cos10cos30cos50cos70 = 3/16
@phi
looks tricky. any ideas ?
no
\[\cos10^{o} \cos30^{o} \cos50^{o} \cos70^{o} = \frac{ 3 }{ 16}\]
\[\cos 30\cos 70 \cos 10 \cos 50=\frac{ \sqrt{3} }{ 2 }*\frac{ 1 }{ 2 }\left( 2\cos 70 \cos 10 \right)\cos 50\] \[=\frac{ \sqrt{3} }{ 4 }\left( \cos \left( 70-10 \right)+\cos \left( 70+10 \right) \right)\cos 50\] \[=\frac{ \sqrt{3} }{ 4 }\left( \cos 60+\cos 80 \right)\cos 50\] \[=\frac{ \sqrt{3} }{ 4 }\left( \frac{ 1 }{ 2 }+\cos 80 \right)\cos 50\] can you proceed further?
\[=\frac{ \sqrt{3} }{ 8 }\left( \cos 50+(2\cos 80 \cos 50 \right))=?\]
\[=\frac{ \sqrt{3} }{ 8 }\left[ \cos 50+\cos \left( 80-50 \right)+\cos \left( 80+50 \right) \right]\] \[=\frac{ \sqrt{3} }{ 8 }\left[ \cos 50+\cos 30+\cos 130 \right]\] \[=\frac{ \sqrt{3} }{ 8 }\left[ \cos 50+\frac{ \sqrt{3} }{ 2 }+\cos \left( 180-50 \right) \right]\] \[=\frac{ \sqrt{3} }{ 8 }\left[ \cos 50+\frac{ \sqrt{3} }{ 2 }-\cos 50 \right]=?\]
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