Exercises — Reference angles

Locate: $180°<232°<270°$, so the terminal side is in Q3. Reference: Q3 rule is ${\theta }^{\prime }=\theta -180°$ (gap to the negative x-axis at $180°$). ${\theta }^{\prime }=232°-180°=52°.$ Check it is acute: $52°$ is between $0°$ and $90°$. ✔

Convert for intuition: $\frac{3\pi }{4}=135°$ (using Radian and Degree Measure). $90°<135°<180°$ → Q2. Reference: Q2 rule ${\theta }^{\prime }=\pi -\theta$ (gap to negative x-axis at $\pi$). ${\theta }^{\prime }=\pi -\frac{3\pi }{4}=\frac{\pi }{4}.$

These sit on the axes, so the terminal side coincides with an axis direction. The gap to the nearest x-axis direction is either $0°$ (already on the x-axis) or $90°$ (on the y-axis).

• $\theta =90°=\frac{\pi }{2}$: terminal side is the positive y-axis. Nearest x-axis direction is $0°$ or $180°$ — either way the gap is $90°$. So ${\theta }^{\prime }=90°=\frac{\pi }{2}$ (the largest a reference angle can be — the tie case).
• $\theta =180°=\pi$: terminal side lies along the negative x-axis. Gap to the x-axis is $0°$. So ${\theta }^{\prime }=0°$.
• $\theta =270°=\frac{3\pi }{2}$: terminal side is the negative y-axis. Nearest x-axis direction is $180°$ or $360°$ — gap is $90°$ either way. So ${\theta }^{\prime }=90°=\frac{\pi }{2}$ (the other tie case).
• $\theta =360°=2\pi$: terminal side lies along the positive x-axis (same as $0°$). Gap to the x-axis is $0°$. So ${\theta }^{\prime }=0°$. Reading: reference angles live in the closed range $0°\le {\theta }^{\prime }\le 90°$ (i.e. $0\le {\theta }^{\prime }\le \frac{\pi }{2}$); the endpoints happen exactly on the axes — $0°$ on the x-axis, $90°$ on the y-axis.

Locate: $180°<240°<270°$ → Q3. Reference: ${\theta }^{\prime }=240°-180°=60°$. Helper value: $\sin 60°=\frac{\sqrt{3}}{2}$ (from Trigonometric Ratios of Standard Angles). Sign: In Q3, $y<0$, so sine is negative (ASTC: only Tan is + in Q3). $\sin 240°=-\frac{\sqrt{3}}{2}.$

Locate: $270°<315°<360°$ → Q4. Reference: Q4 rule ${\theta }^{\prime }=360°-\theta =360°-315°=45°$. Helper: $\cos 45°=\frac{\sqrt{2}}{2}$. Sign: Q4 has $x>0$ → cosine positive (ASTC: Cos + in Q4). $\cos 315°=+\frac{\sqrt{2}}{2}.$

Convert: $\frac{5\pi }{4}=225°$. $180°<225°<270°$ → Q3. Reference: ${\theta }^{\prime }=\frac{5\pi }{4}-\pi =\frac{\pi }{4}$. Helper: $\tan \frac{\pi }{4}=1$. Sign: Q3 → $x<0$ and $y<0$, so $\tan =y/x$ is a negative-over-negative = positive. $\tan \frac{5\pi }{4}=+1.$

Reduce: negative angle → add $360°$ once: $-150°+360°=210°$. Locate: $180°<210°<270°$ → Q3. Reference: ${\theta }^{\prime }=210°-180°=30°$. Helper: $\cos 30°=\frac{\sqrt{3}}{2}$. Sign: Q3 has $x<0$ → cosine negative. $\cos (-150°)=-\frac{\sqrt{3}}{2}.$

Reduce: subtract $360°$ twice: $760°-720°=40°$. So $760°$ is coterminal with $40°$. Locate: $0°<40°<90°$ → Q1. Reference: in Q1, ${\theta }^{\prime }=\theta =40°$. Sign: Q1 → all positive. $\sin 760°=\sin 40°\approx \mathrm{0.6428.}$ (Not a standard angle, so we leave it as $\sin 40°$ or give a decimal.)

${\theta }_1=130°$: Q2, ${\theta }^{\prime }=180°-130°=50°$. ${\theta }_2=230°$: Q3, ${\theta }^{\prime }=230°-180°=50°$. Both have ${\theta }^{\prime }=50°$, so both have $|\cos |=\cos 50°$. Signs: Q2 has $x<0$ (cosine negative); Q3 also has $x<0$ (cosine negative). So here they agree in sign: $\cos 130°=-\cos 50°,\cos 230°=-\cos 50°.$ The point: equal reference angle ⇒ equal magnitude, but the sign is decided independently by each quadrant. (Had we asked for $\sin$: Q2 gives $+\sin 50°$, Q3 gives $-\sin 50°$ — opposite signs from the same reference.)

