A con­di­tional iden­tity is true not for all angles but only when the angles sat­isfy a given con­di­tion: an equa­tion link­ing them, or a rela­tion such as A+B+C=180°A + B + C = 180°. The art is to use the con­di­tion at exactly the right moment. This les­son works through the stan­dard types: deriv­ing one rela­tion from another, iden­ti­ties in a tri­an­gle, and iden­ti­ties when the angles add to some other fixed value.

Deriv­ing one rela­tion from another

Exam­ple 1

If sec⁡(θ+α)+sec⁡(θ−α)=2sec⁡θ\sec(\theta + \alpha) + \sec(\theta - \alpha) = 2\sec\theta, prove that cos⁡θ=±2cos⁡α2\displaystyle \cos\theta = \pm\sqrt{2}\cos\frac{\alpha}{2}.

Write in cosines and add the frac­tions:

1cos⁡(θ+α)+1cos⁡(θ−α)=2cos⁡θcos⁡(θ−α)+cos⁡(θ+α)cos⁡(θ+α)cos⁡(θ−α)=2cos⁡θ2cos⁡θcos⁡αcos⁡2θ−sin⁡2α=2cos⁡θ\displaystyle \begin{aligned} \frac{1}{\cos(\theta + \alpha)} + \frac{1}{\cos(\theta - \alpha)} &= \frac{2}{\cos\theta} \\ \frac{\cos(\theta - \alpha) + \cos(\theta + \alpha)}{\cos(\theta + \alpha)\cos(\theta - \alpha)} &= \frac{2}{\cos\theta} \\ \frac{2\cos\theta\cos\alpha}{\cos^2\theta - \sin^2\alpha} &= \frac{2}{\cos\theta} \end{aligned}

using cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B. Cross-mul­ti­ply:

cos⁡2θcos⁡α=cos⁡2θ−sin⁡2αsin⁡2α=cos⁡2θ(1−cos⁡α)(2sin⁡α2cos⁡α2)2=cos⁡2θ⋅2sin⁡2α24sin⁡2α2cos⁡2α2=2cos⁡2θsin⁡2α2\displaystyle \begin{aligned} \cos^2\theta\cos\alpha &= \cos^2\theta - \sin^2\alpha \\ \sin^2\alpha &= \cos^2\theta(1 - \cos\alpha) \\ \left(2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\right)^2 &= \cos^2\theta \cdot 2\sin^2\frac{\alpha}{2} \\ 4\sin^2\frac{\alpha}{2}\cos^2\frac{\alpha}{2} &= 2\cos^2\theta\sin^2\frac{\alpha}{2} \end{aligned}

Divide by 2sin⁡2α2\displaystyle 2\sin^2\frac{\alpha}{2} (assumed not zero): cos⁡2θ=2cos⁡2α2\displaystyle \cos^2\theta = 2\cos^2\frac{\alpha}{2}, so cos⁡θ=±2cos⁡α2\displaystyle \cos\theta = \pm\sqrt{2}\cos\frac{\alpha}{2}.

Exam­ple 2

If cos⁡x+cos⁡y=45\displaystyle \cos x + \cos y = \frac{4}{5} and cos⁡x−cos⁡y=27\displaystyle \cos x - \cos y = \frac{2}{7}, prove that 14tan⁡x−y2+5cot⁡x+y2=0\displaystyle 14\tan\frac{x - y}{2} + 5\cot\frac{x + y}{2} = 0.

Divide the two given equa­tions:

cos⁡x+cos⁡ycos⁡x−cos⁡y=4/52/7=45×72=145\displaystyle \frac{\cos x + \cos y}{\cos x - \cos y} = \frac{4/5}{2/7} = \frac{4}{5} \times \frac{7}{2} = \frac{14}{5}

Use the sum-to-prod­uct for­mu­lae on the left:

2cos⁡x+y2cos⁡x−y2−2sin⁡x+y2sin⁡x−y2=145  ⟹  cot⁡x+y2tan⁡x−y2=−145\displaystyle \frac{2\cos\frac{x + y}{2}\cos\frac{x - y}{2}}{-2\sin\frac{x + y}{2}\sin\frac{x - y}{2}} = \frac{14}{5} \implies \frac{\cot\frac{x + y}{2}}{\tan\frac{x - y}{2}} = -\frac{14}{5}

Cross-mul­ti­ply: 5cot⁡x+y2=−14tan⁡x−y2\displaystyle 5\cot\frac{x + y}{2} = -14\tan\frac{x - y}{2}, which rearranges to the result.

