Let's salute cos 1

Geometry Level 3

1 A = cos π 11 cos 2 π 11 cos 3 π 11 cos 4 π 11 cos 5 π 11 \dfrac1A = \cos \dfrac{\pi}{11} \cos \dfrac{2\pi}{11} \cos \dfrac{3\pi}{11} \cos \dfrac{4\pi}{11} \cos \dfrac{5\pi}{11}

Find the value of the positive integer A A .


The answer is 32.

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2 solutions

Akshat Sharda
Oct 4, 2015

= cos π 11 cos 2 π 11 cos 3 π 11 cos 4 π 11 cos 5 π 11 = 2 sin π 11 2 sin π 11 cos π 11 cos 2 π 11 cos 3 π 11 cos 4 π 11 cos 5 π 11 Now, 2 sin π 11 cos π 11 = sin 2 π 11 because 2 sin A cos A = sin 2 A Note: We are going to use this identity again and again. = 2 4 sin π 11 sin 2 π 11 cos 2 π 11 cos 3 π 11 cos 4 π 11 cos 5 π 11 = 2 8 sin π 11 sin 4 π 11 cos 4 π 11 cos 3 π 11 cos 5 π 11 = 1 8 sin π 11 sin 8 π 11 cos 3 π 11 cos 5 π 11 Note: sin 8 π 11 = sin ( π 3 π 11 ) = sin 3 π 11 = 2 16 sin π 11 sin 3 π 11 cos 3 π 11 cos 5 π 11 = 1 16 sin π 11 sin 6 π 11 cos 5 π 11 Note: sin 6 π 11 = sin ( π 5 π 11 ) = sin 5 π 11 = 2 32 sin π 11 sin 5 π 11 cos 5 π 11 = 1 32 sin π 11 sin 10 π 11 Note: s i n 10 π 11 = s i n ( π π 11 ) = sin π 11 = 1 32 sin π 11 sin π 11 = 1 32 Therefore, A = 32 =\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{2 \sin \frac{\pi}{11}}{2 \sin \frac{\pi}{11}} \cos \frac{\pi}{11} \cos \frac{2\pi}{11} \cos \frac{3\pi}{11} \cos \frac{4\pi}{11} \cos \frac{5\pi}{11} \\ \text{Now, } 2 \sin \frac{\pi}{11} \cos \frac{\pi}{11}=\sin \frac{2\pi}{11} \text{ because } \color{#D61F06}{2 \sin A \cos A=\sin 2A} \\ \text{Note: We are going to use this identity again and again.} \\ =\frac{2}{4 \sin \frac{\pi}{11}} \sin \frac{2\pi}{11} \cos \frac{2\pi}{11} \cos \frac{3\pi}{11} \cos \frac{4\pi}{11} \cos \frac{5\pi}{11} \\ =\frac{2}{8 \sin \frac{\pi}{11}} \sin \frac{4\pi}{11} \cos \frac{4\pi}{11} \cos \frac{3\pi}{11} \cos \frac{5\pi}{11} \\ =\frac{1}{8 \sin \frac{\pi}{11}} \color{#3D99F6}{ \sin \frac{8\pi}{11}} \cos \frac{3\pi}{11} \cos \frac{5\pi}{11} \\ \text{Note: } \color{#3D99F6}{\sin \frac{8\pi}{11}}=\sin \left(\pi-\frac{3\pi}{11}\right)=\sin \frac{3\pi}{11} \\ =\frac{2}{16 \sin \frac{\pi}{11}} \sin \frac{3\pi}{11} \cos \frac{3\pi}{11} \cos \frac{5\pi}{11} \\ =\frac{1}{16 \sin \frac{\pi}{11}} \color{#EC7300}{\sin \frac{6\pi}{11}} \cos \frac{5\pi}{11} \\ \text{Note: } \color{#EC7300}{\sin \frac{6\pi}{11}}=\sin \left(\pi-\frac{5\pi}{11}\right)=\sin \frac{5\pi}{11} \\ =\frac{2}{32 \sin \frac{\pi}{11}} \sin \frac{5\pi}{11} \cos \frac{5\pi}{11} \\ =\frac{1}{32 \sin \frac{\pi}{11}}\color{#20A900}{ \sin \frac{10\pi}{11}} \\ \text{Note: } \color{#20A900}{sin \frac{10\pi}{11}}=sin \left(\pi-\frac{\pi}{11}\right)=\sin \frac{\pi}{11} \\ =\frac{1}{32 \sin \frac{\pi}{11}} \sin \frac{\pi}{11}=\frac{1}{32} \\ \text{Therefore, }A=\boxed{32}

Nicely done. Congratulations.

Niranjan Khanderia - 5 years, 8 months ago
Chan Lye Lee
Oct 12, 2015

Let P = cos π 11 cos 2 π 11 cos 3 π 11 cos 4 π 11 cos 5 π 11 \displaystyle P=\cos \frac{\pi}{11}\cos \frac{2\pi}{11}\cos \frac{3\pi}{11}\cos \frac{4\pi}{11}\cos \frac{5\pi}{11} and Q = sin π 11 sin 2 π 11 sin 3 π 11 sin 4 π 11 sin 5 π 11 \displaystyle Q=\sin \frac{\pi}{11}\sin \frac{2\pi}{11}\sin \frac{3\pi}{11}\sin \frac{4\pi}{11}\sin \frac{5\pi}{11} .

Then 32 P Q = sin 2 π 11 sin 4 π 11 sin 6 π 11 sin 8 π 11 sin 10 π 11 = Q \displaystyle 32PQ =\sin \frac{2\pi}{11}\sin \frac{4\pi}{11}\sin \frac{6\pi}{11}\sin \frac{8\pi}{11}\sin \frac{10\pi}{11}=Q and hence P = 1 32 \displaystyle P =\frac{1}{32} .

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