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Prove that: `cos18^0-si n 18^0=sqrt(2)sin27^0`

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Prove that: `cos18^0-si n 18^0=sqrt(2)sin27^0`
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`L.H.S. = cos 18^@ - sin 18^@`
`= cos 18^@ - sin (90-72)^@`
`= cos 18^@ - cos72^@`
Using `cosA - cosB = -2sin((A-B)/2)((A+B)/2),`
`= -2sin((18-72)/2)^@sin((18+72)/2)^@`
`= -2sin(-27^@)sin45^@`
`=2sin27^@(1/sqrt2)`
`=sqrt2sin27^@ = R.H.S.`
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