Trigonometry properties

In a right triangle, each acute angle has three trig ratios based on the side lengths.
Relative to the angle, one side is opposite, one is adjacent, and the longest (across from the right angle) is the hypotenuse.

The ratios are remembered as SOH-CAH-TOA.
$\sin = \frac{\text{opposite}}{\text{hypotenuse}}$, $\cos = \frac{\text{adjacent}}{\text{hypotenuse}}$, and $\tan = \frac{\text{opposite}}{\text{adjacent}}$.

The two acute angles of a right triangle always add to $90^\circ$; they are complementary.
The side opposite one angle is adjacent to the other, so $\sin$ of one equals $\cos$ of the other: $\sin A = \cos B$.

Written with angles, this is the co-function identity: $\sin(A) = \cos(90^\circ - A)$.
So whenever you see $\sin(\text{something}) = \cos(\text{something else})$, the two angles inside must add to $90^\circ$.
Setting their sum equal to $90$ lets you solve for the unknown.

The same idea works in radians, where $90^\circ = \frac{\pi}{2}$.
So $\cos\left(\frac{\pi}{2} - x\right) = \sin(x)$, and for example $\sin\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{3}\right)$ because $\frac{\pi}{6}$ and $\frac{\pi}{3}$ add to $\frac{\pi}{2}$.

If $\sin(A^\circ) = \frac{4}{5}$, what is $\cos(90^\circ - A^\circ)$?
By the co-function rule, $\cos(90^\circ - A) = \sin(A)$.
So $\cos(90^\circ - A^\circ) = \frac{4}{5}$.

If $\sin(3x - 12^\circ) = \cos(2y + 7^\circ)$, how are $x$ and $y$ related?
Since $\sin$ equals $\cos$ only when the angles are complementary, the two angles add to $90^\circ$.
So $(3x - 12) + (2y + 7) = 90$, which simplifies to $3x + 2y = 95$.

If $\cos(\theta) = 0.6$, what is $\sin(90^\circ - \theta)$?
By the co-function rule, $\sin(90^\circ - \theta) = \cos(\theta)$.
So $\sin(90^\circ - \theta) = 0.6$.