Trigonometric Functions: Mixed Practice, Worked Solutions

Ques­tion 1: Con­vert­ing between degrees and radi­ans#

The prob­lem#

Con­vert $-37^\circ 30'$ to radi­ans, and $\displaystyle \frac{11\pi}{18}$ radi­ans to degrees.

Under­stand­ing the prob­lem#

$30'$ means $30$ min­utes, and $60' = 1^\circ$. So $-37^\circ 30'$ is a neg­a­tive angle with a frac­tional num­ber of degrees. The sec­ond angle is given in radi­ans.

The idea#

Use $180^\circ = \pi$ radi­ans: degrees $\to$ radi­ans mul­ti­ply by $\displaystyle \tfrac{\pi}{180}$; radi­ans $\to$ degrees mul­ti­ply by $\displaystyle \tfrac{180}{\pi}$.

Step-by-step solu­tion#

Step 1. Change the min­utes to degrees: $\displaystyle 30' = \tfrac{30}{60}{}^\circ = \tfrac12{}^\circ$, so

$$\displaystyle -37^\circ 30' = -37\tfrac{1}{2}{}^\circ = -\tfrac{75}{2}{}^\circ.$$

Step 2. Con­vert to radi­ans.

$$\displaystyle -\frac{75}{2} \times \frac{\pi}{180} = -\frac{75\pi}{360} = -\frac{5\pi}{24}.$$

Step 3. Con­vert $\displaystyle \tfrac{11\pi}{18}$ to degrees.

$$\displaystyle \frac{11\pi}{18} \times \frac{180}{\pi} = 11 \times 10 = 110^\circ.$$

Check­ing the answer#

$\displaystyle \tfrac{5\pi}{24} \approx 0.654$ rad $\approx 37.5^\circ$ ✓. And $110^\circ$ is a lit­tle more than $\displaystyle \tfrac{\pi}{2} = \tfrac{9\pi}{18}$, match­ing $\displaystyle \tfrac{11\pi}{18}$ ✓.

Answer#

$\displaystyle -37^\circ 30' = -\frac{5\pi}{24}$ rad; $\displaystyle \frac{11\pi}{18}$ rad $= 110^\circ$.

Ques­tion 2: Angle turned on a curved track#

The prob­lem#

A rail­way track curves along an arc of a cir­cle of radius $1.5$ km. Through what angle (in degrees) does a train turn in trav­el­ling $500$ m along it? Use $\displaystyle \pi = \tfrac{22}{7}$.

Under­stand­ing the prob­lem#

The train moves along an arc of length $l = 500$ m on a cir­cle of radius $r = 1.5$ km. The angle it turns through equals the angle the arc sub­tends at the cen­tre.

The idea#

Arc length and angle are linked by $l = r\theta$ with $\theta$ in radi­ans. Units must match, so con­vert km to m first. Then change radi­ans to degrees.

Step-by-step solu­tion#

Step 1. Make units match: $r = 1.5$ km $= 1500$ m.

Step 2. Find $\theta$ in radi­ans.

$$\displaystyle \theta = \frac{l}{r} = \frac{500}{1500} = \frac{1}{3}\text{ rad}.$$

Step 3. Con­vert to degrees with $\displaystyle \pi = \tfrac{22}{7}$.

$$\displaystyle \frac13 \times \frac{180}{\pi} = \frac13 \times \frac{180 \times 7}{22} = \frac{1260}{66} = \frac{210}{11} = 19\tfrac{1}{11}{}^\circ.$$

Step 4. Express the frac­tion in min­utes: $\displaystyle \tfrac{1}{11}{}^\circ = \tfrac{60}{11}' \approx 5.45'$, so $\theta \approx 19^\circ 5'$.

Check­ing the answer#

$1$ rad $\approx 57.3^\circ$, and a third of that is about $19.1^\circ$ ✓.

