Доказать B3 с лева 2cos^2beta

(2cos²2β+cos5β-1) / (sin5β+2cos2βsin2β)=ctg (9β/2) 2cos²2β+cos5β-1) / (sin5β+2cos2βsin2β)=(2cos²2β+cos5β-sin²2β-cos²2β) / (sin5β+sin4β)=(cos²2β-sin²2β+cos5β) / (sin5β+sin4β)=(cos4β+cos5β) / (sin5β+sin4β)=2cos (9β/2)*cos (β/2) /2sin (9β/2)*cos (β/2)=cos (9β/2) /sin (9β/2)=ctg (9β/2). ctg (9β/2)=ctg (9β/2) — доказано!