Question

(2*cos^2 x-1):[4tg(π/4-x) ∙ sin^2 (π/4 + x) ]=?

Answer 1

2cos²(x)-1=cos(2x)=cos²(x)-sin²(x)=(cos(x)+sin(x))(cos(x)-sin(x)).

4tan(π/4-x)=4(1-tan(x))/(1+tan(x))=4(cos(x)-sin(x))/(cos(x)+sin(x));

sin²(π/4+x)=½(cos(x)+sin(x))²=½(1+sin(2x)).

4tan(π/4-x)·sin²(π/4+x)=2(cos(x)-sin(x))(1+sin(2x))/(cos(x)+sin(x)).

(2cos²(x)-1):[4tan(π/4-x)·sin²(π/4+x)]=(2cos²(x)-1)/[4tan(π/4-x)·sin²(π/4+x)],

(cos(x)+sin(x))(cos(x)-sin(x))/2(cos(x)-sin(x))(1+sin(2x))/(cos(x)+sin(x))=

(cos(x)+sin(x))²/2(1+sin(2x))=(1+sin(2x))/2(1+sin(2x))=1/2 or 1:2.