Question

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

Answer 1

Answer:

$sin \theta =\frac{-8}{10} =-\frac{4}{5}$

Step-by-step 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:

$c= \sqrt{a^2 +b^2}=\sqrt{(-6)^2 +(-8)^2}=10$

From the definition of sin we know this:

$sin O =\frac{opposite}{hypothenuse}$

And if we replace we got this:

$sin \theta =\frac{-8}{10} =-\frac{4}{5}$

We can aslo find the cos with the following identity:

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

And then:

$cos \theta = \pm \sqrt{1-sin^2 \theta}=\pm \sqrt{1- (-4/5)^2}=\pm \frac{3}{5}$

But since both corrdinates are negative we are on the 3 quadrant and then $cos \theta= -\frac{3}{5}$