Let (-6, -5) be a point on the terminal side of 0. Find the exact value of sin0, sec0, and tan0

Answer 1

Given point (-6, -5).

Since both digits are odd; the point is in the 3rd quad.

The sides of the triangle are: -6, 0; 0, -5; -6, -5.

Label the triangle: side "a"=5.

"b"=6.

"c"=a^2+b^2=c^2.

=25+36=61;^.5=7.81.

I guess angle "O" is α; it's the center.

So: a/sinα=b/sinβ=c/sinγ.

5/sinα=7.81 (sin of a rt angle = 1).

5/7.81=.641; asin=39.807........°

sin39.8=.641; sec39.8=1/cos39.8=1.30137.......; tan39.8=.83337.......

Answer 2

If (-6, -5) is on the terminal side that puts us in the third quadrant. If you plot that point and draw a diagonal to it from the origin, you will be able to draw a right triangle with that diagonal as the hypotenuse. That means the adjacent side is -6 and opposite side it -5.

By Pythagorean theorem, we find that the hypotenuse is sq rt 61 (not pretty).

From there, sin is opposite over hypotenuse (-5/ sqrt 61) but don't forget to rationalize your denominator! (Should get -5sqrt61/61)

sec theta is the inverse (or reciprocal) of cos...so cos would be adjacent over hypotenuse, meaning sec will be hypotenuse over adjacent. Or sqrt 61/-6

tan is the opposite over adjacent or in this case; -5/-6 which = 5/6

Does that make sense?