Scribe post

I'm sorry this is late but I've been having troubles with this posting thing.
To start the class we did a warm up where we had to evaluate

cos^-1(√3/3)
a) you would be looking for an angle and since you are looking for an angle where the inverse cos is a -√(3)/2 then it would be 150˚it is also possible to get an answer of 5π/6 algebraically you could do x=cos
cosx=-√3/2
sin(cos^-1 √5/5)
b) in this one the answer will be a ratio ø=cos^-1 sqrt(5)/5
cosø=sqrt(5)/5 sinø=(√20)/5

we also had to evaluate the graphs of
y=arscin(x-1)
a)y=arcsin(x-1) this is just a sin graph only with the restricted domain to make it a function and the negative 1 makes it move 1 unit to the right and the minimum and maximum is π/2
y=tan(3x-6/π)
b) you can set this up as an inequality because the period of tan is π so -π/2<3x-π/6<π/2>2π/9 y="sec^-1" secy="x" cosy="1/x" y="arccos^-1(1/x"