Answer#

The train turns through $\displaystyle \tfrac13$ rad $\displaystyle = 19\tfrac{1}{11}{}^\circ \approx 19^\circ 5'$.

Com­mon mis­take to avoid#

Divid­ing $500$ by $1.5$ with­out con­vert­ing km to m gives a non­sense angle. $l$ and $r$ must be in the same unit.

Ques­tion 3: The other five func­tions from sec x#

The prob­lem#

If $\displaystyle \sec x = -\tfrac{5}{3}$ and $x$ lies in the third quad­rant, find the val­ues of the other five trigono­met­ric func­tions.

Under­stand­ing the prob­lem#

We know one func­tion and the quad­rant. In the third quad­rant, $\sin$ and $\cos$ are neg­a­tive, while $\tan$ and $\cot$ are pos­i­tive.

The idea#

Use $\displaystyle \cos x = \tfrac{1}{\sec x}$, then $\sin^2 x + \cos^2 x = 1$ to find $\sin x$, choos­ing its sign from the quad­rant; the rest fol­low as ratios and rec­i­p­ro­cals.

Step-by-step solu­tion#

Step 1. $\displaystyle \cos x = \frac{1}{\sec x} = -\frac{3}{5}$.

Step 2. Find $\sin x$.

$$\displaystyle \sin^2 x = 1 - \cos^2 x = 1 - \frac{9}{25} = \frac{16}{25} \;\Longrightarrow\; \sin x = \pm\frac45.$$

In the third quad­rant $\sin x < 0$, so $\displaystyle \sin x = -\tfrac45$.

Step 3. $\displaystyle \tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = \frac43$, and $\displaystyle \cot x = \frac34$.

Step 4. $\displaystyle \csc x = \frac{1}{\sin x} = -\frac54$.

Check­ing the answer#

$\sec^2 x = 1 + \tan^2 x$: $\displaystyle \tfrac{25}{9} = 1 + \tfrac{16}{9}$ ✓. Signs: $\tan$ and $\cot$ pos­i­tive, all oth­ers neg­a­tive, cor­rect for quad­rant III.

Answer#

$\displaystyle \cos x = -\tfrac35$, $\displaystyle \sin x = -\tfrac45$, $\displaystyle \tan x = \tfrac43$, $\displaystyle \cot x = \tfrac34$, $\displaystyle \csc x = -\tfrac54$.

Ques­tion 4: Val­ues at large and neg­a­tive angles#

The prob­lem#

Eval­u­ate: (a) $\sin 765^\circ$; (b) $\displaystyle \cot\left(-\frac{15\pi}{4}\right)$; (c) $\displaystyle \cos\left(-\frac{19\pi}{3}\right)$.

Under­stand­ing the prob­lem#

These angles are out­side $0$ to $360^\circ$ or neg­a­tive. We must bring each to a famil­iar angle.

The idea#

Use peri­od­ic­ity (sub­tract whole turns: $360^\circ = 2\pi$ for $\sin$, $\cos$; $\pi$ for $\cot$), and the even/odd rules $\cos(-\theta) = \cos\theta$, $\cot(-\theta) = -\cot\theta$.

Step-by-step solu­tion#

Part (a)

Step 1. Remove whole turns: $765^\circ = 2 \times 360^\circ + 45^\circ$.

Step 2. $\displaystyle \sin 765^\circ = \sin 45^\circ = \frac{1}{\sqrt2}$.

Part (b)

Step 1. $\cot$ is odd: $\displaystyle \cot\left(-\tfrac{15\pi}{4}\right) = -\cot\tfrac{15\pi}{4}$.

Step 2. Write $\displaystyle \tfrac{15\pi}{4} = 4\pi - \tfrac{\pi}{4}$. Remov­ing $4\pi$ (whole peri­ods): $\displaystyle \cot\tfrac{15\pi}{4} = \cot\left(-\tfrac{\pi}{4}\right) = -\cot\tfrac{\pi}{4} = -1$.

