Solving sin x = k, cos x = k and tan x = k

The calculator only tells you half the story

$\sin^{-1}0.4 = 23.6^\circ$ . Great, but $\sin 156.4^\circ$ is also $0.4$ , and so is $\sin 383.6^\circ$ .

The calculator gives one answer (the principal value). The question wants every answer in the interval (the set of $x$ -values allowed, like $0^\circ \leqslant x \leqslant 360^\circ$ ).

Below, $k$ is a given number (for sin and cos it must be between $-1$ and $1$ ), and $\alpha$ is the principal value.

The recipe:

Principal value $\alpha$ from $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ .
Second value from the symmetry of the graph (table below).
Add or subtract the period ( $360^\circ$ or $2\pi$ ; for tan, $180^\circ$ or $\pi$ ) until you've covered the interval.
Keep only the values inside the interval. Check whether the ends are included.

Where the second solution lives

Equation	Principal value	Second value (degrees)	(radians)	Then repeat every
$\sin x = k$	$\alpha = \sin^{-1}k$	$180^\circ - \alpha$	$\pi - \alpha$	$360^\circ$ ( $2\pi$ )
$\cos x = k$	$\alpha = \cos^{-1}k$	$360^\circ - \alpha$ (or $-\alpha$ )	$2\pi - \alpha$	$360^\circ$ ( $2\pi$ )
$\tan x = k$	$\alpha = \tan^{-1}k$	$180^\circ + \alpha$	$\pi + \alpha$	$180^\circ$ ( $\pi$ )

Why these? They come from the mirror lines of the graphs:

The sine wave is symmetric about $x = 90^\circ$ . So whatever height it has at $\alpha$ , it has again at $180^\circ - \alpha$ .
The cosine wave is symmetric about $x = 0^\circ$ (and $360^\circ$ ). So $-\alpha$ and $360^\circ - \alpha$ give the same value as $\alpha$ .
Tan simply repeats every $180^\circ$ .

These rules work even when $\alpha$ is negative. Or use CAST: the solutions are in the quadrants where the ratio has the sign of $k$ .