# How to solve inverse trig. equations like sin(arctan 2)?

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

$$\sin(\arctan 1/2+\arccos 1/3)$$ $$= \sin(\alpha+\beta) = \sin\alpha\cos\beta +\cos\alpha\sin\beta.$$

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

• Would you look at this example? math.stackexchange.com/questions/3084939/… I've used your method, but somehow I'm getting slightly wrong result. :( – weno Jan 23 at 19:42
• You need to think about what quadrant you're in. In the first example, you should draw the triangle in the 4th quadrant so that the $3$ is really $-3$. In the second example, the triangle is in the 2nd quadrant, so the $1$ is really $-1$, which will change the sign of your answer. – B. Goddard Jan 23 at 20:06
• Thanks. (10 char) – weno Jan 23 at 20:14

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$$\left\{\begin{array}{l}c^2+s^2=1\\\dfrac sc=2\end{array}\right.$$and don't forget the only positive solutions matter here.

• Hm but where the assumptions about the $c$ came from? There's nothing about $cos(arctan2)$ in the task. Or do I need to do it like that? – weno Jan 23 at 11:40
• Which assumptions do you have in mind? I defined $c$ as $\cos(\arctan 2)$. And, no, you don't need to do it like this. If you find any other way, I'm fine with it. – José Carlos Santos Jan 23 at 11:43

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

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

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

You might also know the relation

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

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.)