Question

What is the value of sin0 given that (-6,-8) is a point on the terminal side of 0​

Answer 1

`sin \theta =(-8)/(10) =-(4)/(5)`

Explanation:

For this case we have a point given (-6,-8) and we know that this point is terminal side of 0​

We can assume that the length of th opposite side is given by:

b=-8 and the length for the adjacent side would be a=-6

And we can find the hypothenuse on this way:

`c= √(a^2 +b^2)=√((-6)^2 +(-8)^2)=10`

From the definition of sin we know this:

`sin O =(opposite)/(hypothenuse)`

And if we replace we got this:

`sin \theta =(-8)/(10) =-(4)/(5)`

We can aslo find the cos with the following identity:

`cos^2 \theta + sin^2 \theta = 1`

And then:

`cos \theta = \pm √(1-sin^2 \theta)=\pm √(1- (-4/5)^2)=\pm (3)/(5)`

But since both corrdinates are negative we are on the 3 quadrant and then
`cos \theta= -(3)/(5)`