Step 3. So $\displaystyle \cot\left(-\tfrac{15\pi}{4}\right) = -(-1) = 1$.

Part (c)

Step 1. $\cos$ is even: $\displaystyle \cos\left(-\tfrac{19\pi}{3}\right) = \cos\tfrac{19\pi}{3}$.

Step 2. $\displaystyle \tfrac{19\pi}{3} = 6\pi + \tfrac{\pi}{3}$, and $6\pi$ is three full turns.

Step 3. $\displaystyle \cos\tfrac{19\pi}{3} = \cos\tfrac{\pi}{3} = \tfrac12$.

Check­ing the answer#

(b) another way: $\displaystyle -\tfrac{15\pi}{4} + 4\pi = \tfrac{\pi}{4}$, and $\displaystyle \cot\tfrac{\pi}{4} = 1$ ✓.

Answer#

(a) $\displaystyle \frac{1}{\sqrt2}$; (b) $1$; (c) $\displaystyle \frac12$.

Ques­tion 5: Eval­u­at­ing an expres­sion in stan­dard angles#

The prob­lem#

Find the value of $\displaystyle \sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4}$.

Under­stand­ing the prob­lem#

All angles are stan­dard: $30^\circ$, $60^\circ$, $45^\circ$. $\sin^2\theta$ means $(\sin\theta)^2$.

The idea#

Sub­sti­tute the known val­ues and square.

Step-by-step solu­tion#

Step 1. $\displaystyle \sin\tfrac{\pi}{6} = \tfrac12$, $\displaystyle \cos\tfrac{\pi}{3} = \tfrac12$, $\displaystyle \tan\tfrac{\pi}{4} = 1$.

Step 2.

$$\displaystyle \left(\tfrac12\right)^2 + \left(\tfrac12\right)^2 - 1^2 = \tfrac14 + \tfrac14 - 1 = -\tfrac12.$$

Check­ing the answer#

$\sin 30^\circ = \cos 60^\circ$, so the first two terms are equal, as they are.

Answer#

$\displaystyle -\frac12$.

Ques­tion 6: Prov­ing a numer­i­cal iden­tity#

The prob­lem#

Prove that $\displaystyle 2\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10$.

Under­stand­ing the prob­lem#

Show that the left side equals $10$ by eval­u­at­ing it.

The idea#

Find each value: $\displaystyle \tfrac{3\pi}{4}$ is in the sec­ond quad­rant, where $\sin$ is pos­i­tive, with ref­er­ence angle $\displaystyle \tfrac{\pi}{4}$.

Step-by-step solu­tion#

Step 1. $\displaystyle \sin\tfrac{3\pi}{4} = \sin\left(\pi - \tfrac{\pi}{4}\right) = \sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt2}$, so $\displaystyle \sin^2\tfrac{3\pi}{4} = \tfrac12$.

Step 2. $\displaystyle \cos^2\tfrac{\pi}{4} = \left(\tfrac{1}{\sqrt2}\right)^2 = \tfrac12$.

Step 3. $\displaystyle \sec\tfrac{\pi}{3} = \tfrac{1}{\cos(\pi/3)} = 2$, so $\displaystyle \sec^2\tfrac{\pi}{3} = 4$.

Step 4. Sub­sti­tute.

$$\displaystyle \text{LHS} = 2\cdot\tfrac12 + 2\cdot\tfrac12 + 2\cdot 4 = 1 + 1 + 8 = 10 = \text{RHS}. \quad\blacksquare$$

Check­ing the answer#

Each term is a sim­ple value; the total $10$ matches the state­ment.

Answer#

LHS $= 1 + 1 + 8 = 10$, as required.

Ques­tion 7: cos(x − y) from given val­ues#

The prob­lem#

Find $\cos(x - y)$, given $\displaystyle \cos x = \tfrac45$ and $\displaystyle \sin y = \tfrac{5}{13}$, where $x$ and $y$ are acute angles.

