Cos<0, sinA=1/3, найдите sin2A, cosA/2

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Given: cosA < 0, sinA = 1/3

We know that sin^2A + cos^2A = 1. So, sin^2A = 1 - cos^2A = 1 - (1/9) = 8/9.

We also know that cosA/2 = ±sqrt((1 + cosA)/2)

Since cosA < 0, cosA/2 = -sqrt((1 + cosA)/2) = -sqrt((1 - cos^2A)/2) = -(4/3)/√2 = -2√2/3.

Therefore, sin2A = 2sinAcosA = 2(1/3)(cosA) = 2/3(cosA).

So, sin2A = 2/3(cosA), cosA/2 = -2√2/3.