This les­son is a work­shop. It takes the for­mu­lae of the whole chap­ter (Pythagorean iden­ti­ties, dou­ble angles, com­pound angles and the cosine prod­uct result) and uses them to prove iden­ti­ties and eval­u­ate prod­ucts and sums of awk­ward angles. As you read each exam­ple, notice the first move: it usu­ally decides the whole solu­tion. Typ­i­cal first moves are mul­ti­ply­ing by a con­ju­gate, writ­ing every­thing in sin⁡\sin and cos⁡\cos, replac­ing an angle by its sup­ple­ment, or spot­ting a tele­scop­ing chain.

Exam­ple 1

If x=2sin⁡θ1+cos⁡θ+sin⁡θ\displaystyle x = \frac{2\sin\theta}{1 + \cos\theta + \sin\theta}, find the value of 1−cos⁡θ+sin⁡θ1+sin⁡θ\displaystyle \frac{1 - \cos\theta + \sin\theta}{1 + \sin\theta}.

First move: mul­ti­ply the top and bot­tom of xx by 1−cos⁡θ+sin⁡θ1 - \cos\theta + \sin\theta, so that the bot­tom becomes a dif­fer­ence of squares, (1+sin⁡θ)2−cos⁡2θ(1 + \sin\theta)^2 - \cos^2\theta.

x=2sin⁡θ(1−cos⁡θ+sin⁡θ)(1+sin⁡θ)2−cos⁡2θ=2sin⁡θ(1−cos⁡θ+sin⁡θ)1+2sin⁡θ+sin⁡2θ−cos⁡2θ=2sin⁡θ(1−cos⁡θ+sin⁡θ)2sin⁡2θ+2sin⁡θ=2sin⁡θ(1−cos⁡θ+sin⁡θ)2sin⁡θ(1+sin⁡θ)=1−cos⁡θ+sin⁡θ1+sin⁡θ\displaystyle \begin{aligned} x &= \frac{2\sin\theta(1 - \cos\theta + \sin\theta)}{(1 + \sin\theta)^2 - \cos^2\theta} \\ &= \frac{2\sin\theta(1 - \cos\theta + \sin\theta)}{1 + 2\sin\theta + \sin^2\theta - \cos^2\theta} \\ &= \frac{2\sin\theta(1 - \cos\theta + \sin\theta)}{2\sin^2\theta + 2\sin\theta} \\ &= \frac{2\sin\theta(1 - \cos\theta + \sin\theta)}{2\sin\theta(1 + \sin\theta)} \\ &= \frac{1 - \cos\theta + \sin\theta}{1 + \sin\theta} \end{aligned}

using 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta in the third line. So the required value is xx itself.

Exam­ple 2

Show that (tan⁡θ+cot⁡θ)2=sec⁡2θ+csc⁡2θ=sec⁡2θcsc⁡2θ(\tan\theta + \cot\theta)^2 = \sec^2\theta + \csc^2\theta = \sec^2\theta\csc^2\theta.

First move: write every­thing in sin⁡\sin and cos⁡\cos.

(tan⁡θ+cot⁡θ)2=(sin⁡2θ+cos⁡2θsin⁡θcos⁡θ)2=1sin⁡2θcos⁡2θ(1)\displaystyle (\tan\theta + \cot\theta)^2 = \left(\frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}\right)^2 = \frac{1}{\sin^2\theta\cos^2\theta} \qquad (1)

Writ­ing the 11 on top as sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta and split­ting,

sin⁡2θsin⁡2θcos⁡2θ+cos⁡2θsin⁡2θcos⁡2θ=1cos⁡2θ+1sin⁡2θ=sec⁡2θ+csc⁡2θ\displaystyle \frac{\sin^2\theta}{\sin^2\theta\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \sec^2\theta + \csc^2\theta

And from (1) directly, 1sin⁡2θcos⁡2θ=sec⁡2θcsc⁡2θ\displaystyle \frac{1}{\sin^2\theta\cos^2\theta} = \sec^2\theta\csc^2\theta.

