How do you find the remaining trigonometric functions of #theta# given #costheta=24/25# and #theta# terminates in QIV?

1 Answer
Mar 30, 2017

Answer:

#sintheta=("-"7)/25#
#costheta=24/25#
#tantheta=("-"7)/24#
#cottheta=("-"24)/7#
#sectheta=1/24#
#csctheta=("-"1)/7#

Explanation:

Let the point #P(x,y)# be on a circle of radius #r# centered at the origin and #P# is on the terminal ray of #theta#. Let the angle #alpha# be the the nonreflex angle from the positive x-axis that is coterminal with #theta#.

#theta# terminates in Q#"IV"#, so for point #P#, #x# is positive and #y# is negative. #alpha# is negative..

The line segments from #(0,0)# to #(x,0)# and to #(0,y)# form a right triangle with hypotenuse #r# and angle #alpha# between #x# and #r#. #cosalpha=x/r# and #sinalpha=y/r#.

Since #alpha# and #theta# start and end in the same place, all of their trig functions are the same, so by finding the function values for #alpha# we find them for #theta#. You can say that since #theta=alpha+2pi#, they are effectively the same angle; I will be using #theta# because that is what we're looking for.

#costheta=x/r#. We are given that #costheta=24/25#. Therefore, #x=24# and #r=25#.

#x#, #y#, and #r# are the lengths of the two legs and the hypotenuse of a right triangle, respectively. Therefore, #x^2+y^2=r^2#. Solving for #y#, #y=+-sqrt(r^2-x^2)#. We know the values of #x# and #r#, so we can find that #y=+-sqrt(25^2-24^2)=+-7#. We know that #y# is negative so #y="-"7#.

#sintheta=y/r#, so #sintheta=("-"7)/25#

#tantheta=y/x#, so #tantheta=("-"7)/24#

#cottheta=x/y#, so #cottheta=24/("-"7)=("-"24)/7#

#sectheta=1/x#, so #sectheta=1/24#

#csctheta=1/y#, so #csctheta=1/("-"7)=("-"1)/7#