16b. Review of Trigonometry — Practice

Drill sheet for 16. Review of Trigonometry. Book-style examples above each worksheet; regenerate for new numbers. Proofs use guided choice.

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1–2. Degree ↔ radian conversion

Book 1. $54^\circ=3\pi/10$; $120^\circ=2\pi/3$.

Book 2. $2\pi/5\to 72^\circ$; $5\pi/6\to 150^\circ$; $2$ rad $\to(360/\pi)^\circ$.

Degrees → radians

Multiply by π/180; simplify

(1)Convert to radians: (a) $54^{\circ}$ (b) $120^{\circ}$.

$54^{\circ}=54\cdot\pi/180=3\pi/10$.

$120^{\circ}=120\cdot\pi/180=2\pi/3$.

(a) $3\pi/10$; (b) $2\pi/3$.

(2)Convert $15^{\circ}$ to radians (exact multiple of $\pi$ when possible).

Multiply by $\pi/180$: $15\cdot\pi/180=\dfrac{1}{12}\pi$.

$\dfrac{1}{12}\pi$ radians.

(3)Convert $150^{\circ}$ to radians (exact multiple of $\pi$ when possible).

Multiply by $\pi/180$: $150\cdot\pi/180=\dfrac{5}{6}\pi$.

$\dfrac{5}{6}\pi$ radians.

Radians → degrees

Multiply by 180/π

(1)Convert $\dfrac{3\pi}{2}$ radians to degrees.

Multiply by $180/\pi$: $3\cdot 180/2=270^{\circ}$.

$270^{\circ}$.

(2)Convert $\dfrac{5\pi}{6}$ radians to degrees.

Multiply by $180/\pi$: $5\cdot 180/6=150^{\circ}$.

$150^{\circ}$.

(3)Convert $\dfrac{1\pi}{4}$ radians to degrees.

Multiply by $180/\pi$: $1\cdot 180/4=45^{\circ}$.

$45^{\circ}$.

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3. Arc length $s=r\theta$

Book. (a) $r=3$, $\theta=\pi/6$ ⇒ $s=\pi/2$. (b) $r=4$, $s=8$ ⇒ $\theta=2$.

Arc length / central angle

s = rθ with θ in radians

(1)Arc length on radius $6$ with central angle $1\pi$ radians.

$s=r\theta=6\cdot 1\pi=6\pi$.

$s=6\pi$.

(2)Arc length on radius $5$ with central angle $\dfrac{1}{2}\pi$ radians.

$s=r\theta=5\cdot \dfrac{1}{2}\pi=\dfrac{5}{2}\pi$.

$s=\dfrac{5}{2}\pi$.

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4. Equivalent angles in $[0,2\pi)$

Book. $11\pi/4\to 3\pi/4$; $405^\circ\to 45^\circ$; $-\pi/3\to 5\pi/3$; $-5\pi\to\pi$.

Reduce modulo 2π (or 360°)

Add/subtract full turns until in the principal interval

(1)Reduce $-5\pi$ to an equivalent angle in $[0,2\pi)$.

$-5\pi+6\pi=\pi$.

$\pi$.

(2)Reduce $\dfrac{11\pi}{4}$ to an equivalent angle in $[0,2\pi)$.

$11\pi/4=2\pi+3\pi/4$.

$3\pi/4$.

(3)Reduce $405^{\circ}$ to an equivalent angle in $[0^{\circ},360^{\circ})$.

$405=360+45$.

$45^{\circ}$.

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5. Pythagorean from one ratio

Book. Acute $\theta$ with $\cos\theta=4/5$ ⇒ $\sin\theta=3/5$.

sin / cos from Pythagoras

Watch the quadrant when taking ±

(1)Third-quadrant $\theta$ with $\cos\theta=-4/5$. Find $\sin\theta$ and $\cos 2\theta$.

$\sin\theta=-3/5$ (III).

$\cos 2\theta=2\cos^{2}-1=2\cdot 16/25-1=7/25$.

$\sin\theta=-3/5$; $\cos 2\theta=7/25$.

(2)Third-quadrant $\theta$ with $\cos\theta=-4/5$. Find $\sin\theta$ and $\cos 2\theta$.

$\sin\theta=-3/5$ (III).

$\cos 2\theta=2\cos^{2}-1=2\cdot 16/25-1=7/25$.

$\sin\theta=-3/5$; $\cos 2\theta=7/25$.

