3}$ is located in the third quadrant.The related angle is $dfrac{pi}{3}$, and $tan$ is positive in the third quadrant.

#### (d)

$dfrac{5pi}{6}$ is located in the second quadrant.The related angle is $dfrac{pi}{6}$, and $sin$ is negative in the second quadrant.

#### (e)

$dfrac{2pi}{3}$ is located in the second quadrant.The related angle is $dfrac{pi}{3}$, and $cos$ is negative in the second quadrant.

#### (f)

$dfrac{7pi}{4}$ is located in the fourth quadrant.The related angle is $dfrac{pi}{4}$, and $cot$ is negative in the fourth quadrant.

#### (a)

(ii) $r^2=6^2+8^2=36+64=100 Rightarrow r=10$

(iii) $sin theta=dfrac{4}{5}, cos theta=dfrac{3}{5}, tan theta=dfrac{4}{3}, csc theta=dfrac{5}{4}, sec theta=dfrac{5}{3}, cot theta=dfrac{3}{4}$

(iv) $sin^{-1}(dfrac{4}{5})=0.93 Rightarrow theta=0.93$

(i)

#### (b)

(ii) $r^2=(-12)^2+(-5)^2=144+25=169 Rightarrow r=13$

(iii) $sin theta=-dfrac{5}{13}, cos theta=-dfrac{12}{13}, tan theta=dfrac{5}{12}, csc theta=-dfrac{13}{5}, sec theta=-dfrac{13}{12}$,

$cot theta=dfrac{12}{5}$

(iv) $sin^{-1}(-dfrac{5}{13})=-0.395 Rightarrow theta=pi+0.395=3.54$

(i)

#### (c)

(ii) $r^2=4^2+(-3)^2=16+9=25 Rightarrow r=5$

(iii) $sin theta=-dfrac{3}{5}, cos theta=dfrac{4}{5}, tan theta=-dfrac{3}{4}, csc theta=-dfrac{5}{3}, sec theta=dfrac{5}{4}, cot theta=-dfrac{4}{3}$

(iv) $sin^{-1}(-dfrac{3}{5})=-0.64 Rightarrow theta=2pi-0.64=5.64$

(i)

#### (d)

(ii) $r^2=0^2+5^2=0+25=25 Rightarrow r=5$

(iii) $sin theta=dfrac{5}{5}=1, cos theta=dfrac{0}{5}=0, tan theta=dfrac{5}{0}=$undefined,

$csc theta=dfrac{5}{5}=1, sec theta=dfrac{5}{0}=$undefined, $cot theta=dfrac{0}{5}=0$

(iv) $sin^{-1}(1)=1.57 Rightarrow theta=dfrac{pi}{2}$

(i)

#### (a)

$x=0$, $y=-1, r=1$

$sinleft(-dfrac{pi}{2} right)=-1, cosleft(-dfrac{pi}{2} right)=0$,

$tanleft(-dfrac{pi}{2} right)=undefined, cscleft(-dfrac{pi}{2} right)=-1$

$secleft(-dfrac{pi}{2} right)=undefined, cotleft(-dfrac{pi}{2} right)=0$

#### (b)

$x=-1, y=0, r=1$

$sin(-pi)=0, cos(-pi)=-1, tan(-pi)=0$,

$csc(-pi)=undefined, sec(-pi)=-1$,

$$
cot(-pi)=undefined
$$

#### (c)

$x=1, y=-1, r=sqrt{2}$

$sinleft( dfrac{7pi}{4}right)=-dfrac{sqrt{2}}{2},cosleft(dfrac{7pi}{4} right)=dfrac{sqrt{2}}{2}$,

$tanleft( dfrac{7pi}{4}right)=-1, cscleft(dfrac{7pi}{4} right)=-sqrt{2}$,

$secleft(dfrac{7pi}{4} right)=sqrt{2}, cotleft(dfrac{7pi}{4} right)=-1$

#### (d)

$x=sqrt{3}, y=-1, r=2$

$sinleft(-dfrac{pi}{6} right)=-dfrac{1}{2},cosleft(-dfrac{pi}{6} right)=dfrac{sqrt{3}}{2}$,

$tanleft(-dfrac{pi}{6} right)=-dfrac{sqrt{3}}{3}, cscleft(-dfrac{pi}{6} right)=-2$,

$secleft(-dfrac{pi}{6} right)=dfrac{2sqrt{3}}{3}, cotleft(-dfrac{pi}{6} right)=-sqrt{3}$

#### (a)

This is an the second quadrrant where sine is positive. Sine is also positive in the first quadrant.
So, an equivalent expression would be $sindfrac{pi}{6}$.

#### (b)

This is in the fourth quadrant where cosine is positive.Cosine is also positive in the first quadrant.
So, an equivalent expression would be cos$dfrac{pi}{3}$.

