Radians and the Trigonometric Functions, Worked Solutions

Ques­tion 1: Degrees to radi­ans#

The prob­lem#

Con­vert to radi­ans: (a) $30^\circ$; (b) $135^\circ$; (c) $-210^\circ$; (d) $40^\circ 30'$.

Under­stand­ing the prob­lem#

Each angle is given in degrees (and min­utes in part (d)). You must express it in radi­ans, usu­ally as a frac­tion times $\pi$.

The idea#

Since $180^\circ = \pi$ rad, mul­ti­ply the num­ber of degrees by $\displaystyle \frac{\pi}{180}$ and sim­plify. Min­utes must first be turned into a dec­i­mal or frac­tion of a degree, using $60' = 1^\circ$.

Step-by-step solu­tion#

Step 1. Part (a).

$$\displaystyle 30^\circ = 30 \times \frac{\pi}{180} = \frac{\pi}{6}$$

Step 2. Part (b). Can­cel the com­mon fac­tor $45$.

$$\displaystyle 135^\circ = 135 \times \frac{\pi}{180} = \frac{3\pi}{4}$$

Step 3. Part (c). The minus sign sim­ply car­ries through (clock­wise rota­tion). Can­cel the com­mon fac­tor $30$.

$$\displaystyle -210^\circ = -210 \times \frac{\pi}{180} = -\frac{7\pi}{6}$$

Step 4. Part (d). First con­vert min­utes: $\displaystyle 30' = \frac{30}{60} = \frac{1}{2}$ degree, so $\displaystyle 40^\circ 30' = \frac{81}{2}$ degrees.

$$\displaystyle \frac{81}{2} \times \frac{\pi}{180} = \frac{81\pi}{360} = \frac{9\pi}{40}$$

Check­ing the answer#

Con­vert back: $\displaystyle \frac{9\pi}{40} \times \frac{180}{\pi} = \frac{1620}{40} = 40.5^\circ = 40^\circ 30'$. ✓

Answer#

(a) $\displaystyle \frac{\pi}{6}$; (b) $\displaystyle \frac{3\pi}{4}$; (c) $\displaystyle -\frac{7\pi}{6}$; (d) $\displaystyle \frac{9\pi}{40}$.

Ques­tion 2: Radi­ans to degrees#

The prob­lem#

Con­vert to degrees: (a) $\displaystyle \frac{2\pi}{5}$; (b) $\displaystyle \frac{-7\pi}{6}$; (c) $3$ rad (use $\displaystyle \pi = \frac{22}{7}$).

Under­stand­ing the prob­lem#

Now go the other way. In (a) and (b) $\pi$ appears, so it can­cels exactly. In (c) there is no $\pi$, so you need an approx­i­mate value of $\pi$.

The idea#

Mul­ti­ply by $\displaystyle \frac{180}{\pi}$. In (c) put $\displaystyle \pi = \frac{22}{7}$, and turn the frac­tional part of a degree into min­utes by mul­ti­ply­ing by $60$.

Step-by-step solu­tion#

Step 1. Part (a).

$$\displaystyle \frac{2\pi}{5} \times \frac{180}{\pi} = \frac{360}{5} = 72^\circ$$

Step 2. Part (b).

$$\displaystyle -\frac{7\pi}{6} \times \frac{180}{\pi} = -7 \times 30 = -210^\circ$$

Step 3. Part (c). Use $\displaystyle \frac{180}{\pi} = 180 \times \frac{7}{22}$.

$$\displaystyle 3 \times 180 \times \frac{7}{22} = \frac{3780}{22} = \frac{1890}{11} = 171\frac{9}{11}\ \text{degrees}$$

Step 4. Con­vert the frac­tion $\displaystyle \frac{9}{11}$ of a degree to min­utes.

$$\displaystyle \frac{9}{11} \times 60 = \frac{540}{11} \approx 49.1'$$

So $3$ rad $\approx 171^\circ 49'$.