Exam­ple 3

Prove that sin⁡2(α−45°)+sin⁡2(α+15°)−sin⁡2(α−15°)=12\displaystyle \sin^2(\alpha - 45°) + \sin^2(\alpha + 15°) - \sin^2(\alpha - 15°) = \frac{1}{2}.

Use sin⁡2A=1−cos⁡2A2\displaystyle \sin^2 A = \frac{1 - \cos 2A}{2} on the first term and sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B) on the other two:

LHS=1−cos⁡(2α−90°)2+sin⁡2αsin⁡30°=12−12cos⁡(90°−2α)+12sin⁡2α=12−12sin⁡2α+12sin⁡2α=12\displaystyle \begin{aligned} \text{LHS} &= \frac{1 - \cos(2\alpha - 90°)}{2} + \sin 2\alpha\sin 30° \\ &= \frac{1}{2} - \frac{1}{2}\cos(90° - 2\alpha) + \frac{1}{2}\sin 2\alpha \\ &= \frac{1}{2} - \frac{1}{2}\sin 2\alpha + \frac{1}{2}\sin 2\alpha = \frac{1}{2} \end{aligned}

(This one is true for every α\alpha: it is an iden­tity with no con­di­tion.)

Iden­ti­ties in a tri­an­gle

Through­out, A+B+C=180°A + B + C = 180°, so

cos⁡A+B2=cos⁡(90°−C2)=sin⁡C2\displaystyle \cos\frac{A + B}{2} = \cos\left(90° - \frac{C}{2}\right) = \sin\frac{C}{2}

Exam­ple 4

In a tri­an­gle ABCABC, prove that cos⁡A+cos⁡B−cos⁡C=−1+4cos⁡A2cos⁡B2sin⁡C2\displaystyle \cos A + \cos B - \cos C = -1 + 4\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}.

LHS=2cos⁡A+B2cos⁡A−B2−cos⁡C=2sin⁡C2cos⁡A−B2−(1−2sin⁡2C2)=−1+2sin⁡C2(cos⁡A−B2+sin⁡C2)=−1+2sin⁡C2(cos⁡A−B2+cos⁡A+B2)=−1+2sin⁡C2(2cos⁡A2cos⁡B2)=−1+4cos⁡A2cos⁡B2sin⁡C2\displaystyle \begin{aligned} \text{LHS} &= 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} - \cos C \\ &= 2\sin\frac{C}{2}\cos\frac{A - B}{2} - \left(1 - 2\sin^2\frac{C}{2}\right) \\ &= -1 + 2\sin\frac{C}{2}\left(\cos\frac{A - B}{2} + \sin\frac{C}{2}\right) \\ &= -1 + 2\sin\frac{C}{2}\left(\cos\frac{A - B}{2} + \cos\frac{A + B}{2}\right) \\ &= -1 + 2\sin\frac{C}{2}\left(2\cos\frac{A}{2}\cos\frac{B}{2}\right) \\ &= -1 + 4\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2} \end{aligned}

Exam­ple 5

In a tri­an­gle ABCABC, prove that cos⁡2A2+cos⁡2B2−cos⁡2C2=2cos⁡A2cos⁡B2sin⁡C2\displaystyle \cos^2\frac{A}{2} + \cos^2\frac{B}{2} - \cos^2\frac{C}{2} = 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}.