#### (c)

This is in the fourth quadrant where cotangent is negative. Cotangent is also negative in the second quadrant.So, an equivalent expression would be cot$dfrac{3pi}{4}$

#### (d)

This is in the third quadrant where secant is negative. Secant is also negative in the second quadrant.
So, an equivalent expression would be sec$dfrac{5pi}{6}$.

#### (a)

$$
sin(dfrac{2pi}{3})=dfrac{sqrt{3}}{2}
$$

#### (b)

$$
cos(dfrac{5pi}{4})=-dfrac{1}{sqrt{2}}=-dfrac{sqrt{2}}{2}
$$

#### (c)

$$
tan(dfrac{11pi}{6})=dfrac{-1}{sqrt{3}}=-dfrac{sqrt{3}}{3}
$$

#### (d)

$$
sin(dfrac{7pi}{4})=dfrac{-1}{sqrt{2}}=-dfrac{sqrt{2}}{2}
$$

#### (e)

$$
csc(dfrac{5pi}{6})=dfrac{2}{1}=2
$$

#### (f)

$$
sec(dfrac{5pi}{3})=dfrac{2}{1}=2
$$

#### (a)

$costheta=-dfrac{1}{2}$

$theta=dfrac{2pi}{3}, dfrac{4pi}{3}$

#### (b)

$costheta=dfrac{sqrt{3}}{2}$

$theta=dfrac{pi}{6}, dfrac{11pi}{6}$

#### (c)

$-dfrac{sqrt{2}}{2}=-dfrac{1}{sqrt{2}}$

$costheta=-dfrac{1}{sqrt{2}}$

$theta=dfrac{3pi}{4},dfrac{5pi}{4}$

#### (d)

$costheta=-dfrac{sqrt{3}}{2}$

$theta=dfrac{5pi}{6}, dfrac{7pi}{6}$

#### (e)

$costheta=0$

$theta=dfrac{pi}{2}, dfrac{3pi}{2}$

#### (f)

$costheta=-1$

$$
theta=pi
$$

#### (a)

$(-7,8)$ is in the second quadrant.

$tan^{-1}left(dfrac{8}{-7} right)=-0.852$

$theta=pi-0.852=2.29$

#### (b)

$(12,2)$ is in the first quadrant.

$tan^{-1}left(dfrac{2}{12} right)=0.17$

$theta=0.17$

#### (c)

$(3, 11)$ is in the first quadrant.

$tan^{-1}left(dfrac{11}{3} right)=1.30$

$theta=1.30$

#### (d)

$(-4, -2)$ is in the third quadrant.

$tan^{-1}left(dfrac{-2}{-4} right)=0.464$

$theta=pi+0.464=3.61$

#### (e)

$(9,10)$ is in the first quadrant.

$tan^{-1}left(dfrac{10}{9} right)=0.84$

$theta=0.84$

#### (f)

$(6, -1)$ is in the fourth quadrant.

$tan^{-1}left(dfrac{-1}{6} right)=-0.165$

$theta=2pi-0.165=6.12$

#### (a)

This is in the second quadrant where cosine is negative.Cosine is also negative in the third quadrant.
So, an equivalent expression would be $cosdfrac{5pi}{4}$.

#### (b)

This is in the fourth quadrant where tangent is negative. Tangent is also negative in the second quadrant. So, an equivalent expression would be $tandfrac{5pi}{6}$

#### (c)

This is in the fourth quadrant where cosecant is negative. Cosecant is also negative in the third quadrant. So, an equivalent expression would be $cscdfrac{4pi}{3}$.

#### (d)

This is in the second quadrant where cotangent is negative.Cotangent is also negative in the fourth quadrant. So, an equivalent expression would be $cotdfrac{5pi}{3}$

#### (e)

This is in the fourth quadrant where sine is negative. Sine is also negative in the third quadrant.
So, an equiavalent expression would be $sindfrac{7pi}{6}$.

#### (f)

This is in the fourth quadrant where secant is positive.Secant is also positive in the first quadrant.
So, an equivalent expression would be $secdfrac{pi}{4}$.

$sintheta=dfrac{3.4}{5}$

$theta=sin^{-1}left(dfrac{3.4}{5} right)=0.748$

$pi-0.748=2.39$

$theta=4-pi=0.8584$

$sin(0.8584)=dfrac{4.2}{x}$

$x=dfrac{4.2}{sin(0.8584)}=5.55$cm

Use a proportion to find $theta$.

$dfrac{10 sec}{60 sec}=dfrac{r rad}{2pi rad}$

$r=1.05 rad$

$cos(1.05)=dfrac{x}{9}$

$x=9cos(1.05)$

$x=4.5 cm$

#### (a)

It lies in the second or third quadrant because cosine is negative in these quadrants.