Check­ing the answer#

$1$ rad $\approx 57^\circ 16'$, so $3$ rad $\approx 171^\circ 48'$; our $171^\circ 49'$ agrees to within round­ing. ✓

Answer#

(a) $72^\circ$; (b) $-210^\circ$; (c) $\displaystyle 171\frac{9}{11}^\circ \approx 171^\circ 49'$.

Ques­tion 3: Find­ing a radius from an arc#

The prob­lem#

An arc of $33$ cm sub­tends $\displaystyle \frac{3\pi}{4}$ at the cen­tre. Radius? (Use $\displaystyle \pi = \frac{22}{7}$.)

Under­stand­ing the prob­lem#

You know the arc length $l = 33$ cm and the angle $\displaystyle \theta = \frac{3\pi}{4}$ radi­ans. You need the radius $r$.

The idea#

Use $l = r\theta$, which needs $\theta$ in radi­ans (it already is), so $\displaystyle r = \frac{l}{\theta}$.

Step-by-step solu­tion#

Step 1. Find $\theta$ as a num­ber with $\displaystyle \pi = \frac{22}{7}$.

$$\displaystyle \theta = \frac{3}{4} \times \frac{22}{7} = \frac{66}{28} = \frac{33}{14}$$

Step 2. Divide.

$$\displaystyle r = \frac{l}{\theta} = 33 \div \frac{33}{14} = 33 \times \frac{14}{33} = 14 \text{ cm}$$

Check­ing the answer#

$\displaystyle l = r\theta = 14 \times \frac{33}{14} = 33$ cm. ✓

Answer#

The radius is $14$ cm.

Ques­tion 4: Turn­ing of a cart wheel#

The prob­lem#

The wheels of a cart turn $180$ times a minute. Through how many radi­ans does a wheel turn in one sec­ond?

Under­stand­ing the prob­lem#

"Turn $180$ times" means $180$ com­plete rev­o­lu­tions. You need the angle turned in one sec­ond, in radi­ans.

The idea#

One full rev­o­lu­tion is $2\pi$ radi­ans. Find rev­o­lu­tions per sec­ond, then mul­ti­ply by $2\pi$.

Step-by-step solu­tion#

Step 1. Rev­o­lu­tions per sec­ond: $\displaystyle \frac{180}{60} = 3$.

Step 2. Radi­ans per sec­ond.

$$3 \times 2\pi = 6\pi \text{ rad}$$

Check­ing the answer#

In a minute: $6\pi \times 60 = 360\pi$ rad $= 180 \times 2\pi$, which is $180$ rev­o­lu­tions. ✓

Answer#

$6\pi$ radi­ans in one sec­ond.

Ques­tion 5: Ratio of radii#

The prob­lem#

Two cir­cles have arcs of equal length sub­tend­ing $72^\circ$ and $54^\circ$. Ratio of radii?

Under­stand­ing the prob­lem#

The arcs have the same length $l$, but they sit on cir­cles of dif­fer­ent radii $r_1$ and $r_2$, sub­tend­ing $72^\circ$ and $54^\circ$. You need $r_1 : r_2$.

The idea#

Con­vert both angles to radi­ans and use $l = r\theta$ for each cir­cle; set the two lengths equal.

Step-by-step solu­tion#

Step 1. Con­vert the angles.

$$\displaystyle 72^\circ = 72 \times \frac{\pi}{180} = \frac{2\pi}{5}, \qquad 54^\circ = 54 \times \frac{\pi}{180} = \frac{3\pi}{10}$$

Step 2. Equal arc lengths.

$$\displaystyle r_1 \cdot \frac{2\pi}{5} = r_2 \cdot \frac{3\pi}{10}$$

Step 3. Solve for the ratio. Mul­ti­ply both sides by $\displaystyle \frac{10}{\pi}$.

$$\displaystyle 4r_1 = 3r_2 \quad\Longrightarrow\quad \frac{r_1}{r_2} = \frac{3}{4}$$

Check­ing the answer#

A big­ger angle for the same arc means a smaller cir­cle, so $r_1 < r_2$ makes sense. ✓

Answer#

$r_1 : r_2 = 3 : 4$.