Under­stand­ing the prob­lem#

The for­mula needs $\cos x$, $\sin x$, $\cos y$ and $\sin y$; two are miss­ing. "Acute" means all four val­ues are pos­i­tive.

The idea#

$\cos(x - y) = \cos x\cos y + \sin x\sin y$. Find the miss­ing $\sin x$ and $\cos y$ from $\sin^2 + \cos^2 = 1$.

Step-by-step solu­tion#

Step 1. $\displaystyle \sin x = \sqrt{1 - \tfrac{16}{25}} = \sqrt{\tfrac{9}{25}} = \tfrac35$ (pos­i­tive, $x$ acute).

Step 2. $\displaystyle \cos y = \sqrt{1 - \tfrac{25}{169}} = \sqrt{\tfrac{144}{169}} = \tfrac{12}{13}$ (pos­i­tive, $y$ acute).

Step 3. Sub­sti­tute.

$$\displaystyle \cos(x - y) = \frac45\cdot\frac{12}{13} + \frac35\cdot\frac{5}{13} = \frac{48}{65} + \frac{15}{65} = \frac{63}{65}.$$

Check­ing the answer#

$\displaystyle \tfrac{63}{65} < 1$, as a cosine must be. Numer­i­cally $x \approx 36.87^\circ$, $y \approx 22.62^\circ$, $\displaystyle \cos 14.25^\circ \approx 0.969 \approx \tfrac{63}{65}$ ✓.

Answer#

$\displaystyle \cos(x - y) = \frac{63}{65}$.

Ques­tion 8: Recog­nis­ing a com­pound-angle for­mula#

The prob­lem#

Prove that $\sin(n + 1)x\,\sin(n + 2)x + \cos(n + 1)x\,\cos(n + 2)x = \cos x$.

Under­stand­ing the prob­lem#

The left side is a sum of a prod­uct of sines and a prod­uct of cosines, with angles $(n + 1)x$ and $(n + 2)x$. We must show it equals $\cos x$.

The idea#

The pat­tern $\cos A\cos B + \sin A\sin B$ is exactly $\cos(A - B)$.

Step-by-step solu­tion#

Step 1. Let $A = (n + 2)x$ and $B = (n + 1)x$. The left side is

$$\cos A\cos B + \sin A\sin B.$$

Step 2. By the com­pound-angle for­mula this equals $\cos(A - B)$.

Step 3. Com­pute the angle: $A - B = (n + 2)x - (n + 1)x = x$.

Step 4. So LHS $= \cos x =$ RHS. $\blacksquare$

Check­ing the answer#

Test $n = 0$, $\displaystyle x = \tfrac{\pi}{6}$: $\displaystyle \sin\tfrac{\pi}{6}\sin\tfrac{\pi}{3} + \cos\tfrac{\pi}{6}\cos\tfrac{\pi}{3} = \tfrac{\sqrt3}{4} + \tfrac{\sqrt3}{4} = \tfrac{\sqrt3}{2} = \cos\tfrac{\pi}{6}$ ✓.

Answer#

LHS $= \cos\big((n + 2)x - (n + 1)x\big) = \cos x$.

Ques­tion 9: Dif­fer­ence of two squares of cosines#

The prob­lem#

Prove that $\cos^2 2x - \cos^2 6x = \sin 4x\,\sin 8x$.

Under­stand­ing the prob­lem#

The left side is a dif­fer­ence of squares; the right side is a prod­uct of sines of $4x$ and $8x$. Notice $4x$ and $8x$ are the half-dif­fer­ence and half-sum of $2x$ and $6x$, a strong hint.

The idea#

Fac­tor with $a^2 - b^2 = (a - b)(a + b)$, then use the sum-to-prod­uct for­mu­las

$$\displaystyle \cos C - \cos D = -2\sin\tfrac{C + D}{2}\sin\tfrac{C - D}{2}, \qquad \cos C + \cos D = 2\cos\tfrac{C + D}{2}\cos\tfrac{C - D}{2},$$

and finally $2\sin\theta\cos\theta = \sin 2\theta$.

