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These are some of the tasks I am supposed to be prepared for. I have no idea where to begin when solving them. Below I present what I already know regarding the subject and what I have problems with.

Calculate:

Please note I am not very familiar with inverse trig functions, if you could use beginner-friendly math language that would be perfect. :)

What I actually do know regarding the subject:

$sin(arcsin x) = x$

$cos(arccos x) = x$

$sin(arcsin x) = x$

Not sure about $text{arccot} x$ case, though. I am familiar with very basic concept of trig. and inv. trig. functions like: its domain, range, zeros, how the functions' graphs looks like, values at points like: 0, 30, 45, 60, 90 degree.

What is confusing me is "doing operations on the rectangular triangle". I know how to apply sin/cos/tan/ctg to the triangle, no idea what about inverse trig. functions.

All kinds of tips on how to solve:

$sin{arctan{2} - text{arccot 3}} = ?$

or

$arcsin{sin{100}}$

are appreciated! Thanks.

I think it's best to think of inverse trig functions as angles. For something like $sin(arctan x),$ think of $arctan x = theta$, so that $tan theta = x.$ Then draw a right triangle that tells this story. There are infinitely many, but a good choice is the right triangle with opposite leg $x$ and adjacent leg $1$. Then compute the hypotenuse to be $sqrt{x^2+1}.$ Now it's easy to find $sin(arctan x) = sin theta = x/sqrt{x^2+1}.$

For something like $sin(arctan 1/2+arccos 1/3)$ let $arctan 1/2 =alpha$ and $arccos 1/3 =beta$. Then

$$sin(arctan 1/2+arccos 1/3)$$ $$ = sin(alpha+beta) =
sinalphacosbeta +cosalphasinbeta.$$

Then proceed to evaluate each of $sin alpha,sin beta$ etc.

If $s=sin(arctan 2)$ and $c=cos(arctan 2)$, then $c^2+s^2=1$ and$$frac sc=tan(arctan 2)=2.$$So, solve the system$$left{begin{array}{l}c^2+s^2=1\dfrac sc=2end{array}right.$$and don't forget the only positive solutions matter here.

Ok, so I will try to illustrate how I generally approach these by virtue of some examples.

arccos(sinx) : you want $sinx = cosy$ for some $y$ . Obviously $y = frac{pi}{2} - x$ works so $arccos(sinx) = frac{pi}{2} - x $ (wherever this is properly defined, note!)

cos(arcsinx) : write $cos(arcsinx) = +-sqrt{1-sin(arcsinx)^2} = +-sqrt{1-x^2}$ (the sign depends on your actual value for x and convention for arcsin)

sin(arctanx)): use the fact that $sinx = frac{2tan(frac{x}{2})}{1+tan(frac{x}{2})^2}$

The rest of the combinations should be deducible from these.

$$sin(arctan2)$$

is asking you "the tangent of some angle is $2$, what is the sine ?"

Now, you can use

$$4=tan^2theta=frac{sin^2theta}{cos^2theta}=frac{sin^2theta}{1-sin^2theta}.$$

Solving for $sin^2theta$, you get $dfrac45$.

You might also know the relation

$$sin^2theta=frac{tan^2theta}{tan^2theta+1}.$$

If you want to simplify $sin(arccos x),$ for example, try to use formulas to re-express the $sin$ with $cos,$ and then make use of $cos(arccos x) = x.$

Specifically, in this example, use $sin(x)=sqrt{1-cos^2(x)}$ to arrive at
$$
sin(arccos x) = sqrt{1-cos^2(arccos x)} = sqrt{1-x^2}.
$$
(Skipping over issues of where the function is defined, or multiple values for the root.)