Ques­tion 6: The other five func­tions#

The prob­lem#

Find the other five trigono­met­ric func­tions: (a) $\displaystyle \cos x = \frac{3}{5}$, $x$ in quad­rant IV; (b) $\displaystyle \csc x = \frac{13}{5}$, $x$ in quad­rant II; (c) $\displaystyle \cot x = \frac{4}{3}$, $x$ in quad­rant III.

Under­stand­ing the prob­lem#

In each part one func­tion and the quad­rant are given. The quad­rant fixes the signs of the oth­ers.

The idea#

Use $\sin^2 x + \cos^2 x = 1$ (or $1 + \cot^2 x = \csc^2 x$) to find the miss­ing value, choose its sign from the quad­rant table, then use the rec­i­p­ro­cal and quo­tient rela­tions, as in Exam­ples 5 and 6.

Step-by-step solu­tion#

Part (a)

Step 1. Find $\sin x$.

$$\displaystyle \sin^2 x = 1 - \frac{9}{25} = \frac{16}{25}, \qquad \sin x = \pm\frac{4}{5}$$

Step 2. In quad­rant IV sine is neg­a­tive, so $\displaystyle \sin x = -\frac{4}{5}$.

Step 3. The rest.

$$\displaystyle \tan x = \frac{-4/5}{3/5} = -\frac{4}{3}, \quad \cot x = -\frac{3}{4}, \quad \sec x = \frac{5}{3}, \quad \csc x = -\frac{5}{4}$$

Part (b)

Step 1. $\displaystyle \sin x = \frac{1}{\csc x} = \frac{5}{13}$.

Step 2. Find $\cos x$.

$$\displaystyle \cos^2 x = 1 - \frac{25}{169} = \frac{144}{169}, \qquad \cos x = \pm \frac{12}{13}$$

Step 3. In quad­rant II cosine is neg­a­tive: $\displaystyle \cos x = -\frac{12}{13}$.

Step 4. The rest.

$$\displaystyle \tan x = -\frac{5}{12}, \quad \cot x = -\frac{12}{5}, \quad \sec x = -\frac{13}{12}$$

Part (c)

Step 1. $\displaystyle \tan x = \frac{1}{\cot x} = \frac{3}{4}$.

Step 2. Find $\csc x$.

$$\displaystyle \csc^2 x = 1 + \cot^2 x = 1 + \frac{16}{9} = \frac{25}{9}, \qquad \csc x = \pm\frac{5}{3}$$

Step 3. In quad­rant III sine (and so cose­cant) is neg­a­tive: $\displaystyle \csc x = -\frac{5}{3}$, $\displaystyle \sin x = -\frac{3}{5}$.

Step 4. Cosine from $\cos x = \cot x \cdot \sin x$.

$$\displaystyle \cos x = \frac{4}{3} \times \left(-\frac{3}{5}\right) = -\frac{4}{5}, \qquad \sec x = -\frac{5}{4}$$

Check­ing the answer#

(a) $\displaystyle \left(\tfrac{3}{5}\right)^2 + \left(-\tfrac{4}{5}\right)^2 = 1$ ✓. (b) $\displaystyle \tfrac{25}{169} + \tfrac{144}{169} = 1$ ✓. (c) $\displaystyle \tfrac{9}{25} + \tfrac{16}{25} = 1$ ✓, and in quad­rant III $\tan$ and $\cot$ are pos­i­tive, as found.