Step-by-step solu­tion#

Step 1. Fac­tor the left side.

$$\cos^2 2x - \cos^2 6x = (\cos 2x - \cos 6x)(\cos 2x + \cos 6x).$$

Step 2. First fac­tor, with $C = 2x$, $D = 6x$: $\displaystyle \tfrac{C + D}{2} = 4x$, $\displaystyle \tfrac{C - D}{2} = -2x$.

$$\cos 2x - \cos 6x = -2\sin 4x\sin(-2x) = 2\sin 4x\sin 2x.$$

Step 3. Sec­ond fac­tor:

$$\cos 2x + \cos 6x = 2\cos 4x\cos(-2x) = 2\cos 4x\cos 2x.$$

Step 4. Mul­ti­ply and regroup.

$$\begin{aligned} \text{LHS} &= (2\sin 4x\cos 4x)(2\sin 2x\cos 2x) \\ &= \sin 8x \cdot \sin 4x. \end{aligned}$$

This is the RHS. $\blacksquare$

Check­ing the answer#

At $\displaystyle x = \tfrac{\pi}{24}$: LHS $= \cos^2 15^\circ - \cos^2 45^\circ \approx 0.9330 - 0.5 = 0.4330$; RHS $= \sin 30^\circ\sin 60^\circ = 0.5 \times 0.8660 = 0.4330$ ✓.

Answer#

LHS $= (2\sin 4x\sin 2x)(2\cos 4x\cos 2x) = \sin 8x\sin 4x =$ RHS.

Com­mon mis­take to avoid#

For­get­ting that $\sin(-2x) = -\sin 2x$ in Step 2 leaves a stray minus sign.

Ques­tion 10: A quo­tient that sim­pli­fies to tan 2x#

The prob­lem#

Prove that $\displaystyle \frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x$.

Under­stand­ing the prob­lem#

Numer­a­tor and denom­i­na­tor are both sums of two func­tions of $x$ and $3x$, whose aver­age is $2x$, the angle on the right.

The idea#

Use $\displaystyle \sin C + \sin D = 2\sin\tfrac{C + D}{2}\cos\tfrac{C - D}{2}$ and $\displaystyle \cos C + \cos D = 2\cos\tfrac{C + D}{2}\cos\tfrac{C - D}{2}$, then can­cel the com­mon fac­tor.

Step-by-step solu­tion#

Step 1. Numer­a­tor with $C = 3x$, $D = x$:

$$\sin 3x + \sin x = 2\sin 2x\cos x.$$

Step 2. Denom­i­na­tor:

$$\cos 3x + \cos x = 2\cos 2x\cos x.$$

Step 3. Divide; the fac­tor $2\cos x$ can­cels (where it is non-zero).

$$\displaystyle \text{LHS} = \frac{2\sin 2x\cos x}{2\cos 2x\cos x} = \frac{\sin 2x}{\cos 2x} = \tan 2x = \text{RHS}. \quad\blacksquare$$

Check­ing the answer#

At $\displaystyle x = \tfrac{\pi}{12}$ ($15^\circ$): LHS $\displaystyle = \tfrac{\sin 15^\circ + \sin 45^\circ}{\cos 15^\circ + \cos 45^\circ} \approx \tfrac{0.2588 + 0.7071}{0.9659 + 0.7071} = \tfrac{0.9659}{1.6730} \approx 0.5774 = \tan 30^\circ$ ✓.

Answer#

LHS $\displaystyle = \frac{2\sin 2x\cos x}{2\cos 2x\cos x} = \tan 2x$.

Ques­tion 11: An iden­tity with cot 4x and cot x#

The prob­lem#

Prove that $\cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)$.

