Sign of trigonometric functions

On the unit circle, a point $P(a,b)$ at angle $x$ has $\cos x = a$ and $\sin x = b$ — the coordinates of $P$ are literally the values of cosine and sine. As $x$ sweeps through a full revolution, $P$ visits all four quadrants in turn, and $a$ and $b$ change sign exactly as $x$- and $y$-coordinates do anywhere in coordinate geometry.

• Quadrant I ($0 < x < \tfrac{\pi}{2}$): $a>0,\ b>0$, so $\sin x>0$ and $\cos x>0$ — and since every other function is a ratio or reciprocal of these two, all six functions are positive here.
• Quadrant II ($\tfrac{\pi}{2} < x < \pi$): $a<0,\ b>0$, so $\sin x>0$ and $\operatorname{cosec} x>0$, while $\cos x, \sec x, \tan x, \cot x$ are negative.
• Quadrant III ($\pi < x < \tfrac{3\pi}{2}$): $a<0,\ b<0$, so $\sin x$ and $\cos x$ are both negative; but $\tan x = \dfrac{\sin x}{\cos x}$ is a negative divided by a negative, hence positive, and $\cot x$ is positive too. $\sec x$ and $\operatorname{cosec} x$ stay negative.
• Quadrant IV ($\tfrac{3\pi}{2} < x < 2\pi$): $a>0,\ b<0$, so $\cos x>0$ and $\sec x>0$, while $\sin x, \operatorname{cosec} x, \tan x, \cot x$ are negative.

This pattern is captured by the mnemonic "All Silver Tea Cups" (equivalently "All Sin Tan Cos"), read quadrant by quadrant from Quadrant I onward:

where: each entry lists the two functions that are positive in that quadrant; every other function there is negative. At the quadrantal angles $0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}, 2\pi,\dots$ one of $\sin x$ or $\cos x$ equals $0$, so any function that divides by it becomes undefined — these boundary angles belong to no open quadrant and carry no "sign" at all.

Problem. Given $\tan x = -\dfrac{5}{12}$ with $x$ in Quadrant IV, find $\sin x$ and $\cos x$.

Step 1. Use the identity $1+\tan^2x=\sec^2x$ to bring in $\cos x$. Compute $\sec^2x = 1+\left(-\dfrac{5}{12}\right)^2 = 1+\dfrac{25}{144}=\dfrac{169}{144}$, so $\sec x=\pm\dfrac{13}{12}$.

Step 2. Fix the sign using the quadrant. In Quadrant IV, $\cos x>0$ (from the ASTC pattern above), and $\sec x$ shares cosine's sign, so $\sec x=\dfrac{13}{12}$ and hence $\cos x=\dfrac{12}{13}$.

Step 3. Recover $\sin x$ from $\tan x=\sin x/\cos x$, i.e. $\sin x=\tan x\cdot\cos x$:

Step 4. Check against the quadrant. In Quadrant IV, $\sin x$ must be negative — and $-\tfrac{5}{13}$ is negative, confirming the answer. $\sin x = -\dfrac{5}{13}$ and $\cos x = \dfrac{12}{13}$.

Competitive stretch

If instead only $\sin x<0$ and $\cos x<0$ were given, the quadrant sign chart shows $\sin x<0$ occurs in QIII and QIV, while $\cos x<0$ occurs in QII and QIII — the only quadrant common to both lists is QIII, so $\tan x = \dfrac{\text{negative}}{\text{negative}}>0$ and $\operatorname{cosec} x = \dfrac{1}{\text{negative}}<0$, exactly as Step 3 of the derivation predicts.

• Applying ASTC only to sine, cosine and tangent and forgetting the reciprocal functions: $\operatorname{cosec} x$ always matches $\sin x$'s sign, $\sec x$ always matches $\cos x$'s sign, and $\cot x$ always matches $\tan x$'s sign.
• Picking the wrong quadrant when only one condition is given. A single condition like $\sin x<0$ is true in two quadrants (III and IV); a second condition on $\cos x$ or $\tan x$ is needed to pin down exactly one.
• Concluding "negative times negative is negative" for tangent in Quadrant III, when a ratio of two negatives is actually positive — this is the single most common sign slip in the whole topic.
• Ignoring that quadrantal angles are undefined for some functions. At $x=\tfrac{\pi}{2}$, $\tan x$ and $\sec x$ are undefined; at $x=\pi$, $\cot x$ and $\operatorname{cosec} x$ are undefined — these boundary points have no quadrant and no sign.

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