Answer#

(a) $\displaystyle \sin x = -\tfrac{4}{5}$, $\displaystyle \tan x = -\tfrac{4}{3}$, $\displaystyle \cot x = -\tfrac{3}{4}$, $\displaystyle \sec x = \tfrac{5}{3}$, $\displaystyle \csc x = -\tfrac{5}{4}$; (b) $\displaystyle \sin x = \tfrac{5}{13}$, $\displaystyle \cos x = -\tfrac{12}{13}$, $\displaystyle \tan x = -\tfrac{5}{12}$, $\displaystyle \cot x = -\tfrac{12}{5}$, $\displaystyle \sec x = -\tfrac{13}{12}$; (c) $\displaystyle \tan x = \tfrac{3}{4}$, $\displaystyle \sin x = -\tfrac{3}{5}$, $\displaystyle \cos x = -\tfrac{4}{5}$, $\displaystyle \csc x = -\tfrac{5}{3}$, $\displaystyle \sec x = -\tfrac{5}{4}$.

Com­mon mis­take to avoid#

Tak­ing the pos­i­tive square root auto­mat­i­cally. Always decide the sign from the quad­rant.

Ques­tion 7: Val­ues for large and neg­a­tive angles#

The prob­lem#

Eval­u­ate: (a) $\displaystyle \sin \frac{17\pi}{3}$; (b) $\cos 1125^\circ$; (c) $\displaystyle \tan \frac{29\pi}{6}$; (d) $\csc(-750^\circ)$.

Under­stand­ing the prob­lem#

These angles are big­ger than a full turn or neg­a­tive. You must reduce them to a famil­iar angle.

The idea#

Use the peri­ods: $\sin$, $\cos$, $\csc$ repeat every $2\pi$ ($360^\circ$), $\tan$ every $\pi$. Also $\sin(-x) = -\sin x$, so $\csc(-x) = -\csc x$. Remove whole peri­ods, then use a stan­dard value, as in Exam­ple 7.

Step-by-step solu­tion#

Part (a)

Step 1. Write $\displaystyle \frac{17\pi}{3} = 6\pi - \frac{\pi}{3}$ (since $\displaystyle 6\pi = \frac{18\pi}{3}$). $6\pi$ is three full turns.

$$\displaystyle \sin\frac{17\pi}{3} = \sin\left(6\pi - \frac{\pi}{3}\right) = \sin\left(-\frac{\pi}{3}\right)$$

Step 2. Use $\sin(-x) = -\sin x$.

$$\displaystyle = -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}$$

Part (b)

Step 1. $1125 = 3 \times 360 + 45$, so remove three full turns.

$$\displaystyle \cos 1125^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$$

Part (c)

Step 1. Write $\displaystyle \frac{29\pi}{6} = 5\pi - \frac{\pi}{6}$ (since $\displaystyle 5\pi = \frac{30\pi}{6}$). Tan­gent has period $\pi$, so $5\pi$ can be removed.

$$\displaystyle \tan\frac{29\pi}{6} = \tan\left(-\frac{\pi}{6}\right) = -\tan\frac{\pi}{6} = -\frac{1}{\sqrt{3}}$$

Part (d)

Step 1. Pull out the minus sign: $\csc(-750^\circ) = -\csc 750^\circ$.

Step 2. $750 = 2 \times 360 + 30$, so

$$\displaystyle -\csc 750^\circ = -\csc 30^\circ = -\frac{1}{\sin 30^\circ} = -2$$

Check­ing the answer#

(a) $\displaystyle \frac{17\pi}{3}$ ends at the same place as $\displaystyle -\frac{\pi}{3}$, in quad­rant IV, where sine is neg­a­tive ✓. (c) $\displaystyle -\frac{\pi}{6}$ is in quad­rant IV, where tan­gent is neg­a­tive ✓.

Answer#

(a) $\displaystyle -\frac{\sqrt{3}}{2}$; (b) $\displaystyle \frac{1}{\sqrt{2}}$; (c) $\displaystyle -\frac{1}{\sqrt{3}}$; (d) $-2$.

