Question

Given sin theta is -3 / 5 with theta in quadrant 3 find the exact values of the six trig functions.

Answer 1

Answer:

$cos(theta)= \frac{adjacent }{hypotenuse} =-\frac{4}{5}$

$tan(theta)= \frac{opposite}{adjacent } =\frac{3}{4}$

$sec(theta)= \frac{1}{cos(theta)} =-\frac{5}{4}$

$csc(theta)=\frac{1}{sin(theta)}=\frac{-5}{3}$

$cot(theta)= \frac{1}{tan(theta)} =\frac{4}{3}$

Step-by-step explanation:

$sin(theta)=\frac{-3}{5}$

sin= opposite / hypotensue

opposite of theta = 3  and hypotenuse = 5

the theta is in quadrant 3. Sin is negative in third quadrant. tan is positive and cos is negative in third quadrant.

Make a right angle triangle . and find the adjacent side

$c^2= a^2 + b^2$

$5^2= (3)^2 + b^2$

$25=9 + b^2$

subtract 9 from both sides

$b^2= 16$

b=4, adjacent side = 4

$cos(theta)= \frac{adjacent }{hypotenuse} =-\frac{4}{5}$

$tan(theta)= \frac{opposite}{adjacent } =\frac{3}{4}$

$sec(theta)= \frac{1}{cos(theta)} =-\frac{5}{4}$

$csc(theta)=\frac{1}{sin(theta)}=\frac{-5}{3}$

$cot(theta)= \frac{1}{tan(theta)} =\frac{4}{3}$