Under­stand­ing the prob­lem#

Both sides con­tain the angles $5x$ and $3x$, whose half-sum is $4x$ and half-dif­fer­ence is $x$, the angles in the two cotan­gents.

The idea#

Sim­plify each side sep­a­rately with sum-to-prod­uct for­mu­las and show both reduce to the same expres­sion.

Step-by-step solu­tion#

Step 1. Left side. $\sin 5x + \sin 3x = 2\sin 4x\cos x$, and $\displaystyle \cot 4x = \tfrac{\cos 4x}{\sin 4x}$.

$$\displaystyle \text{LHS} = \frac{\cos 4x}{\sin 4x}\cdot 2\sin 4x\cos x = 2\cos 4x\cos x.$$

Step 2. Right side. $\displaystyle \sin 5x - \sin 3x = 2\cos\tfrac{5x + 3x}{2}\sin\tfrac{5x - 3x}{2} = 2\cos 4x\sin x$, and $\displaystyle \cot x = \tfrac{\cos x}{\sin x}$.

$$\displaystyle \text{RHS} = \frac{\cos x}{\sin x}\cdot 2\cos 4x\sin x = 2\cos 4x\cos x.$$

Step 3. Both sides equal $2\cos 4x\cos x$, so LHS $=$ RHS. $\blacksquare$

Check­ing the answer#

At $\displaystyle x = \tfrac{\pi}{12}$: LHS $= \cot 60^\circ(\sin 75^\circ + \sin 45^\circ) \approx 0.5774 \times 1.6730 = 0.9659$; RHS $= \cot 15^\circ(\sin 75^\circ - \sin 45^\circ) \approx 3.7321 \times 0.2588 = 0.9659$ ✓.

Answer#

Both sides reduce to $2\cos 4x\cos x$.

Ques­tion 12: sin 2x and cos(x/2) from sin x#

The prob­lem#

If $\displaystyle \sin x = \tfrac14$ and $x$ lies in the sec­ond quad­rant, find $\sin 2x$ and $\displaystyle \cos\frac{x}{2}$.

Under­stand­ing the prob­lem#

We need $\cos x$ first, with the sign fixed by quad­rant II (cosine neg­a­tive). For $\displaystyle \cos\tfrac{x}{2}$ we must also decide which quad­rant $\displaystyle \tfrac{x}{2}$ lies in, to choose the sign of the square root.

The idea#

$\sin 2x = 2\sin x\cos x$. For the half angle, $\displaystyle \cos^2\tfrac{x}{2} = \tfrac{1 + \cos x}{2}$.

Step-by-step solu­tion#

Step 1. Find $\cos x$.

$$\displaystyle \cos^2 x = 1 - \frac{1}{16} = \frac{15}{16} \;\Longrightarrow\; \cos x = -\frac{\sqrt{15}}{4} \quad(\text{negative in quadrant II}).$$

Step 2. $\sin 2x$:

$$\displaystyle \sin 2x = 2\cdot\frac14\cdot\left(-\frac{\sqrt{15}}{4}\right) = -\frac{\sqrt{15}}{8}.$$

Step 3. Locate $\displaystyle \tfrac{x}{2}$: since $90^\circ < x < 180^\circ$, we get $\displaystyle 45^\circ < \tfrac{x}{2} < 90^\circ$, first quad­rant, so $\displaystyle \cos\tfrac{x}{2} > 0$.

Step 4. Half-angle for­mula.

$$\displaystyle \cos^2\frac{x}{2} = \frac{1 + \cos x}{2} = \frac{1 - \frac{\sqrt{15}}{4}}{2} = \frac{4 - \sqrt{15}}{8}.$$

$$\displaystyle \cos\frac{x}{2} = \sqrt{\frac{4 - \sqrt{15}}{8}}.$$

Check­ing the answer#

$x = 180^\circ - \sin^{-1}(0.25) \approx 165.52^\circ$. Then $\sin 2x = \sin 331.04^\circ \approx -0.4841$ and $\displaystyle -\tfrac{\sqrt{15}}{8} \approx -0.4841$ ✓; $\cos 82.76^\circ \approx 0.1260$ and $\sqrt{(4 - 3.8730)/8} \approx 0.1260$ ✓.

