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How to proof: $\cos(10°)*\cos(30°)*\cos(50°)*\cos(70°)=\frac{3}{16}$

I tried to: $\frac{8*3^{1/2}\sin(10°)*\cos(10°)*\sin(40°)*\sin(20°)}{16*\sin(10°)} $ and got nowhere

VLC
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  • You will find this question in many places, for example this one (http://www.askiitians.com/forums/Trigonometry/cos10-cos30-cos60-cos70-3-16-solve-it-how-to-sol_136529.htm) which provides a straightforward proof. – Jean Marie Dec 16 '17 at 18:29
  • Why don't you use https://math.stackexchange.com/questions/1145406/prove-that-cos-x-cdot-cosx-60-circ-cdot-cosx60-circ-frac14 like in https://math.stackexchange.com/questions/2569252/proof-expression-cos80-sin70-cos60-sin50-frac116 – lab bhattacharjee Dec 17 '17 at 06:21

2 Answers2

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$$\begin{align}\cos(10°)\cos(30°)\cos(50°)\cos(70°) &= \sin(80°)\cos(30°)\sin(40°)\sin(20°) &\\=& \cos(30°)\sin(80°)\cdot\dfrac{-1}{2}(\cos(60°) - \cos(20°))&\\=& \cos(30°)\cdot\dfrac{-1}{2}(\sin(80°)\cos(60°) - \cos(20°)\sin(80°))&\\=& \cos(30°)\cdot\dfrac{-1}{2}\left(\dfrac{1}{2}(\sin(140°)+\sin(20°)) - \\\dfrac12(\sin(100)+\sin(60°))\right)&\\=&\ \cos(30°)\cdot\dfrac{-1}{4}\left(\sin(40°)+\sin(20°) - \sin(80°)-\sin(60°)\right)&\\=&\ \cos(30°)\cdot\dfrac{-1}{4}(\sin(60°)\cos(20°) - \sin(20°)\cos(60°)\\&+\sin(20°) - \sin(60°)\cos(20°) -\sin(20°)\cos(60°)\\&-\sin(60°))&\\=&\ \cos(30°)\cdot\dfrac{-1}{4}\left( - 2\sin(20°)\cos(60°)+\sin(20°) -\sin(60°)\right)&\\=&\ \cos(30°)\cdot\dfrac{1}{4}\sin(60°) = \dfrac{3}{16}\end{align}$$

user8277998
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use that $$\cos(x)\cos(y)=\frac{1}{2}\left(\cos(x-y)+\cos(x+y)\right)$$

Dr. Sonnhard Graubner
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