Your problem is that the cosine function produces only numbers between -1 and 1. That is another way of saying that acos ONLY works on numbers that are between -1 and 1. Remember that acos is the functional inverse of the cosine function.
If you give it (acos) a number outside of that interval, then you get a complex number as a result. For example:
acos(1.2)
Do you see that I got a complex result? Now we can look at what you did. When you have t2 as:
t2=30:0.01:70;
Now, what is:
acosarg = (2*t2)-80;
min(acosarg)
max(acosarg)
Do you see this produces numbers that are far outside of the interval that acos can handle?
For some reason, you seem to be thinking that acos uses degrees. Even radians are not meaningful here, in terms of the ARGUMENTS to acos. Again, look at the plot below:
fplot(@acos,[-1,1])
As you can see, acos lives on the interval [-1,1]. Go beyond that, and you get complex stuff as a result.
Next, even when you do get the arguments to acos correct, you will need to learn about the use of the ./ operator.
When you want to multiply or divide the elements of two vectors or arrays, you need to use .* to ./ to perform that operation. Otherwise, MATLAB will have a problem.
So if t2 is a vector, then even if this line of code was written correctly for scalars, with a vector t2, this line of code SHOULD have been written:
y2=(120/pi)*acos((2*t2)-80)./(((10-t2)));
Do you see that I used a ./ operator there? You can multiply a scalar times a vector, because MATLAB is smart enough to do that operation. You can even divide the elements of a vector by a scalar. But you CANNOT divide the elements of two vectors in an element-wise fashion without using the ./ operator.
