# Inverse trigonometric functions and equations.

I have been facing a lot difficulty to grasp inverse trigonometric functions and equations. I could know that domains and ranges of basic functions are well defined to restrict them such that these functions are one-one functions. I know how to solve those equations but to be honest I don't know what I am doing.

So basically where I am facung difficulty are indicated below,hopefully I can get help from others.

1. For $$f(x)=arcsin(x)$$ domain is $$[-1,1]$$ and range is $$[-\pi/2,\pi/2]$$. Now if we want to convert this function to a function of $$arccosine$$ then we can consider a triangle and put the values of respective sides and then obtain the other side. After that simply put the value of cosine instead of x. I understood that in order to restrict this function into their princiole values we consider e triangle such that the angle is restricted for a small range of angles. Rest of the thing such as in cartesian coordinate system where does that triangle locate depends on the signed values of sides. But the function of $$arccosine$$ has a different set of domain and range. This isn't making any sense to me because we are just relating two different functions of two different sets of domain and range.

2. If we are given with such an equation $$asin(x)+bcos(x)=c$$,by solving these type of equations we often tend obtain extraneous values. But I was taught that if we divide both sides of these equations with $$\sqrt{a^2+b^2}$$, then we can solve this equation and obtain actual roots. But how do we ensure that this method always gives us actual roots and excludes extraneous roots. What are the chances of not excluding any actual root in this method?

Pardon me for any sort of misunderstanding or missing out any vital points

Edit: I wrote arcsine and arccosine instead of sine and cosine respectively in the equation by mistake.

The functions $$\arcsin$$ and $$\arccos$$ both have domain $$[-1, 1]$$, and they satisfy, for all $$x \in [-1, 1]$$, $$\arcsin x + \arccos x = \frac{\pi}{2}.$$ While they have different ranges, this equation always holds true, for any point in their common domain. It can be proven with right-angled triangles for $$x \in [0, 1]$$ as you state in your question: just create a triangle with one leg length $$x$$, and the hypotenuse length $$1$$. Then each of the smaller angles is $$\arcsin x$$ and $$\arccos x$$, and they must sum to $$\frac{\pi}{2}$$.

Proving it geometrically for $$x \in [-1, 0)$$ is also possible, though it requires a little more work (creating right-angled triangles from complements of obtuse angles, and things like that). But, it's also possible to prove easily enough without geometry. Indeed, $$\arcsin$$ and $$\arccos$$ are defined just to be the inverses of $$\sin$$ and $$\cos$$ restricted to $$[-\pi/2, \pi/2]$$ and $$[0, \pi]$$ respectively. On these intervals, the respective functions are injective, so they are invertible when their codomain is restricted to their common range $$[-1, 1]$$. Moreover, since $$\cos(x) = \sin(\pi/2 - x)$$ for all $$x$$, and the interval $$[-\pi/2, \pi/2]$$ is simply $$[0, \pi]$$ when mapped under $$x \mapsto \pi/2 - x$$, we indeed get the above equality for all $$x \in [-1, 1]$$.

If you're given the equation $$a \arcsin x + b \arccos x = c$$, then you should use the above equality to say $$a \arcsin x + b\left(\frac{\pi}{2} - \arcsin x\right) = c \iff \arcsin x = \frac{c - b\frac{\pi}{2}}{a - b} \implies x = \sin\left(\frac{c - b\frac{\pi}{2}}{a - b}\right).$$ Note the final step is an $$\implies$$, not an $$\iff$$, so there could be extraneous solutions introduced. In particular, if $$\frac{c - b\frac{\pi}{2}}{a - b}$$ does not lie in $$[-\pi/2, \pi/2]$$, then the equation will have no solution, but an extraneous solution will be introduced.

The method you're using would be applicable when you have the equation $$a \sin x + b \cos x = c$$. In this case, dividing by $$\sqrt{a^2 + b^2}$$ would yield $$\frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 + b^2}} \cos x = \frac{c}{\sqrt{a^2 + b^2}}.$$ Unless $$a = b = 0$$ (which makes the equation quite trivial indeed!), then this can always be done and the result is equivalent. Then, there exists some $$y \in \Bbb{R}$$ such that \begin{align*} \cos(y) = \frac{a}{\sqrt{a^2 + b^2}} \\ \sin(y) = \frac{b}{\sqrt{a^2 + b^2}}. \end{align*} Indeed, $$y$$ is the angle that the vector $$(a, b)$$ makes with the positive $$x$$-axis. So, we obtain an equivalent (meaning no extraneous solutions created) expression: $$\cos y \sin x + \sin x \cos x = \frac{c}{\sqrt{a^2 + b^2}},$$ where $$y$$ is the angle given as above. This equation, again, has exactly the same solution set as $$a \sin x + b \cos x = c$$. From here, using the angle sum formula yields another equivalent expression: $$\sin(x + y) = \frac{c}{\sqrt{a^2 + b^2}}.$$ From here, it's now a matter of solving an equation of the form $$\sin x = c$$, which will produce infinitely many solutions, including a solution involving an $$\arcsin$$ term, and the rest involving adding or subtracting various multiples of $$\pi/2$$ (or no solutions, if the term on the right is larger than $$1$$). But, the important fact here is that there are no solutions lost or added up to this point, so you can feel confident that you will obtain no extraneous solutions with this method. Indeed, there will be multiple possible values that $$y$$ can take, but it ultimately won't matter which one you take, as they will always produce the complete solution set.

• To be honest I didn't get how is that latter method proven and how does it ensure that there won't be any actual root left out?
– MSKB
Commented Aug 12, 2021 at 7:52
• Once you find a value of $y$ so that $\cos y$ and $\sin y$ are as above, then $\cos y \sin x + \sin y \cos x$ has the same value as $\frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 + b^2}} \cos x$. You are just rewriting the same expression a different way, except in terms of this $y$ angle. The $y$ is not an unknown in the equation, it's an angle that depends on the specific values of $a$ and $b$. For example, if $a = \sqrt{3}$ and $b = 1$, then $y = \pi/6$ is a valid choice. So long as $a$ and $b$ are not both $0$, there will be such a $y$. Commented Aug 12, 2021 at 10:49