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Concept 02 / 11Trigonometry

Multiple and Sub-Multiple Angles: Double, Triple and Half-Angle Formulas

Formulas for sin 2A, cos 2A, tan 2A and their tan-A forms, triple angle formulas, half-angle formulas with the sign determined by the quadrant, products of cosines of angles in geometric progression, and values such as sin 18 degrees.

Setting B=AB = A in the compound angle formulas gives the double angle formulas. sin⁡2A=2sin⁡Acos⁡A=2tan⁡A1+tan⁡2A\sin 2A = 2\sin A\cos A = \frac{2\tan A}{1 + \tan^{2}A}. cos⁡2A=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2A=1−tan⁡2A1+tan⁡2A\cos 2A = \cos^{2}A - \sin^{2}A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A = \frac{1 - \tan^{2}A}{1 + \tan^{2}A}. tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \frac{2\tan A}{1 - \tan^{2}A}. The forms in tan⁡A\tan A express everything rationally in t=tan⁡At = \tan A; with A=x2A = \frac{x}{2} they are the half-angle substitution used in integration. Rearranged 'power reduction' forms: cos⁡2A=1+cos⁡2A2\cos^{2}A = \frac{1 + \cos 2A}{2}, sin⁡2A=1−cos⁡2A2\sin^{2}A = \frac{1 - \cos 2A}{2}, and so 1+cos⁡2A=2cos⁡2A1 + \cos 2A = 2\cos^{2}A, 1−cos⁡2A=2sin⁡2A1 - \cos 2A = 2\sin^{2}A. These are used constantly to remove square roots: 1+cos⁡2A=2∣cos⁡A∣\sqrt{1 + \cos 2A} = \sqrt{2}|\cos A|. Triple angle formulas: sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4\sin^{3}A, cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A = 4\cos^{3}A - 3\cos A, tan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\tan 3A = \frac{3\tan A - \tan^{3}A}{1 - 3\tan^{2}A}. Related product identities: sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=14sin⁡3A\sin A\sin(60^{\circ} - A)\sin(60^{\circ} + A) = \frac{1}{4}\sin 3A, cos⁡Acos⁡(60∘−A)cos⁡(60∘+A)=14cos⁡3A\cos A\cos(60^{\circ} - A)\cos(60^{\circ} + A) = \frac{1}{4}\cos 3A, and tan⁡Atan⁡(60∘−A)tan⁡(60∘+A)=tan⁡3A\tan A\tan(60^{\circ} - A)\tan(60^{\circ} + A) = \tan 3A. Half-angle formulas: sin⁡A2=±1−cos⁡A2\sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}, cos⁡A2=±1+cos⁡A2\cos\frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}, tan⁡A2=sin⁡A1+cos⁡A=1−cos⁡Asin⁡A\tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{1 - \cos A}{\sin A}. The ±\pm is decided by the quadrant of A2\frac{A}{2}, not of AA. The tan⁡A2\tan\frac{A}{2} forms without roots have no sign ambiguity. Products of cosines in geometric progression. Multiply and divide by sin⁡A\sin A and use 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta repeatedly: cos⁡Acos⁡2Acos⁡4A⋯cos⁡2n−1A=sin⁡2nA2nsin⁡A\cos A\cos 2A\cos 4A\cdots\cos 2^{n-1}A = \frac{\sin 2^{n}A}{2^{n}\sin A}. For example, cos⁡π7cos⁡2π7cos⁡4π7=sin⁡8π78sin⁡π7=−18\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}} = -\frac{1}{8}, since sin⁡8π7=−sin⁡π7\sin\frac{8\pi}{7} = -\sin\frac{\pi}{7}. The value of sin⁡18∘\sin 18^{\circ}. With θ=18∘\theta = 18^{\circ}, 5θ=90∘5\theta = 90^{\circ}, so 2θ=90∘−3θ2\theta = 90^{\circ} - 3\theta and sin⁡2θ=cos⁡3θ\sin 2\theta = \cos 3\theta. That is 2sin⁡θcos⁡θ=4cos⁡3θ−3cos⁡θ2\sin\theta\cos\theta = 4\cos^{3}\theta - 3\cos\theta; dividing by cos⁡θ\cos\theta gives 2sin⁡θ=4(1−sin⁡2θ)−32\sin\theta = 4(1 - \sin^{2}\theta) - 3, i.e. 4sin⁡2θ+2sin⁡θ−1=04\sin^{2}\theta + 2\sin\theta - 1 = 0. The positive root is sin⁡18∘=5−14\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}. Then cos⁡36∘=1−2sin⁡218∘=5+14\cos 36^{\circ} = 1 - 2\sin^{2}18^{\circ} = \frac{\sqrt{5} + 1}{4}. Key Takeaways: - sin⁡2A=2sin⁡Acos⁡A=2t1+t2\sin 2A = 2\sin A\cos A = \frac{2t}{1 + t^{2}}; cos⁡2A=1−t21+t2\cos 2A = \frac{1 - t^{2}}{1 + t^{2}}; tan⁡2A=2t1−t2\tan 2A = \frac{2t}{1 - t^{2}} with t=tan⁡At = \tan A. - 1+cos⁡2A=2cos⁡2A1 + \cos 2A = 2\cos^{2}A; 1−cos⁡2A=2sin⁡2A1 - \cos 2A = 2\sin^{2}A. - sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4\sin^{3}A; cos⁡3A=4cos⁡3A−3cos⁡A\cos 3A = 4\cos^{3}A - 3\cos A. - Half-angle signs come from the quadrant of A2\frac{A}{2}. - ∏k=0n−1cos⁡2kA=sin⁡2nA2nsin⁡A\prod_{k=0}^{n-1}\cos 2^{k}A = \frac{\sin 2^{n}A}{2^{n}\sin A}. - sin⁡18∘=5−14\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}, cos⁡36∘=5+14\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}. - Pitfall: writing 1−cos⁡2A=2sin⁡A\sqrt{1 - \cos 2A} = \sqrt{2}\sin A without the absolute value. Brain-Teaser 1: Evaluate sin⁡20∘sin⁡40∘sin⁡60∘sin⁡80∘\sin 20^{\circ}\sin 40^{\circ}\sin 60^{\circ}\sin 80^{\circ}. Solution: sin⁡20∘sin⁡40∘sin⁡80∘=sin⁡20∘sin⁡(60∘−20∘)sin⁡(60∘+20∘)=14sin⁡60∘=38\sin 20^{\circ}\sin 40^{\circ}\sin 80^{\circ} = \sin 20^{\circ}\sin(60^{\circ} - 20^{\circ})\sin(60^{\circ} + 20^{\circ}) = \frac{1}{4}\sin 60^{\circ} = \frac{\sqrt{3}}{8}. Multiplying by sin⁡60∘=32\sin 60^{\circ} = \frac{\sqrt{3}}{2} gives 316\frac{3}{16}. Spotting the pattern A,60∘−A,60∘+AA, 60^{\circ} - A, 60^{\circ} + A is the key. Brain-Teaser 2: Simplify 2+2+2cos⁡4θ\sqrt{2 + \sqrt{2 + 2\cos 4\theta}} for 0<θ<π40 < \theta < \frac{\pi}{4}. Solution: 2+2cos⁡4θ=4cos⁡22θ2 + 2\cos 4\theta = 4\cos^{2}2\theta, and cos⁡2θ>0\cos 2\theta > 0 on this interval, so the inner root is 2cos⁡2θ2\cos 2\theta. Then 2+2cos⁡2θ=4cos⁡2θ2 + 2\cos 2\theta = 4\cos^{2}\theta, and cos⁡θ>0\cos\theta > 0, giving 2cos⁡θ2\cos\theta. The answer is 2cos⁡θ2\cos\theta. Each step needs the sign check; on other intervals the answer changes. Brain-Teaser 3: Prove that cos⁡20∘cos⁡40∘cos⁡80∘=18\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ} = \frac{1}{8}. Solution: The angles double successively, so use the product formula with A=20∘A = 20^{\circ}, n=3n = 3: sin⁡160∘8sin⁡20∘=sin⁡20∘8sin⁡20∘=18\frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{\sin 20^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}, since sin⁡160∘=sin⁡20∘\sin 160^{\circ} = \sin 20^{\circ}. Alternatively, cos⁡20∘cos⁡40∘cos⁡80∘=cos⁡20∘cos⁡(60∘−20∘)cos⁡(60∘+20∘)=14cos⁡60∘=18\cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ} = \cos 20^{\circ}\cos(60^{\circ} - 20^{\circ})\cos(60^{\circ} + 20^{\circ}) = \frac{1}{4}\cos 60^{\circ} = \frac{1}{8}. Two independent methods confirm the result.

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Multiple and Sub-Multiple Angles: Double, Triple and Half-Angle Formulas — DPP

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