Question

1. given \\( \csc \theta = \frac{8}{5} \\) and \\( \cot \theta > 0 \\), evaluate the six trigonometric functions of \\( \theta \\).

find the six trig functions from the given point that lies on the terminal side of the angle in standard position.

a) \\( (-5, 12) \\) b) \\( (-4, -6) \\)

c) \\( (20, -21) \\) d) \\( (-3, 7) \\)

determine which quadrant \\( \theta \\) must lie in if:
a) \\( \cos \theta > 0 \\) and \\( \cot \theta > 0 \\)
b) \\( \csc \theta < 0 \\) and \\( \cos \theta > 0 \\)
c) \\( \sin \theta > 0 \\) and \\( \tan \theta < 0 \\)
d) \\( \cos \theta < 0 \\) and \\( \cot \theta < 0 \\)

Explanation:

Step1: Analyze $\csc\theta$ and $\cot\theta$ signs

$\csc\theta = \frac{8}{5}>0$, so $\sin\theta=\frac{1}{\csc\theta}=\frac{5}{8}>0$. $\cot\theta>0$ means $\frac{\cos\theta}{\sin\theta}>0$. Since $\sin\theta>0$, then $\cos\theta>0$. So $\theta$ is in Quadrant I.

Step2: Find $r$ (radius)

For a point $(x,y)$, $r = \sqrt{x^2 + y^2}$. But here, from $\csc\theta=\frac{r}{y}=\frac{8}{5}$, so $y = 5$, $r = 8$. Then $x=\sqrt{r^2 - y^2}=\sqrt{8^2 - 5^2}=\sqrt{64 - 25}=\sqrt{39}$ (since $\cos\theta>0$, $x>0$).

Step3: Calculate $\sin\theta$

$\sin\theta=\frac{y}{r}=\frac{5}{8}$

Step4: Calculate $\cos\theta$

$\cos\theta=\frac{x}{r}=\frac{\sqrt{39}}{8}$

Step5: Calculate $\tan\theta$

$\tan\theta=\frac{y}{x}=\frac{5}{\sqrt{39}}=\frac{5\sqrt{39}}{39}$

Step6: Calculate $\csc\theta$

$\csc\theta=\frac{r}{y}=\frac{8}{5}$ (given, verified)

Step7: Calculate $\sec\theta$

$\sec\theta=\frac{r}{x}=\frac{8}{\sqrt{39}}=\frac{8\sqrt{39}}{39}$

Step8: Calculate $\cot\theta$

$\cot\theta=\frac{x}{y}=\frac{\sqrt{39}}{5}$

For part a) $(-5,12)$:

Step1: Find $r$

$r=\sqrt{(-5)^2 + 12^2}=\sqrt{25 + 144}=\sqrt{169}=13$

Step2: $\sin\theta$

$\sin\theta=\frac{y}{r}=\frac{12}{13}$

Step3: $\cos\theta$

$\cos\theta=\frac{x}{r}=\frac{-5}{13}$

Step4: $\tan\theta$

$\tan\theta=\frac{y}{x}=\frac{12}{-5}=-\frac{12}{5}$

Step5: $\csc\theta$

$\csc\theta=\frac{r}{y}=\frac{13}{12}$

Step6: $\sec\theta$

$\sec\theta=\frac{r}{x}=\frac{13}{-5}=-\frac{13}{5}$

Step7: $\cot\theta$

$\cot\theta=\frac{x}{y}=\frac{-5}{12}=-\frac{5}{12}$

For part b) $(-4,-6)$:

Step1: Find $r$

$r=\sqrt{(-4)^2 + (-6)^2}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13}$

Step2: $\sin\theta$

$\sin\theta=\frac{y}{r}=\frac{-6}{2\sqrt{13}}=-\frac{3}{\sqrt{13}}=-\frac{3\sqrt{13}}{13}$

Step3: $\cos\theta$

$\cos\theta=\frac{x}{r}=\frac{-4}{2\sqrt{13}}=-\frac{2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}$

Step4: $\tan\theta$

$\tan\theta=\frac{y}{x}=\frac{-6}{-4}=\frac{3}{2}$

Step5: $\csc\theta$

$\csc\theta=\frac{r}{y}=\frac{2\sqrt{13}}{-6}=-\frac{\sqrt{13}}{3}$

Step6: $\sec\theta$

$\sec\theta=\frac{r}{x}=\frac{2\sqrt{13}}{-4}=-\frac{\sqrt{13}}{2}$

Step7: $\cot\theta$

$\cot\theta=\frac{x}{y}=\frac{-4}{-6}=\frac{2}{3}$

For part c) $(20,-21)$:

Step1: Find $r$

$r=\sqrt{20^2 + (-21)^2}=\sqrt{400 + 441}=\sqrt{841}=29$

Step2: $\sin\theta$

$\sin\theta=\frac{y}{r}=\frac{-21}{29}$

Step3: $\cos\theta$

$\cos\theta=\frac{x}{r}=\frac{20}{29}$

Step4: $\tan\theta$

$\tan\theta=\frac{y}{x}=\frac{-21}{20}$

Step5: $\csc\theta$

$\csc\theta=\frac{r}{y}=\frac{29}{-21}=-\frac{29}{21}$

Step6: $\sec\theta$

$\sec\theta=\frac{r}{x}=\frac{29}{20}$

Step7: $\cot\theta$

$\cot\theta=\frac{x}{y}=\frac{20}{-21}=-\frac{20}{21}$

For part d) $(-3,7)$:

Step1: Find $r$

$r=\sqrt{(-3)^2 + 7^2}=\sqrt{9 + 49}=\sqrt{58}$

Step2: $\sin\theta$

$\sin\theta=\frac{y}{r}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

Step3: $\cos\theta$

$\cos\theta=\frac{x}{r}=\frac{-3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

Step4: $\tan\theta$

$\tan\theta=\frac{y}{x}=\frac{7}{-3}=-\frac{7}{3}$

Step5: $\csc\theta$

$\csc\theta=\frac{r}{y}=\frac{\sqrt{58}}{7}$

Step6: $\sec\theta$

$\sec\theta=\frac{r}{x}=\frac{\sqrt{58}}{-3}=-\frac{\sqrt{58}}{3}$

Step7: $\cot\theta$

$\cot\theta=\frac{x}{y}=\frac{-3}{7}$

Quadrant Determination:

a) $\cos\theta>0$ (Quadrants I, IV) and $\cot\theta>0$ (Quadrants I, III) → Intersection: Quadrant I

b) $\csc\theta<0$ (Quadrants III, IV) and $\cos\theta>0$ (Quadrants I, IV) → Intersection: Quadrant IV

c) $\sin\theta>0$ (Quadrants I, II) and $\tan\theta<0$ (Quadrants II, IV) → Intersection: Quadrant II

d) $\cos\theta<0$ (Quadrants II, III) and $\cot\theta<0$ (Quadrants II, IV) → Intersection: Quadrant II

Answer:

For $\boldsymbol{\csc\theta=\frac{8}{5},\cot\theta>0}$:

$\sin\theta=\boldsymbol{\frac{5}{8}}$, $\cos\theta=\boldsymbol{\frac{\sqrt{39}}{8}}$, $\tan\theta=\boldsymbol{\frac{5\sqrt{39}}{39}}$, $\csc\theta=\boldsymbol{\frac{8}{5}}$, $\sec\theta=\boldsymbol{\frac{8\sqrt{39}}{39}}$, $\cot\theta=\boldsymbol{\frac{\sqrt{39}}{5}}$

a) Point $(-5,12)$:

$\sin\theta=\boldsymbol{\frac{12}{13}}$, $\cos\theta=\boldsymbol{-\frac{5}{13}}$, $\tan\theta=\boldsymbol{-\frac{12}{5}}$, $\csc\theta=\boldsymbol{\frac{13}{12}}$, $\sec\theta=\boldsymbol{-\frac{13}{5}}$, $\cot\theta=\boldsymbol{-\frac{5}{12}}$

b) Point $(-4,-6)$:

$\sin\theta=\boldsymbol{-\frac{3\sqrt{13}}{13}}$, $\cos\theta=\boldsymbol{-\frac{2\sqrt{13}}{13}}$, $\tan\theta=\boldsymbol{\frac{3}{2}}$, $\csc\theta=\boldsymbol{-\frac{\sqrt{13}}{3}}$, $\sec\theta=\boldsymbol{-\frac{\sqrt{13}}{2}}$, $\cot\theta=\boldsymbol{\frac{2}{3}}$

c) Point $(20,-21)$:

$\sin\theta=\boldsymbol{-\frac{21}{29}}$, $\cos\theta=\boldsymbol{\frac{20}{29}}$, $\tan\theta=\boldsymbol{-\frac{21}{20}}$, $\csc\theta=\boldsymbol{-\frac{29}{21}}$, $\sec\theta=\boldsymbol{\frac{29}{20}}$, $\cot\theta=\boldsymbol{-\frac{20}{21}}$

d) Point $(-3,7)$:

$\sin\theta=\boldsymbol{\frac{7\sqrt{58}}{58}}$, $\cos\theta=\boldsymbol{-\frac{3\sqrt{58}}{58}}$, $\tan\theta=\boldsymbol{-\frac{7}{3}}$, $\csc\theta=\boldsymbol{\frac{\sqrt{58}}{7}}$, $\sec\theta=\boldsymbol{-\frac{\sqrt{58}}{3}}$, $\cot\theta=\boldsymbol{-\frac{3}{7}}$

Quadrant Determination:

a) $\boldsymbol{\text{Quadrant I}}$, b) $\boldsymbol{\text{Quadrant IV}}$, c) $\boldsymbol{\text{Quadrant II}}$, d) $\boldsymbol{\text{Quadrant II}}$