Відповіді АЛГЕБРА: 14.45

14.45

1) α = 45°; β = 135°

2sinα+β2 cosα- β2 = 2sin 45° + 135°2cos 45° - 135°2 = 2sin 90° cos 45° =

= 2 • 1 • 22 = 2

sinα + sinβ = sin45° + sin135° = 22 + 22 = 2

2) α = 90°; β = –30°.

2sinα+β2 cosα- β2 = 2sin 90°+30°2cos 90°-30°2 = 2sin 30°cos 60° =

= 2 • 12 • 12 = 12

sinα + sinβ = sin90° + sin(–30°) = 1 – 12 = 12

2sin α+ β2 cosα- β2 = sinα + sinβ