Write cos⁡2B2=1−sin⁡2B2\displaystyle \cos^2\frac{B}{2} = 1 - \sin^2\frac{B}{2}, then use cos⁡2P−sin⁡2Q=cos⁡(P+Q)cos⁡(P−Q)\cos^2 P - \sin^2 Q = \cos(P + Q)\cos(P - Q):

LHS=1+(cos⁡2A2−sin⁡2B2)−cos⁡2C2=1+cos⁡A+B2cos⁡A−B2−(1−sin⁡2C2)=sin⁡C2cos⁡A−B2+sin⁡2C2=sin⁡C2[cos⁡A−B2+cos⁡A+B2]=2cos⁡A2cos⁡B2sin⁡C2\displaystyle \begin{aligned} \text{LHS} &= 1 + \left(\cos^2\frac{A}{2} - \sin^2\frac{B}{2}\right) - \cos^2\frac{C}{2} \\ &= 1 + \cos\frac{A + B}{2}\cos\frac{A - B}{2} - \left(1 - \sin^2\frac{C}{2}\right) \\ &= \sin\frac{C}{2}\cos\frac{A - B}{2} + \sin^2\frac{C}{2} \\ &= \sin\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\frac{A + B}{2}\right] \\ &= 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2} \end{aligned}

Exam­ple 6

In a tri­an­gle ABCABC, prove that sin⁡A2+sin⁡B2+sin⁡C2=1+4sin⁡π−A4sin⁡π−B4sin⁡π−C4\displaystyle \sin\frac{A}{2} + \sin\frac{B}{2} + \sin\frac{C}{2} = 1 + 4\sin\frac{\pi - A}{4}\sin\frac{\pi - B}{4}\sin\frac{\pi - C}{4}.

Start from the right side. Note that π−C4=A+B4\displaystyle \frac{\pi - C}{4} = \frac{A + B}{4}, and π−A4+π−B4=π2−A+B4\displaystyle \frac{\pi - A}{4} + \frac{\pi - B}{4} = \frac{\pi}{2} - \frac{A + B}{4}.

RHS=1+2sin⁡A+B4[2sin⁡π−A4sin⁡π−B4]=1+2sin⁡A+B4[cos⁡B−A4−cos⁡(π2−A+B4)]=1+2sin⁡A+B4cos⁡A−B4−2sin⁡2A+B4\displaystyle \begin{aligned} \text{RHS} &= 1 + 2\sin\frac{A + B}{4}\left[2\sin\frac{\pi - A}{4}\sin\frac{\pi - B}{4}\right] \\ &= 1 + 2\sin\frac{A + B}{4}\left[\cos\frac{B - A}{4} - \cos\left(\frac{\pi}{2} - \frac{A + B}{4}\right)\right] \\ &= 1 + 2\sin\frac{A + B}{4}\cos\frac{A - B}{4} - 2\sin^2\frac{A + B}{4} \end{aligned}

Now 1−2sin⁡2A+B4=cos⁡A+B2\displaystyle 1 - 2\sin^2\frac{A + B}{4} = \cos\frac{A + B}{2}, and 2sin⁡Pcos⁡Q=sin⁡(P+Q)+sin⁡(P−Q)2\sin P\cos Q = \sin(P + Q) + \sin(P - Q) with P=A+B4\displaystyle P = \frac{A + B}{4}, Q=A−B4\displaystyle Q = \frac{A - B}{4} gives sin⁡A2+sin⁡B2\displaystyle \sin\frac{A}{2} + \sin\frac{B}{2}. So

RHS=cos⁡A+B2+sin⁡A2+sin⁡B2=sin⁡C2+sin⁡A2+sin⁡B2=LHS\displaystyle \text{RHS} = \cos\frac{A + B}{2} + \sin\frac{A}{2} + \sin\frac{B}{2} = \sin\frac{C}{2} + \sin\frac{A}{2} + \sin\frac{B}{2} = \text{LHS}

Other con­di­tions on the sum

Exam­ple 7

If A+B+C=2SA + B + C = 2S, prove that cos⁡(S−A)+cos⁡(S−B)+cos⁡(S−C)+cos⁡S=4cos⁡A2cos⁡B2cos⁡C2\displaystyle \cos(S - A) + \cos(S - B) + \cos(S - C) + \cos S = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.