Answer#

$\displaystyle \sin 2x = -\frac{\sqrt{15}}{8}$; $\displaystyle \cos\frac{x}{2} = \sqrt{\frac{4 - \sqrt{15}}{8}}$.

Com­mon mis­take to avoid#

Using $\displaystyle +\tfrac{\sqrt{15}}{4}$ for $\cos x$. Always fix the sign from the quad­rant.

Ques­tion 13: A prod­uct of tan­gents#

The prob­lem#

Prove that $\tan 3x\tan 2x\tan x = \tan 3x - \tan 2x - \tan x$.

Under­stand­ing the prob­lem#

The angles sat­isfy $3x = 2x + x$. That con­nec­tion is the key.

The idea#

Write $\tan 3x = \tan(2x + x)$, expand with $\displaystyle \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$, clear the frac­tion and rearrange.

Step-by-step solu­tion#

Step 1. Use $3x = 2x + x$.

$$\displaystyle \tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x\tan x}.$$

Step 2. Mul­ti­ply both sides by $1 - \tan 2x\tan x$.

$$\tan 3x - \tan 3x\tan 2x\tan x = \tan 2x + \tan x.$$

Step 3. Move terms so the prod­uct stands alone.

$$\tan 3x - \tan 2x - \tan x = \tan 3x\tan 2x\tan x.$$

This is the required iden­tity. $\blacksquare$

Check­ing the answer#

At $x = 20^\circ$: $\tan 60^\circ \tan 40^\circ \tan 20^\circ \approx 1.7321 \times 0.8391 \times 0.3640 \approx 0.5290$; $1.7321 - 0.8391 - 0.3640 = 0.5290$ ✓.

Answer#

From $\displaystyle \tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x\tan x}$, cross-mul­ti­ply­ing gives $\tan 3x\tan 2x\tan x = \tan 3x - \tan 2x - \tan x$.

Ques­tion 14: cos 6x as a poly­no­mial in cos x#

The prob­lem#

Prove that $\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$.

Under­stand­ing the prob­lem#

We must write $\cos 6x$ entirely in pow­ers of $\cos x$. Let $c = \cos x$ to keep the alge­bra tidy.

The idea#

$6x = 2(3x)$, so $\cos 6x = 2\cos^2 3x - 1$, and the triple-angle for­mula gives $\cos 3x = 4c^3 - 3c$.

Step-by-step solu­tion#

Step 1. Dou­ble-angle for­mula with angle $3x$.

$$\cos 6x = 2\cos^2 3x - 1.$$

Step 2. Sub­sti­tute $\cos 3x = 4c^3 - 3c$.

$$\cos 6x = 2(4c^3 - 3c)^2 - 1.$$

Step 3. Expand the square: $(4c^3 - 3c)^2 = 16c^6 - 24c^4 + 9c^2$.

Step 4. Mul­ti­ply by $2$ and sub­tract $1$.

$$\cos 6x = 32c^6 - 48c^4 + 18c^2 - 1 = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1. \quad\blacksquare$$

Check­ing the answer#

$x = 0$: $\cos 0 = 1$ and $32 - 48 + 18 - 1 = 1$ ✓. $\displaystyle x = \tfrac{\pi}{3}$: $\cos 2\pi = 1$; with $\displaystyle c = \tfrac12$: $\displaystyle \tfrac{32}{64} - \tfrac{48}{16} + \tfrac{18}{4} - 1 = 0.5 - 3 + 4.5 - 1 = 1$ ✓.

Answer#

$\cos 6x = 2(4\cos^3 x - 3\cos x)^2 - 1 = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1$.