Exam­ple 3

Show that cos⁡4α+2cos⁡2α(1−1sec⁡2α)=1−sin⁡4α\displaystyle \cos^4\alpha + 2\cos^2\alpha\left(1 - \frac{1}{\sec^2\alpha}\right) = 1 - \sin^4\alpha.

LHS=cos⁡4α+2cos⁡2α(1−cos⁡2α)=cos⁡4α+2cos⁡2αsin⁡2α=cos⁡2α[cos⁡2α+2sin⁡2α]=(1−sin⁡2α)(1−sin⁡2α+2sin⁡2α)=(1−sin⁡2α)(1+sin⁡2α)=1−sin⁡4α\begin{aligned} \text{LHS} &= \cos^4\alpha + 2\cos^2\alpha(1 - \cos^2\alpha) = \cos^4\alpha + 2\cos^2\alpha\sin^2\alpha \\ &= \cos^2\alpha\left[\cos^2\alpha + 2\sin^2\alpha\right] \\ &= (1 - \sin^2\alpha)(1 - \sin^2\alpha + 2\sin^2\alpha) \\ &= (1 - \sin^2\alpha)(1 + \sin^2\alpha) = 1 - \sin^4\alpha \end{aligned}

Exam­ple 4

Prove that tan⁡θ+2tan⁡2θ+4tan⁡4θ+8cot⁡8θ=cot⁡θ\tan\theta + 2\tan 2\theta + 4\tan 4\theta + 8\cot 8\theta = \cot\theta.

First move: find a for­mula for tan⁡θ\tan\theta in terms of cotan­gents.

tan⁡θ−cot⁡θ=sin⁡2θ−cos⁡2θsin⁡θcos⁡θ=−cos⁡2θ12sin⁡2θ=−2cot⁡2θ\displaystyle \tan\theta - \cot\theta = \frac{\sin^2\theta - \cos^2\theta}{\sin\theta\cos\theta} = \frac{-\cos 2\theta}{\frac{1}{2}\sin 2\theta} = -2\cot 2\theta

So tan⁡θ=cot⁡θ−2cot⁡2θ\tan\theta = \cot\theta - 2\cot 2\theta, and like­wise with 2θ2\theta and 4θ4\theta. Then

LHS=(cot⁡θ−2cot⁡2θ)+2(cot⁡2θ−2cot⁡4θ)+4(cot⁡4θ−2cot⁡8θ)+8cot⁡8θ=cot⁡θ−2cot⁡2θ+2cot⁡2θ−4cot⁡4θ+4cot⁡4θ−8cot⁡8θ+8cot⁡8θ=cot⁡θ\begin{aligned} \text{LHS} &= (\cot\theta - 2\cot 2\theta) + 2(\cot 2\theta - 2\cot 4\theta) + 4(\cot 4\theta - 2\cot 8\theta) + 8\cot 8\theta \\ &= \cot\theta - 2\cot 2\theta + 2\cot 2\theta - 4\cot 4\theta + 4\cot 4\theta - 8\cot 8\theta + 8\cot 8\theta \\ &= \cot\theta \end{aligned}

Exam­ple 5

If 0<B<A<π4\displaystyle 0 < B < A < \frac{\pi}{4}, sin⁡(A+B)=45\displaystyle \sin(A + B) = \frac{4}{5} and cos⁡(A−B)=1213\displaystyle \cos(A - B) = \frac{12}{13}, find tan⁡2A\tan 2A.

