Математика 10 класса

найти sin(α+β) и cos(α-β), если
sin α = 4/5, π/2 < α < π
cos β= -15/17, π/2 < β < π
=>cos b
cos^2 a = 1 - sin^2 a = 1 - (4/5)^2 = 1 - 16/25 = 9/25 = (+ - 3/5)^2
π/2 < α < π => это III четверть и знак косинуса там (-) =>
cos a = - 3/5

sin^2 b = 1 - cos^2 b = 1 - (-15/17)^2 = 1 - 15^2/17^2 =
= (17^2 - 15^2)/15^2 = (17+15)(17-15)/15^2 = 32*2/15^2 = (+ - 8/15)^2
π/2 < β < π => это III четверть и знак синуса там (+) =>
sin b = + 8/15

sin (a+b) =sin a ⋅ cos b + cos a ⋅ sin b =
= 4/5 * (-15/17) + (- 3/5) * 8/15 = ...

cos (a-b) = cos a ⋅ cos b + sin a ⋅ sin b =
= (-3/5) * (-15/17) + 4/5 * 8/15 = ...