Pair the first two terms and the last two:

LHS=2cos⁡2S−A−B2cos⁡B−A2+2cos⁡2S−C2cos⁡−C2=2cos⁡C2cos⁡A−B2+2cos⁡A+B2cos⁡C2=2cos⁡C2[cos⁡A−B2+cos⁡A+B2]=4cos⁡A2cos⁡B2cos⁡C2\displaystyle \begin{aligned} \text{LHS} &= 2\cos\frac{2S - A - B}{2}\cos\frac{B - A}{2} + 2\cos\frac{2S - C}{2}\cos\frac{-C}{2} \\ &= 2\cos\frac{C}{2}\cos\frac{A - B}{2} + 2\cos\frac{A + B}{2}\cos\frac{C}{2} \\ &= 2\cos\frac{C}{2}\left[\cos\frac{A - B}{2} + \cos\frac{A + B}{2}\right] \\ &= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} \end{aligned}

using 2S−A−B=C2S - A - B = C and 2S−C=A+B2S - C = A + B.

Exam­ple 8

If A+B+C=270°A + B + C = 270°, prove that cos⁡2A+cos⁡2B−cos⁡2C=−2cos⁡Acos⁡Bsin⁡C\cos^2 A + \cos^2 B - \cos^2 C = -2\cos A\cos B\sin C.

Here cos⁡(A+B)=cos⁡(270°−C)=−sin⁡C\cos(A + B) = \cos(270° - C) = -\sin C.

LHS=1+(cos⁡2A−sin⁡2B)−cos⁡2C=1+cos⁡(A+B)cos⁡(A−B)−(1−sin⁡2C)=−sin⁡Ccos⁡(A−B)+sin⁡2C=−sin⁡C[cos⁡(A−B)−sin⁡C]=−sin⁡C[cos⁡(A−B)+cos⁡(A+B)]=−2cos⁡Acos⁡Bsin⁡C\begin{aligned} \text{LHS} &= 1 + (\cos^2 A - \sin^2 B) - \cos^2 C = 1 + \cos(A + B)\cos(A - B) - (1 - \sin^2 C) \\ &= -\sin C\cos(A - B) + \sin^2 C = -\sin C\left[\cos(A - B) - \sin C\right] \\ &= -\sin C\left[\cos(A - B) + \cos(A + B)\right] = -2\cos A\cos B\sin C \end{aligned}

Exam­ple 9

If sin⁡(α+β)cos⁡(α−β)=1−m1+m\displaystyle \frac{\sin(\alpha + \beta)}{\cos(\alpha - \beta)} = \frac{1 - m}{1 + m}, prove that tan⁡(π4−α)=mtan⁡(π4+β)\displaystyle \tan\left(\frac{\pi}{4} - \alpha\right) = m\tan\left(\frac{\pi}{4} + \beta\right).

Apply com­po­nendo and div­i­dendo, and write cos⁡(α−β)=sin⁡(π2−α+β)\displaystyle \cos(\alpha - \beta) = \sin\left(\frac{\pi}{2} - \alpha + \beta\right):

sin⁡(α+β)+sin⁡(π2−α+β)sin⁡(α+β)−sin⁡(π2−α+β)=(1−m)+(1+m)(1−m)−(1+m)=2−2m=−1m\displaystyle \frac{\sin(\alpha + \beta) + \sin\left(\frac{\pi}{2} - \alpha + \beta\right)}{\sin(\alpha + \beta) - \sin\left(\frac{\pi}{2} - \alpha + \beta\right)} = \frac{(1 - m) + (1 + m)}{(1 - m) - (1 + m)} = \frac{2}{-2m} = -\frac{1}{m}

