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Prove that cos 36^(@) cos 72^(@) cos 108...

Prove that `cos 36^(@) cos 72^(@) cos 108^(@) cos 144^(@)=1//16`.

Text Solution

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`cos 36^(@)cos 72^(@)cos 108^(@)cos 144^(@)`
`=cos 36^(@)sin 18^(@)(-sin 18^(@))(-cos 36^(@))`
`=cos^(2)36^(@)sin^(2)18^(@)`
`=((sqrt(5)+1)/(4))^(2)((sqrt(5)-1)/(4))^(2)`
`=[((sqrt(5)+1)/(4))((sqrt(5)-1)/(4))]^(2)=(1)/(16)`
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