P.T cos 3A/(2cos2A-1)= cos A .find cos 15⁰
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Answer:
RHS = cos3A/(2cos2A - 1)
we know, cos3x = 4cos³x - 3cosx
cos2x = 2cos²x - 1
so, cos3A/(2cos2A - 1) = (4cos³A - 3cosA)/{2(2cos²A - 1) - 1}
= (4cos³A - 3cosA)/(4cos²A - 3 )
= cosA(4cos²A - 3)/(4cos²A - 3)
= cosA = LHS [ hence proved]
cos15° = cos3 × 15°/{2 cos2(15°) - 1}
= cos45°/{2cos30° - 1}
= (1/√2)/{2 × √3/2 - 1}
= (1/√2)/{√3 - 1} × (√3 + 1)/(√3 + 1)
= (√3 + 1)/(√3² - 1²)√2
= (√3 + 1)/2√2
hence,cos15° = (√3 + 1)/2√2
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