The dou­ble angle for­mu­lae lead to a set of famous results: exact val­ues of tan⁡712°\displaystyle \tan 7\frac{1}{2}° and tan⁡3712°\displaystyle \tan 37\frac{1}{2}°, chains of tan­gents that col­lapse to a sin­gle term, prod­ucts of cosines of dou­bling angles, and sums of cosines whose angles are in arith­metic pro­gres­sion. Together they give the beau­ti­ful facts about the angles 2π7\displaystyle \frac{2\pi}{7}, 4π7\displaystyle \frac{4\pi}{7}, 6π7\displaystyle \frac{6\pi}{7}. Each result has a short proof that rests on one key step, and that step is what to remem­ber.

Result 1: tan⁡712°\displaystyle \tan 7\frac{1}{2}° and its rel­a­tives

tan⁡712°=(3−2)(2−1)=6−4−3+2\displaystyle \tan 7\tfrac{1}{2}° = (\sqrt{3} - \sqrt{2})(\sqrt{2} - 1) = \sqrt{6} - \sqrt{4} - \sqrt{3} + \sqrt{2}

cot⁡712°=(3+2)(2+1),tan⁡3712°=(3−2)(2+1),cot⁡3712°=(3+2)(2−1)\displaystyle \cot 7\tfrac{1}{2}° = (\sqrt{3} + \sqrt{2})(\sqrt{2} + 1), \qquad \tan 37\tfrac{1}{2}° = (\sqrt{3} - \sqrt{2})(\sqrt{2} + 1), \qquad \cot 37\tfrac{1}{2}° = (\sqrt{3} + \sqrt{2})(\sqrt{2} - 1)

Proof for tan⁡712°\displaystyle \tan 7\frac{1}{2}°. Use tan⁡A2=2sin⁡2A22sin⁡A2cos⁡A2=1−cos⁡Asin⁡A\displaystyle \tan\frac{A}{2} = \frac{2\sin^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}} = \frac{1 - \cos A}{\sin A} with A=15°A = 15°:

tan⁡712°=1−cos⁡15°sin⁡15°=1−3+1223−122=22−3−13−1\displaystyle \tan 7\tfrac{1}{2}° = \frac{1 - \cos 15°}{\sin 15°} = \frac{1 - \frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} = \frac{2\sqrt{2} - \sqrt{3} - 1}{\sqrt{3} - 1}

Mul­ti­ply top and bot­tom by 3+1\sqrt{3} + 1; the bot­tom becomes 3−1=23 - 1 = 2:

(22−3−1)(3+1)2=26+22−3−3−3−12=26+22−23−42\displaystyle \frac{(2\sqrt{2} - \sqrt{3} - 1)(\sqrt{3} + 1)}{2} = \frac{2\sqrt{6} + 2\sqrt{2} - 3 - \sqrt{3} - \sqrt{3} - 1}{2} = \frac{2\sqrt{6} + 2\sqrt{2} - 2\sqrt{3} - 4}{2}

=6+2−3−2=(3−2)(2−1)= \sqrt{6} + \sqrt{2} - \sqrt{3} - 2 = (\sqrt{3} - \sqrt{2})(\sqrt{2} - 1)

(Mul­ti­ply out (3−2)(2−1)=6−3−2+2(\sqrt{3} - \sqrt{2})(\sqrt{2} - 1) = \sqrt{6} - \sqrt{3} - 2 + \sqrt{2} to check the last step.)

