Question

Вариант 3. Номер 1. Заранее благодарю.

Answer 1

sin²a=(1-cos2a)/2=(1-1/3)/2=1/3⇒sina=1/√3
cos²a=(1+cos2a)/2=(1+1/3)/2=4/3⇒cosx=2/√3
tg(a+π/4)=(sina+cosa)/(cosa-sina)=(1/√3+2/√3)/(2/√3-1/√3)=3/√3:1/√3=3

Answer 2

Значение cosx найдено неверно

Answer 3

$cos2a=frac{1}{3}; ,; ain (0,frac{pi}{2})\\tga=frac{sin2a}{1+cos2a}\\sin2a=+sqrt{1-cos^22a}=sqrt{1-frac{1}{9}}=sqrt{frac{8}{9}}=frac{2sqrt2}{3}\\tga=frac{2sqrt2}{3(1+frac{1}{3})}=frac{2sqrt2}{4}=frac{sqrt2}{2}\\tg(a+frac{pi}{4})=frac{1+tga}{1-tga}=frac{1+frac{sqrt2}{2}}{1-frac{sqrt2}{2}}=frac{2+sqrt2}{2-sqrt2}=frac{sqrt2+1}{sqrt2-1}$