Olympiad Maths Prep

Track / Stage 5 / 373 of 400 #973 of 2000

Problem 973

AIME late
Algebra Difficulty 5.9 Find the answer

3. Find the values of the following expressions:
(1) sin⁡10∘⋅sin⁡30∘⋅sin⁡50∘⋅sin⁡70∘\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ};
(2) sin⁡220∘+cos⁡280∘+3sin⁡20∘⋅cos⁡80∘\sin ^{2} 20^{\circ}+\cos ^{2} 80^{\circ}+\sqrt{3} \sin 20^{\circ} \cdot \cos 80^{\circ};
(3) cos⁡2A+cos⁡2(60∘−A)+cos⁡2(60∘+A)\cos ^{2} A+\cos ^{2}\left(60^{\circ}-A\right)+\cos ^{2}\left(60^{\circ}+A\right);
(4) cos⁡π15⋅cos⁡2π15⋅cos⁡3π15⋅cos⁡4π15⋅cos⁡5π15⋅cos⁡6π15⋅cos⁡7π15\cos \frac{\pi}{15} \cdot \cos \frac{2 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{5 \pi}{15} \cdot \cos \frac{6 \pi}{15} \cdot \cos \frac{7 \pi}{15}.

Official solution

3. (1) Let A=sin⁡10∘⋅sin⁡30∘⋅sin⁡50∘⋅sin⁡70∘,B=cos⁡10∘⋅cos⁡30∘⋅cos⁡50∘⋅cos⁡70∘A=\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}, B=\cos 10^{\circ} \cdot \cos 30^{\circ} \cdot \cos 50^{\circ} \cdot \cos 70^{\circ}, then A⋅B=A \cdot B= 116sin⁡20∘⋅sin⁡60∘⋅sin⁡100∘⋅sin⁡140∘=116cos⁡10∘⋅cos⁡30∘⋅cos⁡50∘⋅cos⁡70∘=116B\frac{1}{16} \sin 20^{\circ} \cdot \sin 60^{\circ} \cdot \sin 100^{\circ} \cdot \sin 140^{\circ}=\frac{1}{16} \cos 10^{\circ} \cdot \cos 30^{\circ} \cdot \cos 50^{\circ} \cdot \cos 70^{\circ}=\frac{1}{16} B, and B≠0B \neq 0, so A=A= 116\frac{1}{16}, that is, sin⁡10∘⋅sin⁡30∘⋅sin⁡50∘⋅sin⁡70∘=116\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}=\frac{1}{16}.
Or by sin⁡3θ=4sin⁡θ⋅sin⁡(60∘−θ)⋅sin⁡(60∘+θ)\sin 3 \theta=4 \sin \theta \cdot \sin \left(60^{\circ}-\theta\right) \cdot \sin \left(60^{\circ}+\theta\right) to find, also get 116\frac{1}{16}.
(2) Let A=sin⁡220∘+cos⁡280∘+3sin⁡20∘⋅cos⁡80∘,B=cos⁡220∘+sin⁡280∘−3cos⁡20∘⋅sin⁡80∘A=\sin ^{2} 20^{\circ}+\cos ^{2} 80^{\circ}+\sqrt{3} \sin 20^{\circ} \cdot \cos 80^{\circ}, B=\cos ^{2} 20^{\circ}+\sin ^{2} 80^{\circ}-\sqrt{3} \cos 20^{\circ} \cdot \sin 80^{\circ}, then A+A+ B=2−3(cos⁡20∘⋅sin⁡80∘−sin⁡20∘⋅cos⁡80∘)=2−3sin⁡60∘=12,B−A=cos⁡40∘−cos⁡160∘−B=2-\sqrt{3}\left(\cos 20^{\circ} \cdot \sin 80^{\circ}-\sin 20^{\circ} \cdot \cos 