How do you determine #sintheta# given #cottheta=-4/3, 270^circ<theta<360^circ#?

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Answer 1 · 2017-11-10T09:45:34

#323.13^o#

#cot^2x = csc^2x-1#

#cscx=1/sinx#

#:.#

#(-4/3)^2=16/9#

So:

#csc^2theta-1=16/9#

#1/sin^2theta-1=16/9#

#sin^2x=9/25=>sintheta=3/5#

#theta= arcsin(sintheta)=arcsin(3/5)=36.87^o#

Required angle for interval is:

#360^o-36.87^o=323.13^o# ( 2 .d.p.)