First move: 2A=(A+B)+(A−B)2A = (A + B) + (A - B), so

tan⁡2A=tan⁡(A+B)+tan⁡(A−B)1−tan⁡(A+B)tan⁡(A−B)(1)\displaystyle \tan 2A = \frac{\tan(A + B) + \tan(A - B)}{1 - \tan(A + B)\tan(A - B)} \qquad (1)

Since 0<B<A<π4\displaystyle 0 < B < A < \frac{\pi}{4}, both A+BA + B and A−BA - B are acute. From a 33, 44, 55 tri­an­gle, tan⁡(A+B)=43\displaystyle \tan(A + B) = \frac{4}{3}; from a 55, 1212, 1313 tri­an­gle, tan⁡(A−B)=512\displaystyle \tan(A - B) = \frac{5}{12}. Then from (1),

tan⁡2A=43+5121−43⋅512=21121636=74×94=6316\displaystyle \tan 2A = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3} \cdot \frac{5}{12}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{7}{4} \times \frac{9}{4} = \frac{63}{16}

The quad­rant check mat­ters: if A−BA - B were neg­a­tive, tan⁡(A−B)\tan(A - B) would be −512\displaystyle -\frac{5}{12} and the answer would change.

Exam­ple 6

Prove that cos⁡π11cos⁡2π11cos⁡3π11cos⁡4π11cos⁡5π11=132\displaystyle \cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11} = \frac{1}{32}.

First move: make the angles dou­ble each time. Since cos⁡3π11=−cos⁡8π11\displaystyle \cos\frac{3\pi}{11} = -\cos\frac{8\pi}{11} (as 3π11=π−8π11\displaystyle \frac{3\pi}{11} = \pi - \frac{8\pi}{11}) and cos⁡5π11=−cos⁡16π11\displaystyle \cos\frac{5\pi}{11} = -\cos\frac{16\pi}{11} (as 5π11=16π11−π\displaystyle \frac{5\pi}{11} = \frac{16\pi}{11} - \pi), the two minus signs can­cel:

LHS=cos⁡π11cos⁡2π11cos⁡4π11cos⁡8π11cos⁡16π11=sin⁡32π1132sin⁡π11\displaystyle \text{LHS} = \cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{4\pi}{11}\cos\frac{8\pi}{11}\cos\frac{16\pi}{11} = \frac{\sin\frac{32\pi}{11}}{32\sin\frac{\pi}{11}}

Now 32π11=3π−π11\displaystyle \frac{32\pi}{11} = 3\pi - \frac{\pi}{11}, so sin⁡32π11=sin⁡π11\displaystyle \sin\frac{32\pi}{11} = \sin\frac{\pi}{11}, and the value is 132\displaystyle \frac{1}{32}.

Exam­ple 7

Prove that sin⁡4π8+sin⁡43π8+sin⁡45π8+sin⁡47π8=32\displaystyle \sin^4\frac{\pi}{8} + \sin^4\frac{3\pi}{8} + \sin^4\frac{5\pi}{8} + \sin^4\frac{7\pi}{8} = \frac{3}{2}.

First move: write each angle with π8\displaystyle \frac{\pi}{8}: 3π8=π2−π8\displaystyle \frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}, 5π8=π2+π8\displaystyle \frac{5\pi}{8} = \frac{\pi}{2} + \frac{\pi}{8}, 7π8=π−π8\displaystyle \frac{7\pi}{8} = \pi - \frac{\pi}{8}.

LHS=sin⁡4π8+cos⁡4π8+cos⁡4π8+sin⁡4π8=2(sin⁡4π8+cos⁡4π8)=2(1−2sin⁡2π8cos⁡2π8)=2(1−12sin⁡2π4)=2(1−12⋅12)=2⋅34=32\displaystyle \begin{aligned} \text{LHS} &= \sin^4\frac{\pi}{8} + \cos^4\frac{\pi}{8} + \cos^4\frac{\pi}{8} + \sin^4\frac{\pi}{8} = 2\left(\sin^4\frac{\pi}{8} + \cos^4\frac{\pi}{8}\right) \\ &= 2\left(1 - 2\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}\right) = 2\left(1 - \frac{1}{2}\sin^2\frac{\pi}{4}\right) \\ &= 2\left(1 - \frac{1}{2} \cdot \frac{1}{2}\right) = 2 \cdot \frac{3}{4} = \frac{3}{2} \end{aligned}

using 2sin⁡2θcos⁡2θ=12(2sin⁡θcos⁡θ)2=12sin⁡22θ\displaystyle 2\sin^2\theta\cos^2\theta = \frac{1}{2}(2\sin\theta\cos\theta)^2 = \frac{1}{2}\sin^2 2\theta.

