How do you find sin, cos, tan, sec, csc, and cot given (-4,-4)?

1 Answer
Mar 2, 2018

Answer:

See below.

Explanation:

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If we are given coordinates of the form #(x,y)#, where #x# and #y# are negative then we are in the III quadrant.

Since #(-4,-4)# are the sides of a right triangle, then the length of the terminal side ( the hypotenuse ) is given by Pythagoras' theorem:

Let the terminal side be #bbr#

#r^2=(-4)^2+(-4)^2#

#=>r=sqrt((-4)^2+(-4)^2)=4sqrt(2)#

So for the right triangle #bb(ABC)#, we have:

#c=4sqrt(2)#

#a=-4#

#b=-4#

#sin(theta)="opposite"/"hypotenuse"=a/c=-4/(4sqrt(2))=color(blue)(-sqrt(2)/2)#

#cos(theta)="adjacent"/"hypotenuse"=b/c=-4/(4sqrt(2))=color(blue)(-sqrt(2)/2)#

#tan(theta)="opposite"/"adjacent"=a/b=(-4)/-4=color(blue)(1)#

Since:

#color(red)bb(csc(theta)=1/sin(theta))#

#color(red)bb(sec(theta)=1/cos(theta))#

#color(red)bb(cot(theta)=1/tan(theta))#

We have:

#csc(theta)=1/(-sqrt(2)/2)=color(blue)(-sqrt(2))#

#sec(theta)=1/(-sqrt(2)/2)=color(blue)(-sqrt(2))#

#cot(theta)=1/1=color(blue)(1)#