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

1 Answer
Nov 29, 2016

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#