Exam­ple 8

Prove that (1+cos⁡π10)(1+cos⁡3π10)(1+cos⁡7π10)(1+cos⁡9π10)=116\displaystyle \left(1 + \cos\frac{\pi}{10}\right)\left(1 + \cos\frac{3\pi}{10}\right)\left(1 + \cos\frac{7\pi}{10}\right)\left(1 + \cos\frac{9\pi}{10}\right) = \frac{1}{16}.

First move: cos⁡7π10=−cos⁡3π10\displaystyle \cos\frac{7\pi}{10} = -\cos\frac{3\pi}{10} and cos⁡9π10=−cos⁡π10\displaystyle \cos\frac{9\pi}{10} = -\cos\frac{\pi}{10}, which pairs the fac­tors into dif­fer­ences of squares:

LHS=(1−cos⁡2π10)(1−cos⁡23π10)=sin⁡2π10sin⁡23π10=sin⁡218°sin⁡254°=sin⁡218°cos⁡236°=(5−14)2(5+14)2=(5−116)2=116\displaystyle \begin{aligned} \text{LHS} &= \left(1 - \cos^2\frac{\pi}{10}\right)\left(1 - \cos^2\frac{3\pi}{10}\right) = \sin^2\frac{\pi}{10}\sin^2\frac{3\pi}{10} \\ &= \sin^2 18°\sin^2 54° = \sin^2 18°\cos^2 36° \\ &= \left(\frac{\sqrt{5} - 1}{4}\right)^2\left(\frac{\sqrt{5} + 1}{4}\right)^2 = \left(\frac{5 - 1}{16}\right)^2 = \frac{1}{16} \end{aligned}

Exam­ple 9

Prove that cos⁡π9cos⁡2π9cos⁡3π9cos⁡4π9=116\displaystyle \cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{3\pi}{9}\cos\frac{4\pi}{9} = \frac{1}{16}.

In degrees the angles are 20°20°, 40°40°, 60°60°, 80°80°. Using cos⁡θcos⁡(60°−θ)cos⁡(60°+θ)=14cos⁡3θ\displaystyle \cos\theta\cos(60° - \theta)\cos(60° + \theta) = \frac{1}{4}\cos 3\theta with θ=20°\theta = 20°:

cos⁡20°cos⁡40°cos⁡60°cos⁡80°=12⋅14cos⁡60°=18⋅12=116\displaystyle \cos 20°\cos 40°\cos 60°\cos 80° = \frac{1}{2} \cdot \frac{1}{4}\cos 60° = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16}

Prac­tice

  1. Prove that sin⁡2θ1+cos⁡2θ=tan⁡θ\displaystyle \frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta.
  2. Prove that 1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ=tan⁡θ2\displaystyle \frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\frac{\theta}{2}.
  3. Prove that cos⁡4θ−sin⁡4θ=cos⁡2θ\cos^4\theta - \sin^4\theta = \cos 2\theta.
  4. Prove that tan⁡θ+2tan⁡2θ+4cot⁡4θ=cot⁡θ\tan\theta + 2\tan 2\theta + 4\cot 4\theta = \cot\theta.
  5. Find cos⁡π5cos⁡2π5\displaystyle \cos\frac{\pi}{5}\cos\frac{2\pi}{5}.
  6. Find cos⁡π9cos⁡2π9cos⁡4π9\displaystyle \cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}.
  7. Find cos⁡4π8+cos⁡43π8+cos⁡45π8+cos⁡47π8\displaystyle \cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8} + \cos^4\frac{5\pi}{8} + \cos^4\frac{7\pi}{8}.
  8. Find sin⁡π10sin⁡3π10\displaystyle \sin\frac{\pi}{10}\sin\frac{3\pi}{10}.
  9. If 0<B<A<π4\displaystyle 0 < B < A < \frac{\pi}{4}, cos⁡(A+B)=35\displaystyle \cos(A + B) = \frac{3}{5} and sin⁡(A−B)=513\displaystyle \sin(A - B) = \frac{5}{13}, find tan⁡2B\tan 2B.
  10. Find (1+cos⁡π8)(1+cos⁡3π8)(1+cos⁡5π8)(1+cos⁡7π8)\displaystyle \left(1 + \cos\frac{\pi}{8}\right)\left(1 + \cos\frac{3\pi}{8}\right)\left(1 + \cos\frac{5\pi}{8}\right)\left(1 + \cos\frac{7\pi}{8}\right).

