Question

Evaluate cos² $112\frac{1}{2}^{\textdegree}$ - sin² $52\frac{1}{2}^{\textdegree}$

Answer 1

cos²112 1/2° - sin²52 1/2°

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

= (cos112 1/2° - sin52 1/2°)(cos112 1/2° + sin52 1/2°)

now, sin52 1/2° = sin(90° - 37 1/2°) = cos37 1/2°

= (cos112 1/2° - cos37 1/2°)(cos112 1/2° + cos37 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(75°) sin(-37 1/2°) ] [2cos(75°) cos(37 1/2°)]

= -[2sin75° cos75° ] [2sin(37 1/2°) cos(37 1/2°)]

use formula, sin2x = 2sinx cosx

= - [sin150° ] [sin75° ]

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

= - sin30° × [sin45° cos30° + cos45° sin30° ]

= -1/2 × (√3 + 1)/2√2

= - (√3 + 1)/4√2

Answer 2

HELLO DEAR,



cos²112 1/2° - sin²52 1/2°

[as, a² - b² = (a - b)(a + b) ]

=> (cos112 1/2° - sin52 1/2°)(cos112 1/2° + sin52 1/2°)

now,
sin52 1/2° = sin(90° - 37 1/2°) = cos37 1/2°

=> (cos112 1/2° - cos37 1/2°)(cos112 1/2° + cos37 1/2°)

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

=> [2sin(75°) sin(-37 1/2°) ] [2cos(75°) cos(37 1/2°)]

=> -[2sin75° cos75° ] [2sin(37 1/2°) cos(37 1/2°)]

[as, sin2x = 2sinx cosx ]

=> - [sin150° ] [sin75° ]

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

=> - sin30° * [sin45° cos30° + cos45° sin30° ]

=> -1/2 * (√3 + 1)/2√2

=> - (√3 + 1)/4√2


I HOPE IT'S HELP YOU DEAR,
THANKS