Question

35 19 2. Given the point is the terminal point fort, please answer the following (15 points) a. Please show the point is on the unit circle. b. Find each of the following. sin(t) csc(t) sin(-t) csc(-1) sec(t) cos(-+) soc(-) cot(t) tan(-+) cot(-1)

Answer 1

2. a. We have the terminal point of t as ,

$(-\frac{3\sqrt5}{8},\frac{\sqrt {19}}{8})$

A unit circle is of the form , [unit radius]

$x^2+y^2=1$

giving our terminal point to this equation we get ,

,

(3/3)2 + (19) = = = 6 = 1 +

Since this satisfy the condition, the point is on a unit circle .

b. The point lies in 2nd quadrant , consider a right triangle with hypotenuse 1 , base as x coordinate and height as y coordinate ,

$\sin t=\frac{y}{1}=\frac{\sqrt{19}}{8}$

$\csc t=\frac{1}{\sin t}=\frac{8}{\sqrt{19}}$

$\sin (-t)=-\sin t=-\frac{\sqrt{19}}{8}$

$\csc (-t)=\frac{1}{\sin (-t)}=-\frac{8}{\sqrt{19}}$

$\cos (t)=\frac{x}{1}=-\frac{3\sqrt5}{8}$

$\sec (t)=\frac{1}{\cos t}=-\frac{8}{3\sqrt5}$

$\cos (-t)= \cos t= -\frac{3\sqrt5}{8}$

$\sec (-t)= \frac{1}{ \cos (-t) }= -\frac{8}{3\sqrt5}$

$\tan t=\frac{y}{x}=-\frac{\sqrt{19}/8}{3\sqrt5/8}=-\frac{\sqrt{95}}{15}$

$\cot t=\frac{1}{\tan t}=-\frac{15}{\sqrt{95}}$

$\tan(-t) =-\tan t=\frac{\sqrt{95}}{15}$

$\cot(-t) =\frac{1}{\tan (-t)}=\frac{15}{\sqrt{95}}$