Exercises — Trig functions for angles beyond 90° — ASTC rule (All, Sin, Tan, Cos)

WHAT: locate $200°$. WHY: the quadrant alone fixes the sign of $x=\cos \theta$. $180°<200°<270°$, so the ray is in Quadrant III (look at the bottom-left region of the s01 figure). In QIII, both $x$ and $y$ are negative. Since $\cos 200°=x<0$, the sign is negative. ASTC check: QIII is the "Tan" letter → only $\tan$ positive, so $\cos$ is negative. ✓

$270°<300°<360°$ → Quadrant IV (bottom-right of s01). Here $x>0$, $y<0$. So $\cos =x>0$ (positive), $\sin =y<0$ (negative), $\tan =y/x<0$ (negative). Only $\boxed{\cos }$ is positive — matching the "C" letter of ASTC in QIV. ✓

WHAT — quadrant: $180°<225°<270°$ → QIII. WHY: past the $180°$ line but before straight-down $270°$, so the tip sits bottom-left. WHAT — reference angle ${\theta }^{\prime }=225°-180°=45°$. WHY: the reference angle is the acute gap from the ray to the nearest $x$-axis, and in QIII the nearest half is the negative $x$-axis at $180°$; the gap is $\theta -180°$. WHAT IT LOOKS LIKE: the small red arc at the bottom-left ray in s02. WHAT — sign: QIII is "Tan" positive → $\sin$ is negative. WHY: in QIII the height $y=\sin \theta$ is below the axis, so $y<0$. Value: $\sin 225°=-\sin 45°=-\frac{\sqrt{2}}{2}.$

WHAT — quadrant: $90°<150°<180°$ → QII. WHY: past straight-up $90°$ but before $180°$, so the tip is upper-left. WHAT — reference angle ${\theta }^{\prime }=180°-150°=30°$. WHY: in QII the nearest $x$-axis half is the negative one at $180°$; the gap is $180°-\theta$. WHAT IT LOOKS LIKE: the small red arc at the upper-left ray in s02. WHAT — sign: QII is "Sin" positive → $\cos$ is negative. WHY: in QII the left–right position $x=\cos \theta$ is to the left of centre, so $x<0$. Value: $\cos 150°=-\cos 30°=-\frac{\sqrt{3}}{2}.$

WHAT — quadrant: $180°<240°<270°$ → QIII. WHY: past $180°$, tip bottom-left. WHAT — reference angle ${\theta }^{\prime }=240°-180°=60°$. WHY: QIII gap to the $180°$ line is $\theta -180°$. WHAT — sign: QIII is "Tan" positive → $\tan$ is positive. WHY: $\tan =y/x$, and in QIII both $y<0$ and $x<0$, so their ratio is positive. Value: $\tan 240°=+\tan 60°=\sqrt{3}.$

WHAT: a negative angle means sweep clockwise. WHY: the sign of the angle sets sweep direction. $-120°$ is the same terminal ray as $-120°+360°=240°$ → QIII. Reference angle: $240°-180°=60°$. Sign: QIII → $\cos$ negative. $\cos (-120°)=-\cos 60°=-\frac{1}{2}.$ Cross-check with the even property $\cos (-\theta )=\cos \theta$: $\cos (-120°)=\cos 120°=-\frac{1}{2}$. ✓

WHAT: strip whole turns. WHY: trig repeats every $360°$ ($P$ returns to the same point). $480°-360°=120°$ → QII. Reference angle: $180°-120°=60°$. Sign: QII → $\sin$ positive. $\sin 480°=+\sin 60°=\frac{\sqrt{3}}{2}.$

$200°$ is in QIII (from L1.1). There: $\sin <0$, $\cos <0$, $\tan >0$. Product of signs: $(-)\times (-)\times (+)=(+)$. So the product is positive. Slick check: whenever $\cos \theta \ne 0$ (so that $\tan \theta$ is even defined), $\sin \theta \cos \theta \tan \theta =\sin \theta \cos \theta \cdot \frac{\sin \theta }{\cos \theta }={\sin }^2\theta \ge 0$. So on its whole domain the product is non-negative — consistent with our sign reading. (At $\theta =90°,270°$ the expression is undefined because $\tan$ is, so those points are simply excluded.) ✓

WHAT: use the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$. WHY: it links $\sin$ and $\cos$ so we get magnitude, then ASTC fixes the sign. ${\cos }^2\theta =1-{\left(-\frac{3}{5}\right)}^2=1-\frac{9}{25}=\frac{16}{25}\Rightarrow |\cos \theta |=\frac{4}{5}.$ Sign: QIII → $\cos$ negative, so $\cos \theta =-\frac{4}{5}$. $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{-3/5}{-4/5}=\frac{3}{4}.$ Sign check: QIII → $\tan$ positive. $+\frac{3}{4}>0$. ✓

WHAT: find the reference angle from the magnitude, then place it in the right quadrants. Magnitude: $|\cos \theta |=\frac{1}{2}\Rightarrow {\theta }^{\prime }=60°$ (since $\cos 60°=\frac{1}{2}$). WHY these quadrants: $\cos \theta =x<0$ where $x$ is negative — that's QII and QIII.