By the sum-to-prod­uct for­mu­lae the left side is

2sin⁡(π4+β)cos⁡(α−π4)2cos⁡(π4+β)sin⁡(α−π4)=tan⁡(π4+β)tan⁡(α−π4)=tan⁡(π4+β)−tan⁡(π4−α)\displaystyle \frac{2\sin\left(\frac{\pi}{4} + \beta\right)\cos\left(\alpha - \frac{\pi}{4}\right)}{2\cos\left(\frac{\pi}{4} + \beta\right)\sin\left(\alpha - \frac{\pi}{4}\right)} = \frac{\tan\left(\frac{\pi}{4} + \beta\right)}{\tan\left(\alpha - \frac{\pi}{4}\right)} = \frac{\tan\left(\frac{\pi}{4} + \beta\right)}{-\tan\left(\frac{\pi}{4} - \alpha\right)}

So tan⁡(π4+β)tan⁡(π4−α)=1m\displaystyle \frac{\tan\left(\frac{\pi}{4} + \beta\right)}{\tan\left(\frac{\pi}{4} - \alpha\right)} = \frac{1}{m}, that is, tan⁡(π4−α)=mtan⁡(π4+β)\displaystyle \tan\left(\frac{\pi}{4} - \alpha\right) = m\tan\left(\frac{\pi}{4} + \beta\right).

Prac­tice

  1. In a tri­an­gle ABCABC, prove that sin⁡A+sin⁡B−sin⁡C=4sin⁡A2sin⁡B2cos⁡C2\displaystyle \sin A + \sin B - \sin C = 4\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}.
  2. In a tri­an­gle ABCABC, prove that cos⁡2A+cos⁡2B+cos⁡2C=−1−4cos⁡Acos⁡Bcos⁡C\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A\cos B\cos C.
  3. In a tri­an­gle ABCABC, prove that sin⁡2A+sin⁡2B−sin⁡2C=2sin⁡Asin⁡Bcos⁡C\sin^2 A + \sin^2 B - \sin^2 C = 2\sin A\sin B\cos C.
  4. If cos⁡x+cos⁡y=12\displaystyle \cos x + \cos y = \frac{1}{2} and sin⁡x+sin⁡y=13\displaystyle \sin x + \sin y = \frac{1}{3}, find tan⁡x+y2\displaystyle \tan\frac{x + y}{2}.
  5. If cos⁡x+cos⁡y=35\displaystyle \cos x + \cos y = \frac{3}{5} and cos⁡x−cos⁡y=15\displaystyle \cos x - \cos y = \frac{1}{5}, find tan⁡x+y2tan⁡x−y2\displaystyle \tan\frac{x + y}{2}\tan\frac{x - y}{2}.
  6. Check the iden­tity of Exam­ple 8 for A=100°A = 100°, B=80°B = 80°, C=90°C = 90°.
  7. In a tri­an­gle ABCABC, prove that cos⁡2A+cos⁡2B+cos⁡2C=1−2cos⁡Acos⁡Bcos⁡C\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A\cos B\cos C.
  8. If sec⁡(θ+α)+sec⁡(θ−α)=2sec⁡θ\sec(\theta + \alpha) + \sec(\theta - \alpha) = 2\sec\theta and α=120°\alpha = 120°, find cos⁡θ\cos\theta.
  9. If sin⁡(α+β)cos⁡(α−β)=13\displaystyle \frac{\sin(\alpha + \beta)}{\cos(\alpha - \beta)} = \frac{1}{3}, find the value of mm in Exam­ple 9.

