`sin(pi/18)*sin(5pi/18)*sin(7pi/18)`
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by calculating`pi /18 = 10^@`
we have to find `sin 10^@, sin 50^@, sin 70^@`
`1/2 sin 10^@ [cos(70^@-50^@) + cos(70^@ +50^@)]`
`1/2 sin 10^@[cos 20^@ – cos 120^@]`
`1/2sin10^@[1-2sin^2 10^@ + 1/2]`
`1/2[(3-4sin^2 10^@)Sin 10^@/2]`
`(3*sin10^@ – 4sin^3 10^@)/4`
`1/4sin 30^@ = 1/8`