Question

Вычислить значения функции y=f(x) в точке х=х0. y=sin²x+cos²2x, x0=-π/12 y=sin5xcos3x, x0=π/8

Answer 1

$y=sin^2x+cos^22x=sin^2x+(cos^2x-sin^2x)^2=sin^2x+cos^4x+sin^4x-2sin^2xcos^2x=sin^2x+(1-sin^2x)^2+sin^4x-0.5sin^22x=1-sin^2x+0.5(cos2x-1)^2-0.5sin^22x=0.5(1+cos2x+cos4x-2cos2x+1)=0.5(cos4x-cos2x+2)$

$y(-\pi/12)=0.5(cos(\pi/3)-cos(\pi/6)+2)=(5/4)-(\sqrt{3}/4)$

$y=sin5xcos3x=0.5(sin(5x+3x)+sin(5x-3x))=0.5(sin8x+sin2x)\\\\y(\pi/8)=0.5(sin\pi+sin(\pi/4))=\sqrt{2}/4$