Work each term with the machine.

• $\sin 210°$: Q3, ${\theta }^{\prime }=30°$, sine negative → $-\frac{1}{2}$.
• $\cos 300°$: Q4, ${\theta }^{\prime }=360°-300°=60°$, cosine positive → $+\cos 60°=+\frac{1}{2}$.
• $\tan 135°$: Q2, ${\theta }^{\prime }=180°-135°=45°$, tangent negative → $-\tan 45°=-1$. Combine: $E=\left(-\frac{1}{2}\right)+\left(\frac{1}{2}\right)-\left(-1\right)=-\frac{1}{2}+\frac{1}{2}+1=1.$

Magnitude → reference: $|\sin \theta |=\frac{1}{2}$, and $\sin {\theta }^{\prime }=\frac{1}{2}$ gives ${\theta }^{\prime }=30°$ (standard angle). Which quadrants give $\sin <0$? Below the x-axis: Q3 and Q4 (ASTC — sine is negative there). Both lie inside the allowed range $(180°,360°)$.

• Q3: $\theta =180°+{\theta }^{\prime }=180°+30°=210°$.
• Q4: $\theta =360°-{\theta }^{\prime }=360°-30°=330°$. Answer: $\theta \in \{210°,330°\}$, both with reference angle $30°$. Check: $\sin 210°=-\frac{1}{2}$ ✔, $\sin 330°=-\frac{1}{2}$ ✔.

First term $\cos \frac{17\pi }{6}$: reduce by subtracting $2\pi =\frac{12\pi }{6}$: $\frac{17\pi }{6}-\frac{12\pi }{6}=\frac{5\pi }{6}=150°$ → Q2. ${\theta }^{\prime }=\pi -\frac{5\pi }{6}=\frac{\pi }{6}$; helper $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$; Q2 cosine negative ⇒ $-\frac{\sqrt{3}}{2}$. Second term $\tan (-\frac{7\pi }{6})$: add $2\pi =\frac{12\pi }{6}$: $-\frac{7\pi }{6}+\frac{12\pi }{6}=\frac{5\pi }{6}=150°$ → Q2. ${\theta }^{\prime }=\frac{\pi }{6}$; helper $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$; Q2 tangent negative ⇒ $-\frac{1}{\sqrt{3}}$. Combine: $\cos \frac{17\pi }{6}+\tan \left(-\frac{7\pi }{6}\right)=-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{3}}.$ Rationalise the second term. Why bother? A radical sitting in the denominator ($\frac{1}{\sqrt{3}}$) is awkward to add to another fraction and is the conventional "unsimplified" form — we want a rational denominator so both terms share clean denominators. The trick is to multiply by $1$ written cleverly as $\frac{\sqrt{3}}{\sqrt{3}}$, because $\sqrt{3}\cdot \sqrt{3}=3$ clears the root downstairs: $\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{(\sqrt{3}{)}^2}=\frac{\sqrt{3}}{3}.$ Now both terms carry a factor $\sqrt{3}$: $-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{3}=-\sqrt{3}\left(\frac{1}{2}+\frac{1}{3}\right)=-\sqrt{3}\cdot \frac{3+2}{6}=-\sqrt{3}\cdot \frac{5}{6}=-\frac{5\sqrt{3}}{6}.$

Set-up (see the figure below): Let $\theta$ be in Q2, so its terminal side hits the unit circle at point $P=(\cos \theta ,\sin \theta )$ with $\cos \theta <0$. Reflect $P$ across the y-axis to $P^{\prime }=(-\cos \theta ,\sin \theta )$. The angle of $P^{\prime }$ from the positive x-axis is exactly ${\theta }^{\prime }=180°-\theta$, which lands in Q1. Coordinates of $P^{\prime }$: by definition of the unit circle, $P^{\prime }=(\cos {\theta }^{\prime },\sin {\theta }^{\prime })$. Match x-coordinates: the reflection sent $\cos \theta \to -\cos \theta$, and this must equal $\cos {\theta }^{\prime }$: $\cos {\theta }^{\prime }=-\cos \theta ⟹\cos \theta =-\cos {\theta }^{\prime }.$ Match y-coordinates (bonus): reflection fixes the $y$-value, so $\sin {\theta }^{\prime }=\sin \theta$, i.e. $\sin \theta =+\sin {\theta }^{\prime }$ — the Q2 sign for sine. Both signs fell straight out of one reflection. ∎