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Prove that :`16 cos 2pi/15 cos 4pi/15 cos 8pi/15 cos16pi/15 = 1`

Answer» Here, we will use,
`2sinxcosx = sin2x`
Now,
`L.H.S. = 16cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)cos((16pi)/15)`
`=16cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)cos(pi+pi/15)`
`=16cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)(-cos(pi/15))`
`=-16cos(pi/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)`
`=8/sin(pi/15)*2sin(pi/15)cos(pi/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)`
`=8/sin(pi/15)*sin((2pi)/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)`
`=4/sin(pi/15)*2sin((2pi)/15)cos((2pi)/15)cos((4pi)/15)cos((8pi)/15)`
`=4/sin(pi/15)*sin((4pi)/15)cos((4pi)/15)cos((8pi)/15)`
`=2/sin(pi/15)*2sin((4pi)/15)cos((4pi)/15)cos((8pi)/15)`
`=2/sin(pi/15)*sin((8pi)/15)cos((8pi)/15)`
`=sin((16pi)/15)/(sin((pi)/15))`
`=sin(pi+pi/15)/(sin((pi)/15))`
`=(sin((pi)/15))/(sin((pi)/15))`
`=1 = R.H.S.`


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