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6. Identities $\sin(\pi-\theta)$, $\cos(\pi-\theta)$

π − θ identities

Expand with addition formulas

(1)Show $\sin(\pi-\theta)=\sin\theta$ and $\cos(\pi-\theta)=-\cos\theta$.

$\sin(\pi-\theta)=\sin\pi\cos\theta-\cos\pi\sin\theta=\sin\theta$.

$\cos(\pi-\theta)=\cos\pi\cos\theta+\sin\pi\sin\theta=-\cos\theta$.

Follows from addition formulas with $\sin\pi=0$, $\cos\pi=-1$.

(2)Show $\sin(\pi-\theta)=\sin\theta$ and $\cos(\pi-\theta)=-\cos\theta$.

$\sin(\pi-\theta)=\sin\pi\cos\theta-\cos\pi\sin\theta=\sin\theta$.

$\cos(\pi-\theta)=\cos\pi\cos\theta+\sin\pi\sin\theta=-\cos\theta$.

Follows from addition formulas with $\sin\pi=0$, $\cos\pi=-1$.

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7. Evaluate special angles

Book samples. $\sin(2\pi/3)=\sqrt{3}/2$; $\cos(\pi/12)=(\sqrt{6}+\sqrt{2})/4$; $\sin(\pi/8)=\sqrt{2-\sqrt{2}}/2$; …

Book slip on 7(d)

The prompt lists $\sin 420^\circ$ (equals $\sin 60^\circ=\sqrt{3}/2$). Some answer keys print $390^\circ\to\sin 30^\circ=1/2$ — use the angle in the question.

Evaluate sin / cos

Reduce, use π−θ, addition, or half-angle

(1)Evaluate $\sin(\pi/8)$ (positive root).

$\sin^{2}(\pi/8)=(1-\cos(\pi/4))/2=(2-\sqrt{2})/4$.

$\sin(\pi/8)=\sqrt{2-\sqrt{2}}/2$.

$\sqrt{2-\sqrt{2}}/2$.

(2)Evaluate $\sin(\pi/8)$ (positive root).

$\sin^{2}(\pi/8)=(1-\cos(\pi/4))/2=(2-\sqrt{2})/4$.

$\sin(\pi/8)=\sqrt{2-\sqrt{2}}/2$.

$\sqrt{2-\sqrt{2}}/2$.

(3)Evaluate $\cos(\pi/12)$ using $\cos(u-v)$.

$\pi/12=\pi/3-\pi/4$.

$\cos=\tfrac12\cdot\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{2}}{2}=(\sqrt{6}+\sqrt{2})/4$.

$(\sqrt{6}+\sqrt{2})/4$.

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8. Prove polar → rectangular

Fig. 16-12 — unit point F and point A on the same ray

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9–10. Polar ↔ rectangular numbers

Book 9. $(r,\theta)=(3,\pi/6)$ ⇒ $\bigl(3\sqrt{3}/2,\,3/2\bigr)$.

Book 10. $(1,\sqrt{3})$ ⇒ $(r,\theta)=(2,\pi/3)$.

Book slip on 9

$y=r\sin(\pi/6)=3\cdot\tfrac12=3/2$ (not $\sin(\pi/2)$).

Polar ↔ Cartesian

x=r cos θ, y=r sin θ; r=√(x²+y²)

(1)Polar coords of $(1,\sqrt{3})$ (with $r>0$, $\theta\in[0,2\pi)$).

$r=\sqrt{1+3}=2$.

$\cos\theta=1/2$, $\sin\theta=\sqrt{3}/2$ ⇒ $\theta=\pi/3$.

$(r,\theta)=(2,\pi/3)$.

(2)Polar coords of $(1,\sqrt{3})$ (with $r>0$, $\theta\in[0,2\pi)$).

$r=\sqrt{1+3}=2$.

$\cos\theta=1/2$, $\sin\theta=\sqrt{3}/2$ ⇒ $\theta=\pi/3$.

$(r,\theta)=(2,\pi/3)$.

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11. Prove laws of cosines and sines

Fig. 16-13 — area via ½ a c sin B

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12. Prove $\cos(u-v)$

Fig. 16-14 — law of cosines on the unit circle

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Mixed review (core)

Trig review — mixed set

Convert · arc · reduce · evaluate · polar

(1)Acute $\theta$ with $\cos\theta=4/5$. Find $\sin\theta$.

$\sin^{2}=1-16/25=9/25$; acute ⇒ $\sin\theta=3/5$.

$\sin\theta=3/5$.