Answers

Show answers
  1. sin⁡A+sin⁡B=2cos⁡C2cos⁡A−B2\displaystyle \sin A + \sin B = 2\cos\frac{C}{2}\cos\frac{A - B}{2} and sin⁡C=2sin⁡C2cos⁡C2=2cos⁡C2cos⁡A+B2\displaystyle \sin C = 2\sin\frac{C}{2}\cos\frac{C}{2} = 2\cos\frac{C}{2}\cos\frac{A + B}{2}. The dif­fer­ence is 2cos⁡C2[cos⁡A−B2−cos⁡A+B2]=4sin⁡A2sin⁡B2cos⁡C2\displaystyle 2\cos\frac{C}{2}\left[\cos\frac{A - B}{2} - \cos\frac{A + B}{2}\right] = 4\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}.
  2. cos⁡2A+cos⁡2B=2cos⁡(A+B)cos⁡(A−B)=−2cos⁡Ccos⁡(A−B)\cos 2A + \cos 2B = 2\cos(A + B)\cos(A - B) = -2\cos C\cos(A - B) and cos⁡2C=2cos⁡2C−1\cos 2C = 2\cos^2 C - 1. Together: −1−2cos⁡C[cos⁡(A−B)+cos⁡(A+B)]=−1−4cos⁡Acos⁡Bcos⁡C-1 - 2\cos C\left[\cos(A - B) + \cos(A + B)\right] = -1 - 4\cos A\cos B\cos C.
  3. sin⁡2A−sin⁡2C=sin⁡(A+C)sin⁡(A−C)=sin⁡Bsin⁡(A−C)\sin^2 A - \sin^2 C = \sin(A + C)\sin(A - C) = \sin B\sin(A - C), and sin⁡2B=sin⁡Bsin⁡(A+C)\sin^2 B = \sin B\sin(A + C). Adding: sin⁡B[sin⁡(A−C)+sin⁡(A+C)]=2sin⁡Asin⁡Bcos⁡C\sin B\left[\sin(A - C) + \sin(A + C)\right] = 2\sin A\sin B\cos C.
  4. sin⁡x+sin⁡ycos⁡x+cos⁡y=2sin⁡x+y2cos⁡x−y22cos⁡x+y2cos⁡x−y2=tan⁡x+y2\displaystyle \frac{\sin x + \sin y}{\cos x + \cos y} = \frac{2\sin\frac{x + y}{2}\cos\frac{x - y}{2}}{2\cos\frac{x + y}{2}\cos\frac{x - y}{2}} = \tan\frac{x + y}{2}, so tan⁡x+y2=1/31/2=23\displaystyle \tan\frac{x + y}{2} = \frac{1/3}{1/2} = \frac{2}{3}.
  5. cos⁡x−cos⁡ycos⁡x+cos⁡y=−tan⁡x+y2tan⁡x−y2=13\displaystyle \frac{\cos x - \cos y}{\cos x + \cos y} = -\tan\frac{x + y}{2}\tan\frac{x - y}{2} = \frac{1}{3}, so the prod­uct is −13\displaystyle -\frac{1}{3}.
  6. A+B+C=270°A + B + C = 270°. Left: cos⁡2100°+cos⁡280°−0=2cos⁡280°\cos^2 100° + \cos^2 80° - 0 = 2\cos^2 80°. Right: −2cos⁡100°cos⁡80°⋅1=2cos⁡280°-2\cos 100°\cos 80° \cdot 1 = 2\cos^2 80°, since cos⁡100°=−cos⁡80°\cos 100° = -\cos 80°. They agree.
  7. cos⁡2A+cos⁡2B=1+cos⁡(A+B)cos⁡(A−B)=1−cos⁡Ccos⁡(A−B)\cos^2 A + \cos^2 B = 1 + \cos(A + B)\cos(A - B) = 1 - \cos C\cos(A - B). Adding cos⁡2C=−cos⁡Ccos⁡(A+B)\cos^2 C = -\cos C\cos(A + B) gives 1−cos⁡C[cos⁡(A−B)+cos⁡(A+B)]=1−2cos⁡Acos⁡Bcos⁡C1 - \cos C\left[\cos(A - B) + \cos(A + B)\right] = 1 - 2\cos A\cos B\cos C.
  8. cos⁡2θ=2cos⁡260°=12\displaystyle \cos^2\theta = 2\cos^2 60° = \frac{1}{2}, so cos⁡θ=±12\displaystyle \cos\theta = \pm\frac{1}{\sqrt{2}}.
  9. 1−m1+m=13\displaystyle \frac{1 - m}{1 + m} = \frac{1}{3} gives 3−3m=1+m3 - 3m = 1 + m, so m=12\displaystyle m = \frac{1}{2}.