• QII: $\theta =180°-60°=120°$.
• QIII: $\theta =180°+60°=240°$. Solutions: $\theta =120°$ and $\theta =240°$. Check: $\cos 120°=-\frac{1}{2}$, $\cos 240°=-\frac{1}{2}$. ✓

WHAT: a $90°$ shift. WHY it's special: the "same ratio" shortcut kept the function name because reference-angle reduction uses only $180°$/$360°$ flips, which never swap $x$ and $y$. But $90°$ is a quarter turn to the $y$-axis, so the geometry swaps the roles of $x$ and $y$ — the function changes from $\sin$ to $\cos$ (a co-function swap). For $\theta =40°$: $90°+40°=130°$, which is in QII, where $\sin$ is positive. Reference angle of $130°$ is $180°-130°=50°$, and $50°=90°-40°$, so $\sin 130°=\sin 50°=\cos 40°$. Numerically $\sin 130°\approx 0.7660$ and $\cos 40°\approx 0.7660$. ✓ If you had blindly applied the shortcut as "same ratio, just fix the sign", you'd have written $\sin 130°=+\sin 50°$ and then wrongly kept it as sine of $40°$. The correct statement swaps the function: $\sin (90°+\theta )=\cos \theta$.

WHAT: compare the terminal points of $\theta$ and $180°-\theta$. WHY the picture: the two rays are mirror images across the $y$-axis (vertical line $x=0$). Let $P=(\cos \theta ,\sin \theta )$ be the tip for angle $\theta$. The angle $180°-\theta$ has terminal point $P^{\prime }$ which is $P$ reflected across the $y$-axis: reflection sends $(x,y)\mapsto (-x,y)$. So $P^{\prime }=(-\cos \theta ,\sin \theta )$, giving: $\cos (180°-\theta )=-\cos \theta ,\sin (180°-\theta )=+\sin \theta .$ The $y$-coordinate (the height) is unchanged by a horizontal flip — that is exactly why the sine stays the same. See the mirror pair in the figure below.

WHAT: convert to a quadrant. WHY radians: $\pi$ radians $=180°$, so $\frac{5\pi }{4}=\frac{5}{4}\times 180°=225°$. $225°$ → QIII (from L2.1). Reference angle $=225°-180°=45°=\frac{\pi }{4}$. Sign: QIII → $\tan$ positive. $\tan \frac{5\pi }{4}=+\tan \frac{\pi }{4}=1.$

WHAT: $90°$ lands the ray straight up the $y$-axis, so $P=(0,1)$. WHY it breaks: $\tan \theta =\frac{y}{x}=\frac{1}{0}$, and dividing by zero is undefined. Geometrically, $\tan$ is the slope of the terminal ray (rise/run). A vertical ray has zero run, so its slope is infinite — there is no finite number. Hence $\tan 90°$ is undefined. Same reasoning at $270°$: $P=(0,-1)$, so $\tan 270°=\frac{-1}{0}$ is also undefined. Limiting behaviour: as $\theta$ approaches $90°$ from below, $x\to 0^{+}$ and $y\to 1$, so $\tan \theta \to +\infty$; from above, $x\to 0^{-}$, so $\tan \theta \to -\infty$. The function leaps across a vertical asymptote — see Graphs of Sin Cos Tan.

WHAT: build a reference right triangle from $\tan {\theta }^{\prime }=\frac{5}{12}$ (magnitude). WHY: $|\tan |=\frac{\text{opp}}{\text{adj}}$, so take opp $=5$, adj $=12$, hypotenuse $=\sqrt{5^2+12^2}=13$. Reference-angle ratios: $\sin {\theta }^{\prime }=\frac{5}{13}$, $\cos {\theta }^{\prime }=\frac{12}{13}$. Signs (QIV): $\cos >0$, $\sin <0$, $\tan <0$ — consistent with the given negative $\tan$. So: $\sin \theta =-\frac{5}{13},\cos \theta =+\frac{12}{13}.$ Check: $\tan \theta =\frac{-5/13}{12/13}=-\frac{5}{12}$. ✓ And ${\sin }^2+{\cos }^2=\frac{25+144}{169}=1$. ✓

Sign of $\cos 200°$? ::: Negative (QIII, $x<0$). $\sin 225°$? ::: $-\frac{\sqrt{2}}{2}$. $\tan 240°$? ::: $+\sqrt{3}$. $\cos (-120°)$? ::: $-\frac{1}{2}$. $\sin 480°$? ::: $\frac{\sqrt{3}}{2}$. Solutions of $\cos \theta =-\frac{1}{2}$ in $[0°,360°)$? ::: $120°$ and $240°$. $\tan \frac{5\pi }{4}$? ::: $1$. Is $\tan 90°$ a number? ::: No — undefined ($1/0$). Across a $90°$ shift, what happens to the function name? ::: It swaps to the co-function ($\sin \leftrightarrow \cos$). Definition of $\cot \theta$? ::: $\cot \theta =1/\tan \theta =x/y$ — the reciprocal slope.