Ques­tion 8: Find­ing the quad­rant#

The prob­lem#

In which quad­rant is $x$ if $\sin x < 0$ and $\tan x > 0$?

Under­stand­ing the prob­lem#

You are told the signs of two func­tions and must find the quad­rant where both signs hold together.

The idea#

Read the sign table from the les­son.

Step-by-step solu­tion#

Step 1. $\sin x < 0$ in quad­rants III and IV.

Step 2. $\tan x > 0$ in quad­rants I and III.

Step 3. The only quad­rant in both lists is III.

Check­ing the answer#

In quad­rant III both $\sin$ and $\cos$ are neg­a­tive, so $\displaystyle \tan = \frac{\sin}{\cos}$ is pos­i­tive. ✓

Answer#

Quad­rant III.

Ques­tion 9: Prov­ing an iden­tity#

The prob­lem#

Prove $\displaystyle \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2\csc x$.

Under­stand­ing the prob­lem#

You must show the left side sim­pli­fies to the right side for every $x$ where both are defined ($\sin x \ne 0$, $\cos x \ne -1$).

The idea#

Add the two frac­tions over a com­mon denom­i­na­tor, expand, and use $\sin^2 x + \cos^2 x = 1$.

Step-by-step solu­tion#

Step 1. Com­mon denom­i­na­tor $\sin x\,(1 + \cos x)$.

$$\displaystyle \text{LHS} = \frac{\sin^2 x + (1 + \cos x)^2}{\sin x\,(1 + \cos x)}$$

Step 2. Expand the square in the numer­a­tor.

$$\sin^2 x + 1 + 2\cos x + \cos^2 x$$

Step 3. Use $\sin^2 x + \cos^2 x = 1$.

$$1 + 1 + 2\cos x = 2 + 2\cos x = 2(1 + \cos x)$$

Step 4. Can­cel the com­mon fac­tor $(1 + \cos x)$.

$$\displaystyle \text{LHS} = \frac{2(1 + \cos x)}{\sin x\,(1 + \cos x)} = \frac{2}{\sin x} = 2\csc x = \text{RHS}$$

Check­ing the answer#

Take $\displaystyle x = \frac{\pi}{2}$: LHS $\displaystyle = \frac{1}{1} + \frac{1}{1} = 2$, RHS $= 2 \times 1 = 2$. ✓

Answer#

LHS $\displaystyle = \frac{2(1 + \cos x)}{\sin x(1 + \cos x)} = 2\csc x$ = RHS, so the iden­tity is proved.

Ques­tion 10: Peri­ods of $\sin 2x$ and $\tan 3x$#

The prob­lem#

What is the least pos­i­tive period of $\sin 2x$? Of $\tan 3x$?

Under­stand­ing the prob­lem#

The period is the small­est pos­i­tive $T$ such that the func­tion repeats: $f(x + T) = f(x)$ for all $x$.

The idea#

$\sin u$ repeats when $u$ increases by $2\pi$; $\tan u$ repeats when $u$ increases by $\pi$. Find how much $x$ must increase for $2x$ (or $3x$) to increase by that amount.

Step-by-step solu­tion#

Step 1. For $\sin 2x$ we need $2(x + T) = 2x + 2\pi$.

$$2T = 2\pi \quad\Longrightarrow\quad T = \pi$$

Step 2. For $\tan 3x$ we need $3(x + T) = 3x + \pi$.

$$\displaystyle 3T = \pi \quad\Longrightarrow\quad T = \frac{\pi}{3}$$

Check­ing the answer#

$\sin 2(x + \pi) = \sin(2x + 2\pi) = \sin 2x$ ✓, and $\displaystyle \tan 3\left(x + \tfrac{\pi}{3}\right) = \tan(3x + \pi) = \tan 3x$ ✓.

Answer#

$\sin 2x$ has period $\pi$; $\tan 3x$ has period $\displaystyle \frac{\pi}{3}$.