If #cot theta = - 2# and #cos theta < 0#, how do you find #sintheta#?

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Answer 1 · 2018-03-30T21:28:08

Start with the identity:

#1+ cot^2(theta)=csc^2(theta)#

Substitute #cot^2(theta) = (-2)^2#:

#1+ (-2)^2=csc^2(theta)#

#5 = csc^2(theta)#

Substitute #csc^2(theta) = 1/sin^2(theta)#

#5 = 1/sin^2(theta)#

#sin^2(theta) = 1/5#

#sin(theta) = +-sqrt5/5#

Because we are told that #cos(theta) < 0# and #cot(theta) = -2#, we know that the sine function must be positive in this quadrant:

#sin(theta) = sqrt5/5#