#### (b)

$x=-5, r=13, y=?$

$13^2-(-5)^2=y^2$

$169-25=y^2$

$144=y^2$

$12=y$

$sintheta=dfrac{12}{13}$ or $-dfrac{12}{13}$,

$tantheta=dfrac{12}{5}$ or $-dfrac{12}{5}$,

$sectheta=-dfrac{13}{5}$,

$csctheta=dfrac{13}{12}$ or $-dfrac{13}{12}$

$cottheta=dfrac{5}{12}$ or $-dfrac{5}{12}$

#### (c)

$sintheta=dfrac{12}{13}$

$theta=sin^{-1}left(dfrac{12}{13} right)$

$theta=1.176$

In the second quadrant, $pi-1.176=1.97.$

In the second quadrant, $pi+1.176=4.32.$

$0^{circ}=0^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=0 radians$;

$30^{circ}=30^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{pi}{6} radians$;

$45^{circ}=45^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{pi}{4} radians$;

$60^{circ}=60^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{pi}{3} radians$;

$90^{circ}=90^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{pi}{2} radians$;

$120^{circ}=120^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{2pi}{3} radians$;

$135^{circ}=135^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{3pi}{4} radians$;

$150^{circ}=150^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{5pi}{6} radians$;

$180^{circ}=180^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=pi radians$;

$210^{circ}=210^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{7pi}{6} radians$;

$225^{circ}=225^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{5pi}{4} radians$;

$240^{circ}=240^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{4pi}{3} radians$;

$270^{circ}=270^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{3pi}{2} radians$

$300^{circ}=300^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{5pi}{3} radians;$

$315^{circ}=315^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{7pi}{4} radians$;

$330^{circ}=330^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=dfrac{11pi}{6} radians;$

$360^{circ}=360^{circ}timesleft( dfrac{pi radians}{180^{circ}}right)=2pi radians$

$2left(sin^2left(dfrac{11pi}{6} right) right)-1$

$=2left(-dfrac{1}{2} right)^2 -1$

$=2left(dfrac{1}{4} right)-1=-dfrac{1}{2}$

$left(sin^2dfrac{11pi}{6} right)-left( cos^2dfrac{11pi}{6}right)$

$=left(-dfrac{1}{2} right)^2-left(dfrac{sqrt{3}}{2} right)^2=dfrac{1}{4}-dfrac{3}{4}=-dfrac{1}{2}$

$2left(sin^2left(dfrac{11pi}{6} right) right)-1=left(sin^2dfrac{11pi}{6} right)-left(cos^2dfrac{11pi}{6} right)$

$sinleft(dfrac{pi}{6} right)=dfrac{8}{AB}$

$AB=dfrac{8}{sinleft( dfrac{pi}{6}right)}$

$AB=16$

$(AD)^2=8^2+8^2$

$(AD)^2=64+64$

$(AD)^2=128$

$(AD)=sqrt{128}=8sqrt{2}$

$sin D=dfrac{8}{8sqrt{2}}=dfrac{sqrt{2}}{2}$;

$cos D=dfrac{8}{8sqrt{2}}=dfrac{sqrt{2}}{2}$;

$tan D=dfrac{8}{8}=1$

$tan a =dfrac{3}{3sqrt{3}}$

$a=tan^{-1}left(dfrac{3}{3sqrt{3}} right)$

$a=30^{circ}$

$tan c=dfrac{4sqrt{3}}{4}$

$c=tan^{-1}left(dfrac{4sqrt{3}}{4} right)$

$c=60^{circ}$

$b=180^{circ}-30^{circ}-60^{circ}=90^{circ}$

$sin 90^{circ}=1$

$tan a=dfrac{6sqrt{2}}{6sqrt{2}}$

$a=tan^{-1}(dfrac{1}{1})$

$a=45^circ$

$tan c=dfrac{7}{7sqrt{3}}$

$c=tan^{-1}(dfrac{7}{7sqrt{3}})$

$c=30^circ$

$b=180^circ-30^circ=105^circ$

$cos 150^circ=-0.26$

$cos 150^circ=-0.26$

The ranges of the cosecant and secant functions are both $left{yinBbb{R}|-1geq y or ygeq1 right}$.In other words, the values of these functions can never be between $-1$ and $1$. For the values of these functions to be betwen $-1$ and $1$, the values of the sine and cosine functions would have to be greater than $1$ and less than $-1$, which is never the case.

This is a special triangle, and the hypotenuse is $2$.

$sin theta cot theta-cos^2 theta=-dfrac{1}{2}(-sqrt{3})-(-dfrac{sqrt{3}}{2})^2=dfrac{sqrt{3}}{2}-dfrac{3}{4}=dfrac{2sqrt{3}-3}{4}$

$dfrac{2sqrt{3}-3}{4}$