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