У выражение cos^2(π+x)+cos^2(π/2+x)
sin(π+x)cos(π/2+x)-cos(2π+x)sin(3π/2-x)

1) 1

2) 1

Пошаговое объяснение:

1)

cos(π+x) = -cos(x), cos^2(x) = cos(x)*cos(x) => -cos(x) * -cos(x) = cos^2(x)

cos(π/2+x) = -sin(x), sin^2(x) = sin(x)*sin(x) => -sin(x) * -sin(x) = sin^2(x)

cos^2(π+x)+cos^2(π/2+x) = cos^2(x)+sin^2(x) = 1

2)

sin(π+x) = -sin(x), cos(π/2+x) = -sin(x) => -sin(x) * -sin(x) = sin^2(x)

cos(2π+x) = cos(x), sin(3π/2-x) = -cos(x) => cos(x) * -cos(x) = -cos^2(x)

sin(π+x)cos(π/2+x)-cos(2π+x)sin(3π/2-x) = sin^2(x) + cos^2(x) = 1