Question

Помогите понять: у=10cos^2x-6sinx*cosx+2sin^2x
у=10cos^2x+2sin^2x-3sin2x
y=10cos^2x+2-2cos^2x-3sin2x
Откуда берётся 2-2cos^2x?

Answer 1

$y=10cos^2x-underbrace {6, sinx, cosx}_{3cdot 2, sinx, cosx}+2sin^2x\\y=10cos^2x-3cdot underbrace {2, sinx, cosx}_{sin2x}+2sin^2x; ; ,; ; (sin^2x+cos^2x=1; )\\y=10cos^2x-3cdor sin2x+2underbrace {sin^2x}_{1-cos^2x}\\y=10cos^2x-3sin2x+2(1-cos^2x)\\y=10cos^2x-3sin2x+2-2cos^2x\\underline {y=10cos^2x+2-2cos^2x+2sin2x}\\y=8cos^2x+2+2sin2x$