### Trigonometry: Triangles to Functions

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Imagine you were told that in triangle $ABC$, the side opposite $C$ has length $\quantity{5}{cm}$, the side opposite $A$ has length $\quantity{3}{cm}$ and angle $A=\tfrac{\pi}{6}$. Think about how you would calculate angle $C$.

You may have heard of inverse trigonometric functions, such as $\arcsin x$ or $\sin^{-1}x$ or the phrase “inverse sine”. In principle, inverse trigonometric functions should be functions that undo the effects of functions such as $\sin x$, $\cos x$ and $\tan x$. But we need to be careful because these functions are are not one-to-one; in fact they are periodic, so for instance there are infinitely many values of $x$ that give the same value of $\sin x$.

Although we started with a situation where we were thinking about an inverse for $\sin x$, we’ll think briefly about $\tan x$ first before moving back to $\sin x$ and $\cos x$. Some of the issues that arise for an inverse of $\tan x$ also arise for $\sin x$ and $\cos x$, but we’ll see that we have to be even more careful for these functions.

So $\arctan a$, $\arcsin a$ and $\arccos a$ give the principal values for the equations $\tan x=a$, $\sin x=a$ and $\cos x=a$ respectively, and we can work from these to find all possible solutions. You can see more about this in General solutions and in Inverse or not?.

Going back to our original triangle problem, we had $\sin C=\tfrac{5}{6}$. The inverse function will give us the principal value (in radians) as $C=\arcsin\tfrac{5}{6}\approx0.985.$ But there is more than one value of $C$ that would satisfy our equation. We must always consider other possibilities when using inverse trigonometric functions.

Since we are talking about an angle in a triangle we need only consider values in the interval $[0,\pi]$. There is one more possible value which is $\pi-\arcsin\tfrac{5}{6}\approx2.156.$