найдите sin альфа-cos альфа, если tg альфа = - 3/4 и пи/2 < альфа < пи

1. Вычислим ctgα:

tgα = - 3/4; ctgα = 1/tgα = 1 / (-3/4) = - 4/3.

2. Выразим функции sinα и cosα через tgα и ctgα, для α ∈ (π/2; π):

sin^2 (α) + cos^2 (α) = 1; sin^2 (α) / cos^2 (α) + cos^2 (α) / cos^2 (α) = 1/cos^2 (α); tg^2 (α) + 1 = 1/cos^2 (α); cos^2 (α) = 1 / (1 + tg^2 (α)); cosα = - 1/√ (1 + tg^2 (α)) = - 1/√ (1 + 9/16) = - 1/√ (25/16) = - 1 / (5/4) = - 4/5; sin^2 (α) + cos^2 (α) = 1; sin^2 (α) / sin^2 (α) + cos^2 (α) / sin^2 (α) = 1/sin^2 (α); 1 + ctg^2 (α) = 1/sin^2 (α); sin^2 (α) = 1 / (1 + ctg^2 (α)); sinα = 1/√ (1 + ctg^2 (α)) = 1/√ (1 + 16/9) = 1/√ (25/9) = 1 / (5/3) = 3/5.

3. sinα - cosα = 3/5 - (-4/5) = 3/5 + 4/5 = 7/5 = 1,4.