80^{\circ}\right)=2-\sqrt{3} \sin 60^{\circ}=\frac{1}{2}, B-A=\cos 40^{\circ}-\cos 160^{\circ}- 3(cos⁡20∘⋅sin⁡80∘+sin⁡20∘⋅cos⁡80∘)=2cos⁡30∘⋅cos⁡10∘−3sin⁡100∘=0\sqrt{3}\left(\cos 20^{\circ} \cdot \sin 80^{\circ}+\sin 20^{\circ} \cdot \cos 80^{\circ}\right)=2 \cos 30^{\circ} \cdot \cos 10^{\circ}-\sqrt{3} \sin 100^{\circ}=0. Thus 2A=12,A=142 A=\frac{1}{2}, A=\frac{1}{4}, that is, sin⁡220∘+cos⁡280∘+3⋅sin⁡20∘⋅cos⁡80∘=14\sin ^{2} 20^{\circ}+\cos ^{2} 80^{\circ}+\sqrt{3} \cdot \sin 20^{\circ} \cdot \cos 80^{\circ}=\frac{1}{4}.
(3) Let m=cos⁡2A+cos⁡2(60∘−A)+cos⁡2(60∘+A),n=sin⁡2A+sin⁡2(60∘−A)+sin⁡2(60∘+A)m=\cos ^{2} A+\cos ^{2}\left(60^{\circ}-A\right)+\cos ^{2}\left(60^{\circ}+A\right), n=\sin ^{2} A+\sin ^{2}\left(60^{\circ}-A\right)+\sin ^{2}\left(60^{\circ}+A\right), then x+y=3,y−x=cos⁡2A+cos⁡2(60∘−A)+cos⁡2(60∘+A)=cos⁡2A+2cos⁡120∘⋅cos⁡2A=0x+y=3, y-x=\cos 2 A+\cos 2\left(60^{\circ}-A\right)+\cos 2\left(60^{\circ}+A\right)=\cos 2 A+2 \cos 120^{\circ} \cdot \cos 2 A=0. Thus y=x=y=x= 32\frac{3}{2}, that is, cos⁡2A+cos⁡2(60∘−A)+cos⁡2(60∘+A)=32\cos ^{2} A+\cos ^{2}\left(60^{\circ}-A\right)+\cos ^{2}\left(60^{\circ}+A\right)=\frac{3}{2}.
(4) Let M=cos⁡π15⋅cos⁡2π15⋅cos⁡3π15⋅cos⁡4π15⋅cos⁡5π15⋅cos⁡6π15⋅cos⁡7π15,N=sin⁡π15⋅sin⁡2π15M=\cos \frac{\pi}{15} \cdot \cos \frac{2 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{5 \pi}{15} \cdot \cos \frac{6 \pi}{15} \cdot \cos \frac{7 \pi}{15}, N=\sin \frac{\pi}{15} \cdot \sin \frac{2 \pi}{15}. sin⁡3π15⋅sin⁡4π15⋅sin⁡5π15⋅sin⁡6π15⋅sin⁡7π15\sin \frac{3 \pi}{15} \cdot \sin \frac{4 \pi}{15} \cdot \sin \frac{5 \pi}{15} \cdot \sin \frac{6 \pi}{15} \cdot \sin \frac{7 \pi}{15}, then 27M⋅N=sin⁡2π15⋅sin⁡4π15⋅sin⁡6π15⋅sin⁡8π15⋅sin⁡10π15⋅2^{7} M \cdot N=\sin \frac{2 \pi}{15} \cdot \sin \frac{4 \pi}{15} \cdot \sin \frac{6 \pi}{15} \cdot \sin \frac{8 \pi}{15} \cdot \sin \frac{10 \pi}{15} \cdot sin⁡12π15⋅sin⁡14π15=N\sin \frac{12 \pi}{15} \cdot \sin \frac{14 \pi}{15}=N, and N≠0N \neq 0, so M=127M=\frac{1}{2^{7}}, that is, the original expression =1128=\frac{1}{128}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.