Result 2: cot⁡A\cot A and tan⁡A\tan A

  1. cot⁡A−tan⁡A=2cot⁡2A\cot A - \tan A = 2\cot 2A
  2. cot⁡A+tan⁡A=2csc⁡2A\cot A + \tan A = 2\csc 2A
  3. cot⁡A−tan⁡A−2tan⁡2A−4tan⁡4A=8cot⁡8A\cot A - \tan A - 2\tan 2A - 4\tan 4A = 8\cot 8A

Proof of 1.

cot⁡A−tan⁡A=cos⁡Asin⁡A−sin⁡Acos⁡A=cos⁡2A−sin⁡2Asin⁡Acos⁡A=cos⁡2A12sin⁡2A=2cot⁡2A\displaystyle \cot A - \tan A = \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} = \frac{\cos^2 A - \sin^2 A}{\sin A\cos A} = \frac{\cos 2A}{\frac{1}{2}\sin 2A} = 2\cot 2A

Proof of 3. Rewrite result 1 as tan⁡A=cot⁡A−2cot⁡2A\tan A = \cot A - 2\cot 2A. The same is true with 2A2A or 4A4A in place of AA: tan⁡2A=cot⁡2A−2cot⁡4A\tan 2A = \cot 2A - 2\cot 4A and tan⁡4A=cot⁡4A−2cot⁡8A\tan 4A = \cot 4A - 2\cot 8A. Sub­sti­tute:

cot⁡A−(cot⁡A−2cot⁡2A)−2(cot⁡2A−2cot⁡4A)−4(cot⁡4A−2cot⁡8A)=cot⁡A−cot⁡A+2cot⁡2A−2cot⁡2A+4cot⁡4A−4cot⁡4A+8cot⁡8A=8cot⁡8A\begin{aligned} &\cot A - (\cot A - 2\cot 2A) - 2(\cot 2A - 2\cot 4A) - 4(\cot 4A - 2\cot 8A) \\ &= \cot A - \cot A + 2\cot 2A - 2\cot 2A + 4\cot 4A - 4\cot 4A + 8\cot 8A \\ &= 8\cot 8A \end{aligned}

Every term can­cels with its neigh­bour except the last. A sum that col­lapses like this is called a tele­scop­ing sum.

Result 3: a chain with secants

  1. tan⁡θ2(1+sec⁡θ)=tan⁡θ\displaystyle \tan\frac{\theta}{2}(1 + \sec\theta) = \tan\theta
  2. tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)=tan⁡2θ\displaystyle \tan\frac{\theta}{2}(1 + \sec\theta)(1 + \sec 2\theta) = \tan 2\theta
  3. tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)(1+sec⁡4θ)=tan⁡4θ\displaystyle \tan\frac{\theta}{2}(1 + \sec\theta)(1 + \sec 2\theta)(1 + \sec 4\theta) = \tan 4\theta

Proof of 1.

tan⁡θ2(1+sec⁡θ)=sin⁡θ2cos⁡θ2⋅1+cos⁡θcos⁡θ=sin⁡θ2cos⁡θ2⋅2cos⁡2θ2cos⁡θ=2sin⁡θ2cos⁡θ2cos⁡θ=sin⁡θcos⁡θ=tan⁡θ\displaystyle \tan\frac{\theta}{2}(1 + \sec\theta) = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} \cdot \frac{1 + \cos\theta}{\cos\theta} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} \cdot \frac{2\cos^2\frac{\theta}{2}}{\cos\theta} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta

Results 2 and 3 fol­low by apply­ing result 1 again with 2θ2\theta and 4θ4\theta.

Result 4: prod­ucts of cosines of dou­bling angles

  1. cos⁡θcos⁡2θcos⁡4θ=sin⁡8θ8sin⁡θ\displaystyle \cos\theta\cos 2\theta\cos 4\theta = \frac{\sin 8\theta}{8\sin\theta}
  2. cos⁡θcos⁡2θcos⁡4θcos⁡8θ=sin⁡16θ16sin⁡θ\displaystyle \cos\theta\cos 2\theta\cos 4\theta\cos 8\theta = \frac{\sin 16\theta}{16\sin\theta}
  3. cos⁡θcos⁡2θcos⁡22θ⋯cos⁡2nθ=sin⁡2n+1θ2n+1sin⁡θ\displaystyle \cos\theta\cos 2\theta\cos 2^2\theta\cdots\cos 2^n\theta = \frac{\sin 2^{n+1}\theta}{2^{n+1}\sin\theta}

