Melody

I don't have time to answer now. (even for myself) but I would probably start by letting 

theta = acos (-1/9)

I will get back to it later/

Oh sorry.

I didn't notice the brackets properly.  My fault.

No it wouldn't... 

cos theta = adj / hyp = ( -1/9)/2       [which would simplify to -1/18]

If    \(cos\theta = (-1/9)/2\)

then what is 

\(acos ((-1/9)/2)\)    ?

Maybe it might help to think of it as an equation

\(\qquad cos \theta = ((-1/9)/2)\\ so\\ acos[cos \theta] = acos ((-1/9)/2)\\ \qquad \qquad\theta= acos ((-1/9)/2)\\ \qquad \text{[since acos and cos cancel each other out.]}\)

You have my right angle triangle diagram with theta on it.  

Work out the third side using pythagoras.

Then read sin(theta) off it.

I have hidden my answer and after I write this I will block further answering.

It was a good question.

I am sorry your students are cheating.

I do not believe it is possible to give an internet test for maths.

Maybe in the furure there may be some way around it but at present Maths tests need to be fully supervised in a classroom.  

This is what I think.

sorry, I don't know.

When you get an answer please share it with the forum :)

ok fair enough.

theta is in the second quadrant so cos(theta) will be negative

Now  tell me what cos(theta) is.

Answer quickly if you can. 

You can just put it straight into a calculator...

Most calculators with just give you an approximations of course.

I got an approximation of 0.425 radian.     II would do it over and make sure I had to identical answers before I accepted it though.

Are you looking for an exact answer?

Hint:

Thanks Ticon6  ;)