Question

Show that cos A = $\frac{cos 3A}{(2 cos 2A - 1)}$. Hence find the value of cos 15°.

Answer 1

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

Answer 2

HELLO DEAR,



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


now,

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

I HOPE IT'S HELP YOU DEAR,
THANKS