Question

sin^2(15Pi/4-2x)-cos^2(17Pi/4-2x)= упростить. с решением

Answer 1

sin²(15π/4 -2x)-cos²(17π4 -2x)=sin²(16π/4 -π/4 -2x)-cos²(16π/4+π/4 -2x)=
=sin²(4π-(π/4+2x))-cos²(4π+π/4-2x)=sin²(π/4+2x)-cos²(π/4-2x)=
=(sinπ/4·cos2x+cosπ/4·sin2x)²-(cosπ/4·cos2x+sinπ/4·sin2x)²=
=(√2/2·cos2x+√2/2·sin2x)²-(√2/2·cos2x+√2/2·sin2x)²=
=[√2/2·(cos2x+sin2x)]²-[√2/2·(cos2x+sin2x)]²=0;