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Concept 07 / 11Trigonometry

Compound Angle Formulas and Their Applications

Derivation and use of sin(A +/- B), cos(A +/- B) and tan(A +/- B), useful corollaries such as sin(A + B) sin(A - B) = sin^2 A - sin^2 B, tan of a sum of three angles, and conditional identities when A + B + C = pi.

Figure 1
The compound angle formulas are the foundation of all trigonometric identities: sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B, cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B, tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}, cot⁡(A±B)=cot⁡Acot⁡B∓1cot⁡B±cot⁡A\cot(A \pm B) = \frac{\cot A\cot B \mp 1}{\cot B \pm \cot A}. Note the sign pattern: sine keeps the sign, cosine flips it. A geometric proof for acute angles is shown in the figure. Take a segment OPOP of length 11 at angle A+BA + B. Drop from PP a perpendicular PQPQ onto the line at angle AA; then OQ=cos⁡BOQ = \cos B and QP=sin⁡BQP = \sin B. The height of PP is the height of QQ (cos⁡Bsin⁡A\cos B\sin A) plus the vertical rise from QQ to PP (sin⁡Bcos⁡A\sin B\cos A, since QPQP makes angle AA with the vertical). So sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A\cos B + \cos A\sin B. An even shorter proof uses rotation: in complex numbers ei(A+B)=eiAeiBe^{i(A + B)} = e^{iA}e^{iB}, and comparing real and imaginary parts gives both cos⁡(A+B)\cos(A + B) and sin⁡(A+B)\sin(A + B) at once. Corollaries worth memorising. sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B=cos⁡2B−cos⁡2A\sin(A + B)\sin(A - B) = \sin^{2}A - \sin^{2}B = \cos^{2}B - \cos^{2}A. cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B\cos(A + B)\cos(A - B) = \cos^{2}A - \sin^{2}B. tan⁡(A+B+C)=tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}. tan⁡(π4+θ)=1+tan⁡θ1−tan⁡θ\tan(\frac{\pi}{4} + \theta) = \frac{1 + \tan\theta}{1 - \tan\theta}. If A+B=π4A + B = \frac{\pi}{4}, then (1+tan⁡A)(1+tan⁡B)=2(1 + \tan A)(1 + \tan B) = 2. (Expand tan⁡(A+B)=1\tan(A + B) = 1.) So (1+tan⁡1∘)(1+tan⁡44∘)=2(1 + \tan 1^{\circ})(1 + \tan 44^{\circ}) = 2. Conditional identities in a triangle. If A+B+C=πA + B + C = \pi: tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A\tan B\tan C (from tan⁡(A+B+C)=0\tan(A + B + C) = 0, so its numerator vanishes). cot⁡Acot⁡B+cot⁡Bcot⁡C+cot⁡Ccot⁡A=1\cot A\cot B + \cot B\cot C + \cot C\cot A = 1. sin⁡2A+sin⁡2B+sin⁡2C=4sin⁡Asin⁡Bsin⁡C\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C. 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. sin⁡2A+sin⁡2B+sin⁡2C=2+2cos⁡Acos⁡Bcos⁡C\sin^{2}A + \sin^{2}B + \sin^{2}C = 2 + 2\cos A\cos B\cos C. tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1\tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1. The method in every case: use C=π−(A+B)C = \pi - (A + B), so sin⁡C=sin⁡(A+B)\sin C = \sin(A + B) and cos⁡C=−cos⁡(A+B)\cos C = -\cos(A + B), then apply compound angle or sum-to-product formulas. Evaluating non-standard angles: sin⁡75∘=sin⁡(45∘+30∘)=6+24\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}; tan⁡105∘=tan⁡(60∘+45∘)=3+11−3=−(2+3)\tan 105^{\circ} = \tan(60^{\circ} + 45^{\circ}) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = -(2 + \sqrt{3}). Key Takeaways: - sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B; cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B. - tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}. - sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin(A + B)\sin(A - B) = \sin^{2}A - \sin^{2}B. - A+B=π4⇒(1+tan⁡A)(1+tan⁡B)=2A + B = \frac{\pi}{4} \Rightarrow (1 + \tan A)(1 + \tan B) = 2. - A+B+C=πA + B + C = \pi: ∑tan⁡A=∏tan⁡A\sum\tan A = \prod\tan A; ∑cot⁡Acot⁡B=1\sum\cot A\cot B = 1. - For triangle identities, replace CC by π−(A+B)\pi - (A + B). - Pitfall: cos⁡(A+B)=cos⁡A+cos⁡B\cos(A + B) = \cos A + \cos B. The formula has products, not sums. Brain-Teaser 1: Evaluate (1+tan⁡1∘)(1+tan⁡2∘)⋯(1+tan⁡44∘)(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})\cdots(1 + \tan 44^{\circ}). Solution: Pair k∘k^{\circ} with (45−k)∘(45 - k)^{\circ}: each pair's product is 22. The angles 1∘1^{\circ} to 44∘44^{\circ} form 2222 pairs, so the product is 2222^{22}. The identity for angles summing to π4\frac{\pi}{4} turns a daunting product into a power of 22. Brain-Teaser 2: If tan⁡A=12\tan A = \frac{1}{2} and tan⁡B=13\tan B = \frac{1}{3}, with A,BA, B acute, find A+BA + B. Solution: tan⁡(A+B)=12+131−16=5/65/6=1\tan(A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{5/6}{5/6} = 1. Since A,BA, B are acute, 0<A+B<π0 < A + B < \pi, and tan⁡(A+B)=1>0\tan(A + B) = 1 > 0 puts it in (0,π2)(0, \frac{\pi}{2}), so A+B=π4A + B = \frac{\pi}{4}. The range argument is needed because tan⁡\tan alone determines an angle only up to multiples of π\pi. Brain-Teaser 3: In a triangle, tan⁡A\tan A, tan⁡B\tan B and tan⁡C\tan C are in the ratio 1:2:31 : 2 : 3. Find the angles. Solution: Let the tangents be k,2k,3kk, 2k, 3k. The identity ∑tan⁡A=∏tan⁡A\sum\tan A = \prod\tan A gives 6k=6k36k = 6k^{3}, so k2=1k^{2} = 1 and k=1k = 1 (all tangents must be positive; a negative kk would make all three angles obtuse). So tan⁡A=1\tan A = 1, tan⁡B=2\tan B = 2, tan⁡C=3\tan C = 3: A=45∘A = 45^{\circ}, B=tan⁡−12≈63.4∘B = \tan^{-1}2 \approx 63.4^{\circ}, C=tan⁡−13≈71.6∘C = \tan^{-1}3 \approx 71.6^{\circ}, which do sum to 180∘180^{\circ}.

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Compound Angle Formulas and Their Applications — DPP

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