Question

Evaluate cos² $52\frac{1}{2}^{\textdegree}$ - sin² $22\frac{1}{2}^{\textdegree}$

Answer 1

cos²52 1/2° - sin²22 1/2°

use algebraic formula, a² - b² = (a - b)(a + b)

= (cos52 1/2° - sin22 1/2°)(cos52 1/2° + sin22 1/2°)

now, sin22 1/2° = sin(90° - 67 1/2°) = cos67 1/2°

= (cos52 1/2° - cos67 1/2°)(cos52 1/2° + cos67 1/2°)

use formula, cosC - cosD = 2sin(C + D)/2.sin(D - C)/2
cosC + cosD = 2cos(C + D)/2 cos(C - D)/2

= [2sin(60°) sin(7 1/2°) ] [2cos(60°) cos(7 1/2°)]

= [2sin60° cos60° ] [2sin(7 1/2°) cos(7 1/2°)]

use formula, sin2x = 2sinx cosx

= [sin120° ] [sin15° ]

= [ sin(180° - 60°) ] [ sin(45° - 30°)

= sin60° [ sin45° cos30° - cos45° sin30° ]

= √3/2 × (√3 - 1)/2√2

= (3 - √3)/4√2

Answer 2

HELLO DEAR,



cos²52 1/2° - sin²22 1/2°

[as, a² - b² = (a - b)(a + b) ]

=> (cos52 1/2° - sin22 1/2°)(cos52 1/2° + sin22 1/2°)

now, sin22 1/2° = sin(90° - 67 1/2°) = cos67 1/2°

=> (cos52 1/2° - cos67 1/2°)(cos52 1/2° + cos67 1/2°)

[as, cosC - cosD = 2sin(C + D)/2.sin(D - C)/2
cosC + cosD = 2cos(C + D)/2 cos(C - D)/2 ]

=> [2sin(60°) sin(7 1/2°) ] [2cos(60°) cos(7 1/2°)]

=> [2sin60° cos60° ] [2sin(7 1/2°) cos(7 1/2°)]

[as, sin2x = 2sinx cosx ]

=> [sin120° ] [sin15° ]

=> [ sin(180° - 60°) ] [ sin(45° - 30°)

=> sin60° [as, sin45° cos30° - cos45° sin30° ]

=> √3/2 × (√3 - 1)/2√2

=> (3 - √3)/4√2


I HOPE IT'S HELP YOU DEAR,
THANKS