If #tan(theta) = 3/4# and #sin(theta) < 0#, how do you find #cos(theta)#?

1 Answer
NickTheTurtle
May 5, 2017

#-4/5#

Explanation:

Since #tan(theta)=3/4>0# and #sin(theta)<0#, #theta# is in quadrant 3 (since #tan(theta)>0# iff #theta# is in quadrant 1 or 3 and #sin(theta)<0# iff #theta# is in quadrant 3 or 4).

#cos(theta)# must then be negative.

Remember that #sin^2(theta)+cos^2(theta)=1#. Divide both sides by #cos^2(theta)# to get #sin^2(theta)/cos^2(theta)+1=1/cos^2(theta)#. Since #tan(theta)=sin(theta)/cos(theta)#, this is simply #tan^2(theta)+1=1/cos^2(theta)#. Thus, #cos^2(theta)=1/(tan^2(theta)+1)#.

Since #tan(theta)=3/4#, #cos^2(theta)=1/((3/4)^2+1)=16/25#. Since #cos(theta)# is negative, #cos(theta)=-sqrt(16/25)=-4/5#.

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