(2)Rectangular coords of polar $(r,\theta)=(3,\pi/6)$.

$x=3\cos(\pi/6)=3\sqrt{3}/2$.

$y=3\sin(\pi/6)=3/2$.

$\bigl(3\sqrt{3}/2,\,3/2\bigr)$.

(3)Sides $a=3$, $b=5$ include angle $\theta$ with $\cos\theta=1/2$. Find the opposite side $c$.

$c^{2}=a^{2}+b^{2}-2ab\cos\theta=9+25-2\cdot3\cdot5\cdot\tfrac12$.

Simplify and take the positive square root.

$c=\sqrt{19}$.

(4)Sides $a=3$, $b=7$ include angle $\theta$ with $\cos\theta=1/2$. Find the opposite side $c$.

$c^{2}=a^{2}+b^{2}-2ab\cos\theta=9+49-2\cdot3\cdot7\cdot\tfrac12$.

Simplify and take the positive square root.

$c=\sqrt{37}$.

(5)Arc length on radius $5$ with central angle $\dfrac{3}{4}\pi$ radians.

$s=r\theta=5\cdot \dfrac{3}{4}\pi=\dfrac{15}{4}\pi$.

$s=\dfrac{15}{4}\pi$.

(6)Third-quadrant $\theta$ with $\cos\theta=-4/5$. Find $\sin\theta$ and $\cos 2\theta$.

$\sin\theta=-3/5$ (III).

$\cos 2\theta=2\cos^{2}-1=2\cdot 16/25-1=7/25$.

$\sin\theta=-3/5$; $\cos 2\theta=7/25$.

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Bonus · Supplementary problems

13–16. More convert / reduce / arc

Book keys. $4$ rad $\to(720/\pi)^\circ$; $9^\circ\to\pi/20$; $s=\pi/2$ when $r=7$, $\theta=\pi/14$; $17\pi/4\to\pi/4$; etc.

Supp: convert and reduce

Problems 13–16 style

(1)(a) $r=7$, $\theta=\pi/14$: find $s$. (b) $\theta=30^{\circ}$, $s=2$: find $r$.

$s=\pi/2$.

Convert $30^{\circ}=\pi/6$; $r=s/\theta=2/(\pi/6)=12/\pi$.

(a) $\pi/2$; (b) $12/\pi$.

(2)To radians: (a) $9^{\circ}$ (b) $75^{\circ}$ (c) $(90/\pi)^{\circ}$.

$\pi/20$; $5\pi/12$; $1/2$.

(a) $\pi/20$; (b) $5\pi/12$; (c) $1/2$.

(3)To radians: (a) $9^{\circ}$ (b) $75^{\circ}$ (c) $(90/\pi)^{\circ}$.

$\pi/20$; $5\pi/12$; $1/2$.

(a) $\pi/20$; (b) $5\pi/12$; (c) $1/2$.

17–19. Evaluate and double/half angle

Book 18. Acute $\sin\theta=1/4$ ⇒ $\cos=\sqrt{15}/4$, $\sin 2\theta=\sqrt{15}/8$, $\cos 2\theta=7/8$.

Book 19. QIII $\cos\theta=-4/5$ ⇒ $\sin=-3/5$, $\cos 2\theta=7/25$.

Supp: evaluate / double angle

Quadrant signs + double/half formulas

(1)Acute $\theta$ with $\sin\theta=1/4$. Find $\cos\theta$, $\sin 2\theta$, $\cos 2\theta$.

$\cos\theta=\sqrt{15}/4$.

$\sin 2\theta=2\sin\cos=\sqrt{15}/8$; $\cos 2\theta=1-2\sin^{2}=7/8$.

$\cos=\sqrt{15}/4$; $\sin 2\theta=\sqrt{15}/8$; $\cos 2\theta=7/8$.

(2)Third-quadrant $\theta$ with $\cos\theta=-4/5$. Find $\sin\theta$ and $\cos 2\theta$.

$\sin\theta=-3/5$ (III).

$\cos 2\theta=2\cos^{2}-1=2\cdot 16/25-1=7/25$.

$\sin\theta=-3/5$; $\cos 2\theta=7/25$.

(3)Third-quadrant $\theta$ with $\cos\theta=-4/5$. Find $\sin\theta$ and $\cos 2\theta$.

$\sin\theta=-3/5$ (III).

$\cos 2\theta=2\cos^{2}-1=2\cdot 16/25-1=7/25$.

$\sin\theta=-3/5$; $\cos 2\theta=7/25$.

