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

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