Question

Let (-6,-5) be a point on the terminal side of theta

Answer 1

Answer:

Part 1) $sin(\theta)=-\frac{5\sqrt{61}}{61}$  or $sin(\theta)=-\frac{5}{\sqrt{61}}$

Part 2) $csc(\theta)=-\frac{\sqrt{61}}{5}$

Part 3) $cot(\theta)=\frac{6}{5}$

Step-by-step explanation:

we know that

The point (-6,-5) lies on the III Quadrant

so

sin(theta) is negative

csc(theta) is negative

cot(theta) is positive

Part 1) Find the sine of angle theta

we know that

The function sine is equal to divide the opposite side to the angle theta by the hypotenuse

$sin(\theta)=\frac{y}{r}$

we have

$x=6\ units,y=5\ units$

Applying the Pythagoras Theorem

$r^{2}=x^{2}+y^{2}$

substitute

$r^{2}=6^{2}+5^{2}$

$r^{2}=61$

$r=\sqrt{61}\ units$ -----> the hypotenuse

substitute

$sin(\theta)=\frac{5}{\sqrt{61}}$

Simplify

$sin(\theta)=\frac{5\sqrt{61}}{61}$

Remember that the sin(theta) is negative

so

$sin(\theta)=-\frac{5\sqrt{61}}{61}$

Part 2) Find the cosecant of angle theta

we know that

The function cosecant is equal to divide the hypotenuse by the opposite  side to the angle theta

$csc(\theta)=1/sin(\theta)=\frac{r}{y}$

we have

$sin(\theta)=-\frac{5}{\sqrt{61}}$

so

$csc(\theta)=-\frac{\sqrt{61}}{5}$

Part 3) Find the cotangent of angle theta

we know that

The function cotangent is equal to divide the adjacent side to the angle theta by the opposite side to the angle theta

$cot(\theta)=\frac{x}{y}$

we have

$x=6\ units,y=5\ units$

substitute

$cot(\theta)=\frac{6}{5}$