Question

How do you use the point (7,-2) on the terminal side of the angle to evaluate the six trigonometric functions?

Answer 1

See explanation.

Explanation:

The trigonometric functions are defined as quotients of specified coordinates `x` and `y` and the distance between the point and the origin `r`. So first we have to calculate `r`:

`r = sqrt(x^2+y^2)=sqrt(7^2+(-2)^2)=sqrt(49+4)=sqrt(53)`

Now we can calculate the functions according to their definitions:

`sinalpha=y/r=(-2)/sqrt(53)=(-2sqrt(53))/53`

`cosalpha=x/r=7/sqrt(53)=(7sqrt(53))/sqrt(53)`

`tanalpha=y/x=(-2)/7=-2/7`

`cotalpha=x/y=7/(-2)=-7/2`

`secalpha=r/x=sqrt(53)/7`

`csc alpha=r/y=sqrt(53)/-2=-sqrt(53)/2`