How do you find the exact value of `arccos(sin(3*pi/2))`?

You need to covert the expression involving the `\sin` inside the brackets into one involving a `\cos` because `\arccos(\cos x) =x`.

There are always several ways to manipulate trig functions, however one of the most straight forward ways to covert an expression involving sine into one for cosine is to use the fact that they are the SAME FUNCTION just shifted over by `90^o` or `pi/2` radians, recall

`\sin(x) = \cos(pi/2 - x)` .

So we replace `\sin({3 \pi }/2)` with `\cos(pi/2-{3 \pi }/2)`
or `= \cos (-{2pi}/2)=\cos (-pi)`

`\arccos( \sin({3 \pi }/2))=\arccos( \cos(-\pi ))=-pi`.

There is the odd issue with multiple solutions to many expressions involving inverse trig functions. The most obvious relates to `cos(x)= cos(-x)`, so you can replace `\cos(-pi)` with `\cos(pi)` and repeat the above end up with `\arccos( \sin({3 \pi }/2))=pi` . Why?

Because of the periodicity of the cosine function with have `cos(pi)=cos(2pi*k+pi)` , so there are even more answers! Infinity of them, `\pm (2*k+1)pi`, positive or negative odd multiples of `pi`.

The real issue here is the inverse cosine, cosine is a function with has multiple y values so when you reverse it you actually get an infinite number of possible answers, when we use it we RESTRICT the values to a window of `pi` size, `0 <= x <= pi` is a typical one (calculator often use this one). Others use `- pi <= x <= 0` and `pi <= x <= 2 pi` is also valid. In each of these "windows" we only have one solution. I'm going to go with the calculator's answer for above.

We have, `sin3pi/2=-1.`

Hence, reqd. value `= arccos(sin3pi/2)=arccos(-1) = theta,` say.

Then, by defn. of `arccos, costheta=-1=cos pi,` where of course, `theta in [0,pi].`
`:. theta=pi,` as cos fun. is one-one in `[0,pi].`