# Ask Uncle Colin: Trigonometric inverses and picking the correct quadrant

Dear Uncle Colin,

When I have an angle in the second quadrant, I can find it just fine using $\cos^{-1}$ - but using $\sin^{-1}$ or $\tan^{-1}$ gives me an angle in the fourth quadrant. I don’t understand why this is!

-- I Need Verbose Explanations; Radians Seem Excellent

Hi, INVERSE, and thanks for your message!

The short answer is, convention! Because sine, cosine and tangent are periodic functions, when they’re defined over all $x$, they don’t have inverse functions. However, we can restrict their domains so that they do.

To do that, we need to make sure that each output ($y$) value only appears once. For cosine, that’s an easy choice: when $x=0$, then $y=\cos(x)$ has a maximum, so picking the domain $0 \le x \le \pi$ ensures every $y$ value turns up exactly once. That means the natural inverse, $\cos^{-1}$ returns values in the first (for positive arguments) and second (for negative) quadrants.

Sine is also a fairly easy choice: if we go forwards and backwards from $x=0$, we get to the extrema at $\pm \frac{1}{2}\pi$, which means the natural domain for $\sin(x)$ is $-\frac{1}{2}\pi \le x \le \frac{1}{2}\pi$. Its inverse gives values in the first (for positive arguments) and fourth (for negative) quadrants.

Tangent is the only one where there’s really a choice: we could either go from 0 to $\pi$, or from $-\frac{1}{2}\pi$ to $\frac{1}{2}\pi$. The first of those is sort-of OK - only it covers the discontinuity at $\frac{1}{2}\pi$, which makes for a slightly awkward function. If we pick the second option, we get a nice, continuous one - so we do. The natural domain for $\tan(x)$ is $-\frac{1}{2}\pi < x < \frac{1}{2}\pi$ ((Note that the inequality is strict for $\tan$ - it isn’t defined when $x$ is an odd multiple of $\frac{1}{2}\pi$.)), and its inverse returns values in the first quadrant (for positive arguments) or the fourth (for negative).

Hope that clears it up!

-- Uncle Colin

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