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cos""(2pi)/(15)cos""(4pi)/(15)cos""(8pi)...

`cos""(2pi)/(15)cos""(4pi)/(15)cos""(8pi)/(15)cos""(16pi)/(15)=(1)/(16)`

Text Solution

Verified by Experts

We have
`LHS=cos ""(2pi)/(15)cos""(4pi)/(15)cos ""(8pi)/(15)cos""(pi-(pi)/(15))`
`=(cos ""(2pi)/(15cos""(4pi)/(15)cos ""(8pi)/(15))(-cos""(pi)/(15))`
`=-cos ""(pi)/(15)cos"" (2pi)/(15) cos ""(4pi)/(15)cos ""(8pi)/(15)`
`=-cos A cos 2A cos 2^(2)Acos 2^(3)A`, where `A=pi//15`
`=-[(sin2^(4)A)/(2^(4)sinA)]=-(sin16A)/(2^(4)sinA)`
`-(sin(15A+A))/(16sinA)=(-sin(pi+A))/(16sinA)[because 15A]`
`=(sin A)/(16sinA)=(1)/(16)=RHS`.
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