Given #sin(theta) = -1/5# and #tan(theta) > 0#, find #cos(theta)# (draw angle)?

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Answer 1 · 2017-11-06T02:58:11

#cos(theta) = -(2sqrt6)/5#

Given #sin(theta) = -1/5# and #tan(theta)>0# means that #theta# is in the third quadrant, where the cosine function is negative:

#cos(theta) = -sqrt(1 - sin^2(theta))#

#cos(theta) = -sqrt(1-(-1/5)^2)#

#cos(theta) = -sqrt(1-1/25)#

#cos(theta) = -sqrt(24/25)#

#cos(theta) = -(2sqrt6)/5#

Here is the angle drawn within the unit circle:

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