Proof of 1. Mul­ti­ply and divide by 8sin⁡θ8\sin\theta, then use 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x three times:

cos⁡θcos⁡2θcos⁡4θ=18sin⁡θ[4(2sin⁡θcos⁡θ)cos⁡2θcos⁡4θ]=18sin⁡θ[2(2sin⁡2θcos⁡2θ)cos⁡4θ]=18sin⁡θ(2sin⁡4θcos⁡4θ)=sin⁡8θ8sin⁡θ\displaystyle \begin{aligned} \cos\theta\cos 2\theta\cos 4\theta &= \frac{1}{8\sin\theta}\left[4(2\sin\theta\cos\theta)\cos 2\theta\cos 4\theta\right] \\ &= \frac{1}{8\sin\theta}\left[2(2\sin 2\theta\cos 2\theta)\cos 4\theta\right] \\ &= \frac{1}{8\sin\theta}(2\sin 4\theta\cos 4\theta) \\ &= \frac{\sin 8\theta}{8\sin\theta} \end{aligned}

Exam­ple 1

Prove that cos⁡2π7cos⁡4π7cos⁡6π7=18\displaystyle \cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7} = \frac{1}{8}.

The angles do not dou­ble, so first change cos⁡6π7=cos⁡(π−π7)=−cos⁡π7\displaystyle \cos\frac{6\pi}{7} = \cos\left(\pi - \frac{\pi}{7}\right) = -\cos\frac{\pi}{7}:

LHS=−cos⁡π7cos⁡2π7cos⁡4π7=−sin⁡8π78sin⁡π7\displaystyle \text{LHS} = -\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7} = -\frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}}

Now sin⁡8π7=sin⁡(π+π7)=−sin⁡π7\displaystyle \sin\frac{8\pi}{7} = \sin\left(\pi + \frac{\pi}{7}\right) = -\sin\frac{\pi}{7}, so

LHS=−−sin⁡π78sin⁡π7=18\displaystyle \text{LHS} = -\frac{-\sin\frac{\pi}{7}}{8\sin\frac{\pi}{7}} = \frac{1}{8}

Result 5: sums of cosines and sines in A.P.

If a1,a2,…,ana_1, a_2, \ldots, a_n are in arith­metic pro­gres­sion with com­mon dif­fer­ence dd, then

cos⁡a1+cos⁡a2+⋯+cos⁡an=cos⁡a1+an2sin⁡nd2sin⁡d2\displaystyle \cos a_1 + \cos a_2 + \cdots + \cos a_n = \frac{\cos\frac{a_1 + a_n}{2}\sin\frac{nd}{2}}{\sin\frac{d}{2}}

sin⁡a1+sin⁡a2+⋯+sin⁡an=sin⁡a1+an2sin⁡nd2sin⁡d2\displaystyle \sin a_1 + \sin a_2 + \cdots + \sin a_n = \frac{\sin\frac{a_1 + a_n}{2}\sin\frac{nd}{2}}{\sin\frac{d}{2}}

Key step. Mul­ti­ply and divide the sum by 2sin⁡d2\displaystyle 2\sin\frac{d}{2}. Each term 2cos⁡aksin⁡d2\displaystyle 2\cos a_k\sin\frac{d}{2} becomes a dif­fer­ence of two sines, and the sum tele­scopes.

Exam­ple 2

Find cos⁡2π7+cos⁡4π7+cos⁡6π7\displaystyle \cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}.

