Найдите sin 2а, если cosa = sqrt(7/8), α∈(3π/2; 2π)

We are given that cos α = sqrt(7/8) and α is in the fourth quadrant (3π/2 < α < 2π). In the fourth quadrant, sin α is negative. We use the identity sin^2 α + cos^2 α = 1 to find sin α.

sin^2 α = 1 - cos^2 α = 1 - (sqrt(7/8))^2 = 1 - 7/8 = 1/8.

Since α is in the fourth quadrant, sin α = -sqrt(1/8) = -1/sqrt(8) = -1/(2*sqrt(2)) = -sqrt(2)/4.

Now we use the double angle formula for sine: sin 2α = 2 sin α cos α.

sin 2α = 2 * (-sqrt(2)/4) * (sqrt(7/8)) = 2 * (-sqrt(2)/4) * (sqrt(7)/(2*sqrt(2))) = -sqrt(7)/4.