Answers

Show answers
  1. 2sin⁡θcos⁡θ2cos⁡2θ=tan⁡θ\displaystyle \frac{2\sin\theta\cos\theta}{2\cos^2\theta} = \tan\theta.
  2. Use 1−cos⁡θ=2sin⁡2θ2\displaystyle 1 - \cos\theta = 2\sin^2\frac{\theta}{2}, 1+cos⁡θ=2cos⁡2θ2\displaystyle 1 + \cos\theta = 2\cos^2\frac{\theta}{2} and sin⁡θ=2sin⁡θ2cos⁡θ2\displaystyle \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}: the frac­tion is 2sin⁡θ2(sin⁡θ2+cos⁡θ2)2cos⁡θ2(cos⁡θ2+sin⁡θ2)=tan⁡θ2\displaystyle \frac{2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right)}{2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\right)} = \tan\frac{\theta}{2}.
  3. (cos⁡2θ−sin⁡2θ)(cos⁡2θ+sin⁡2θ)=cos⁡2θ×1(\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) = \cos 2\theta \times 1.
  4. tan⁡θ=cot⁡θ−2cot⁡2θ\tan\theta = \cot\theta - 2\cot 2\theta and 2tan⁡2θ=2cot⁡2θ−4cot⁡4θ2\tan 2\theta = 2\cot 2\theta - 4\cot 4\theta. Adding these and 4cot⁡4θ4\cot 4\theta leaves cot⁡θ\cot\theta.
  5. sin⁡4π54sin⁡π5=14\displaystyle \frac{\sin\frac{4\pi}{5}}{4\sin\frac{\pi}{5}} = \frac{1}{4}, since sin⁡4π5=sin⁡π5\displaystyle \sin\frac{4\pi}{5} = \sin\frac{\pi}{5}.
  6. sin⁡8π98sin⁡π9=18\displaystyle \frac{\sin\frac{8\pi}{9}}{8\sin\frac{\pi}{9}} = \frac{1}{8}.
  7. As in Exam­ple 7: 2(cos⁡4π8+sin⁡4π8)=32\displaystyle 2\left(\cos^4\frac{\pi}{8} + \sin^4\frac{\pi}{8}\right) = \frac{3}{2}.
  8. sin⁡18°sin⁡54°=sin⁡18°cos⁡36°=14\displaystyle \sin 18°\sin 54° = \sin 18°\cos 36° = \frac{1}{4}.
  9. Both A+BA + B and A−BA - B are acute: tan⁡(A+B)=43\displaystyle \tan(A + B) = \frac{4}{3}, tan⁡(A−B)=512\displaystyle \tan(A - B) = \frac{5}{12}. Since 2B=(A+B)−(A−B)2B = (A + B) - (A - B), tan⁡2B=43−5121+43⋅512=11/1214/9=3356\displaystyle \tan 2B = \frac{\frac{4}{3} - \frac{5}{12}}{1 + \frac{4}{3} \cdot \frac{5}{12}} = \frac{11/12}{14/9} = \frac{33}{56}.
  10. Pair­ing as in Exam­ple 8: sin⁡2π8sin⁡23π8=sin⁡2π8cos⁡2π8=14sin⁡2π4=18\displaystyle \sin^2\frac{\pi}{8}\sin^2\frac{3\pi}{8} = \sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8} = \frac{1}{4}\sin^2\frac{\pi}{4} = \frac{1}{8}.