20. Law of cosines application

Book. $AB=5$, $AC=7$, $\cos(\angle ABC)=3/5$ ⇒ $BC=4\sqrt{2}$.

Supp: law of cosines numeric

c² = a² + b² − 2ab cos θ

(1)In $\triangle ABC$, $AB=5$, $AC=7$, $\cos(\angle ABC)=3/5$. Find $BC$.

Law of cosines at $B$: $AC^{2}=AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot\cos B$ — or label carefully.

Book setup: side opposite $A$ etc. With given numbers: $BC=4\sqrt{2}$.

Using $c^{2}=a^{2}+b^{2}-2ab\cos C$ with the matching included angle yields $4\sqrt{2}$.

$BC=4\sqrt{2}$.

(2)Sides $a=4$, $b=6$ include angle $\theta$ with $\cos\theta=1/2$. Find the opposite side $c$.

$c^{2}=a^{2}+b^{2}-2ab\cos\theta=16+36-2\cdot4\cdot6\cdot\tfrac12$.

Simplify and take the positive square root.

$c=\sqrt{28}$.

21. Prove $\dfrac{\sin\theta}{\cos\theta}=\dfrac{1-\cos 2\theta}{\sin 2\theta}$

Supp: prove tan identity

Use 1−cos 2θ = 2 sin²θ and sin 2θ = 2 sin θ cos θ

(1)Prove $\dfrac{\sin\theta}{\cos\theta}=\dfrac{1-\cos 2\theta}{\sin 2\theta}$.

RHS: $(1-(1-2\sin^{2}\theta))/(2\sin\theta\cos\theta)=2\sin^{2}\theta/(2\sin\theta\cos\theta)=\sin\theta/\cos\theta$.

Or use $1-\cos 2\theta=2\sin^{2}\theta$ and $\sin 2\theta=2\sin\theta\cos\theta$.

Both sides equal $\tan\theta$ (where defined).

(2)Prove $\dfrac{\sin\theta}{\cos\theta}=\dfrac{1-\cos 2\theta}{\sin 2\theta}$.

RHS: $(1-(1-2\sin^{2}\theta))/(2\sin\theta\cos\theta)=2\sin^{2}\theta/(2\sin\theta\cos\theta)=\sin\theta/\cos\theta$.

Or use $1-\cos 2\theta=2\sin^{2}\theta$ and $\sin 2\theta=2\sin\theta\cos\theta$.

Both sides equal $\tan\theta$ (where defined).

22. Derive $\pi/4$, $\pi/6$, $\pi/3$ values

Bonus mixed

Supplementary mixed set

Problems 13–22 style

(1)Evaluate $\cos(\pi/12)$ using $\cos(u-v)$.

$\pi/12=\pi/3-\pi/4$.

$\cos=\tfrac12\cdot\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{2}}{2}=(\sqrt{6}+\sqrt{2})/4$.

$(\sqrt{6}+\sqrt{2})/4$.

(2)Convert to radians: (a) $54^{\circ}$ (b) $120^{\circ}$.

$54^{\circ}=54\cdot\pi/180=3\pi/10$.

$120^{\circ}=120\cdot\pi/180=2\pi/3$.

(a) $3\pi/10$; (b) $2\pi/3$.

(3)Convert $36^{\circ}$ to radians (exact multiple of $\pi$ when possible).

Multiply by $\pi/180$: $36\cdot\pi/180=\dfrac{1}{5}\pi$.

$\dfrac{1}{5}\pi$ radians.

(4)Sides $a=4$, $b=6$ include angle $\theta$ with $\cos\theta=1/2$. Find the opposite side $c$.

$c^{2}=a^{2}+b^{2}-2ab\cos\theta=16+36-2\cdot4\cdot6\cdot\tfrac12$.

Simplify and take the positive square root.

$c=\sqrt{28}$.

(5)Polar coords of $(1,\sqrt{3})$ (with $r>0$, $\theta\in[0,2\pi)$).

$r=\sqrt{1+3}=2$.

$\cos\theta=1/2$, $\sin\theta=\sqrt{3}/2$ ⇒ $\theta=\pi/3$.

$(r,\theta)=(2,\pi/3)$.

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Self-check

Quiz

Trig practice review

1 / 4

120° in radians is…

Pick an answer to see if you are right.

Further Reading

• 16. Review of Trigonometry — theory and explorers
• 17. Differentiation of Trig Functions — next: derivatives and graphs
• 15b. Curve Sketching — Practice — prior chapter drills