How do you find the values of the six trigonometric functions given #costheta=-4/5# and #theta# lies in Quadrant III?

1 Answer
May 20, 2017

Answer:

#"see explanation"#

Explanation:

#"in the third quadrant only the ratios for " tantheta" and hence " cottheta" are positive"#

#costheta=-4/5tocolor(red)(1)#

#• sectheta=1/costheta#

#rArrsectheta=1/(-4/5)=-5/4tocolor(red)(2)#

#• sintheta=+-sqrt(1-cos^2theta)#

#rArrsintheta=-sqrt(1-16/25)=-sqrt(9/25)=-3/5tocolor(red)(3)#

#• csctheta=1/sintheta#

#rArrcsctheta=1/(-3/5)=-5/3tocolor(red)(4)#

#• tantheta=sintheta/costheta#

#rArrtantheta=(-3/5)/(-4/5)=3/4tocolor(red)(5)#

#• cottheta=1/tantheta#

#rArrcottheta=1/(3/4)=4/3tocolor(red)(6)#