The angles are in A.P. with d=2π7\displaystyle d = \frac{2\pi}{7}, so d2=π7\displaystyle \frac{d}{2} = \frac{\pi}{7}. Mul­ti­ply and divide by 2sin⁡π7\displaystyle 2\sin\frac{\pi}{7} and use 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B = \sin(A + B) - \sin(A - B):

12sin⁡π7[2cos⁡2π7sin⁡π7+2cos⁡4π7sin⁡π7+2cos⁡6π7sin⁡π7]=12sin⁡π7[(sin⁡3π7−sin⁡π7)+(sin⁡5π7−sin⁡3π7)+(sin⁡π−sin⁡5π7)]=−sin⁡π72sin⁡π7=−12\displaystyle \begin{aligned} &\frac{1}{2\sin\frac{\pi}{7}}\left[2\cos\frac{2\pi}{7}\sin\frac{\pi}{7} + 2\cos\frac{4\pi}{7}\sin\frac{\pi}{7} + 2\cos\frac{6\pi}{7}\sin\frac{\pi}{7}\right] \\ &= \frac{1}{2\sin\frac{\pi}{7}}\left[\left(\sin\frac{3\pi}{7} - \sin\frac{\pi}{7}\right) + \left(\sin\frac{5\pi}{7} - \sin\frac{3\pi}{7}\right) + \left(\sin\pi - \sin\frac{5\pi}{7}\right)\right] \\ &= \frac{-\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}} = -\frac{1}{2} \end{aligned}

The three cosines of sev­enths

Let a=cos⁡2π7\displaystyle a = \cos\frac{2\pi}{7}, b=cos⁡4π7\displaystyle b = \cos\frac{4\pi}{7}, c=cos⁡6π7\displaystyle c = \cos\frac{6\pi}{7}. Then

  1. a+b+c=−12\displaystyle a + b + c = -\frac{1}{2} (Exam­ple 2)
  2. ab+bc+ca=−12\displaystyle ab + bc + ca = -\frac{1}{2}
  3. abc=18\displaystyle abc = \frac{1}{8} (Exam­ple 1)

Proof of 2. Use 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B = \cos(A + B) + \cos(A - B):

ab+bc+ca=12[cos⁡6π7+cos⁡2π7+cos⁡10π7+cos⁡2π7+cos⁡8π7+cos⁡4π7]\displaystyle ab + bc + ca = \frac{1}{2}\left[\cos\frac{6\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{10\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{8\pi}{7} + \cos\frac{4\pi}{7}\right]

Since cos⁡8π7=cos⁡(2π−6π7)=cos⁡6π7\displaystyle \cos\frac{8\pi}{7} = \cos\left(2\pi - \frac{6\pi}{7}\right) = \cos\frac{6\pi}{7} and cos⁡10π7=cos⁡4π7\displaystyle \cos\frac{10\pi}{7} = \cos\frac{4\pi}{7}, the bracket is 2(a+b+c)2(a + b + c). So ab+bc+ca=a+b+c=−12\displaystyle ab + bc + ca = a + b + c = -\frac{1}{2}.

Exam­ple 3

Prove that sin⁡π7sin⁡2π7sin⁡3π7=78\displaystyle \sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7} = \frac{\sqrt{7}}{8}.

Let kk be the prod­uct. Square it and use sin⁡2A=1−cos⁡2A2\displaystyle \sin^2 A = \frac{1 - \cos 2A}{2}:

k2=1−cos⁡2π72⋅1−cos⁡4π72⋅1−cos⁡6π72=18(1−a)(1−b)(1−c)\displaystyle k^2 = \frac{1 - \cos\frac{2\pi}{7}}{2} \cdot \frac{1 - \cos\frac{4\pi}{7}}{2} \cdot \frac{1 - \cos\frac{6\pi}{7}}{2} = \frac{1}{8}(1 - a)(1 - b)(1 - c)

From (x−a)(x−b)(x−c)=x3−(a+b+c)x2+(ab+bc+ca)x−abc(x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc with x=1x = 1:

k2=18[1+12−12−18]=18⋅78=764\displaystyle k^2 = \frac{1}{8}\left[1 + \frac{1}{2} - \frac{1}{2} - \frac{1}{8}\right] = \frac{1}{8} \cdot \frac{7}{8} = \frac{7}{64}

All three sines are pos­i­tive, so k=78\displaystyle k = \frac{\sqrt{7}}{8}.

