1.

`cos (pi/11) cos(2 pi/11) cos(3 pi/11) cos(4 pi/11) cos(5pi/11)`

Answer» Here, we will use,
`2sinxcosx = sin2x`
Now,
`cos(pi/11)cos((2pi)/11)cos((3pi)/11)cos((4pi)/11)cos((5pi)/11)`
`=1/(2sin(pi/11))(2sin(pi/11)cos(pi/11)cos((2pi)/11)cos((3pi)/11)cos((4pi)/11)cos((5pi)/11))`
`=1/(2sin(pi/11))(sin((2pi)/11)cos((2pi)/11)cos((3pi)/11)cos((4pi)/11)cos((5pi)/11))`
`=1/(4sin(pi/11))(2sin((2pi)/11)cos((2pi)/11)cos(pi-(8pi)/11))cos((4pi)/11)cos((5pi)/11))`
`=1/(4sin(pi/11))(sin((4pi)/11)cos((4pi)/11)cos(-(8pi)/11)cos((5pi)/11))`
`=-1/(8sin(pi/11))(2sin((4pi)/11)cos((4pi)/11)cos((8pi)/11)cos((5pi)/11))`
`=-1/(8sin(pi/11))(sin((8pi)/11)cos((8pi)/11)cos((5pi)/11))`
`=1/(16sin(pi/11))(2sin((8pi)/11)cos((8pi)/11)cos(pi+(5pi)/11))`
`=1/(16sin(pi/11))(sin((16pi)/11)cos((16pi)/11))`
`=1/(32sin(pi/11))(2sin((16pi)/11)cos((16pi)/11))`
`=1/(32sin(pi/11))(sin((32pi)/11))`
`=1/(32sin(pi/11))(sin(3pi-(pi)/11))`
`=1/(32sin(pi/11))(sin(pi/11))`
`=1/32`
so, value of the given expression is `1/32.`


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