Prac­tice

  1. Find cot⁡15°−tan⁡15°\cot 15° - \tan 15°.
  2. Find cot⁡2212°+tan⁡2212°\displaystyle \cot 22\frac{1}{2}° + \tan 22\frac{1}{2}°.
  3. Find tan⁡15°(1+sec⁡30°)\tan 15°(1 + \sec 30°).
  4. Find cos⁡20°cos⁡40°cos⁡80°\cos 20°\cos 40°\cos 80°.
  5. Find cos⁡π7cos⁡2π7cos⁡4π7\displaystyle \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}.
  6. Find cos⁡π15cos⁡2π15cos⁡4π15cos⁡8π15\displaystyle \cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}.
  7. Find cos⁡2π5+cos⁡4π5\displaystyle \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5}.
  8. Use Result 5 to find sin⁡30°+sin⁡60°+sin⁡90°\sin 30° + \sin 60° + \sin 90°, and check directly.
  9. Expand cot⁡712°=(3+2)(2+1)\displaystyle \cot 7\frac{1}{2}° = (\sqrt{3} + \sqrt{2})(\sqrt{2} + 1) as a sum of four surds.
  10. Sim­plify cot⁡10°−tan⁡10°−2tan⁡20°−4tan⁡40°\cot 10° - \tan 10° - 2\tan 20° - 4\tan 40°.

Answers

Show answers
  1. 2cot⁡30°=232\cot 30° = 2\sqrt{3}. (Check: (2+3)−(2−3)=23(2 + \sqrt{3}) - (2 - \sqrt{3}) = 2\sqrt{3}.)
  2. 2csc⁡45°=222\csc 45° = 2\sqrt{2}.
  3. By Result 3 with θ=30°\theta = 30°: tan⁡30°=13\displaystyle \tan 30° = \frac{1}{\sqrt{3}}.
  4. sin⁡160°8sin⁡20°=sin⁡20°8sin⁡20°=18\displaystyle \frac{\sin 160°}{8\sin 20°} = \frac{\sin 20°}{8\sin 20°} = \frac{1}{8}.
  5. sin⁡8π78sin⁡π7=−sin⁡π78sin⁡π7=−18\displaystyle \frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}} = \frac{-\sin\frac{\pi}{7}}{8\sin\frac{\pi}{7}} = -\frac{1}{8}.
  6. sin⁡16π1516sin⁡π15\displaystyle \frac{\sin\frac{16\pi}{15}}{16\sin\frac{\pi}{15}}, and sin⁡16π15=−sin⁡π15\displaystyle \sin\frac{16\pi}{15} = -\sin\frac{\pi}{15}, so the value is −116\displaystyle -\frac{1}{16}.
  7. With n=2n = 2, d=2π5\displaystyle d = \frac{2\pi}{5}: cos⁡3π5sin⁡2π5sin⁡π5=2cos⁡3π5cos⁡π5=−2sin⁡18°cos⁡36°=−2⋅14=−12\displaystyle \frac{\cos\frac{3\pi}{5}\sin\frac{2\pi}{5}}{\sin\frac{\pi}{5}} = 2\cos\frac{3\pi}{5}\cos\frac{\pi}{5} = -2\sin 18°\cos 36° = -2 \cdot \frac{1}{4} = -\frac{1}{2}.
  8. With n=3n = 3, d=30°d = 30°: sin⁡60°sin⁡45°sin⁡15°=33−1=3+32\displaystyle \frac{\sin 60°\sin 45°}{\sin 15°} = \frac{\sqrt{3}}{\sqrt{3} - 1} = \frac{3 + \sqrt{3}}{2}. Directly, 12+32+1=3+32\displaystyle \frac{1}{2} + \frac{\sqrt{3}}{2} + 1 = \frac{3 + \sqrt{3}}{2}.
  9. 6+3+2+2\sqrt{6} + \sqrt{3} + 2 + \sqrt{2}.
  10. By Result 2(3) with A=10°A = 10°: 8cot⁡80°=8tan⁡10°8\cot 80° = 8\tan 10°.