Question

Докажите тождество

cosx=1-tg^2*x/2 дробь 1+tg^2*x/2

Заранее спасибо♥

Answer 1

$cos(x)=frac{1-tg^2(frac{x}{2})}{1+tg^2(frac{x}{2})}$

$cos(x)(1+tg^2(frac{x}{2}))=1-tg^2(frac{x}{2})$

$cos(x)(1+(frac{sin(frac{x}{2})}{cos(frac{x}{2})})^2)=1-(frac{sin(frac{x}{2})}{cos(frac{x}{2})})^2$

$cos(x)(1+frac{sin^2(frac{x}{2})}{cos^2(frac{x}{2})})=1-frac{sin^2(frac{x}{2})}{cos^2(frac{x}{2})}$

$cos(x)frac{cos^2(frac{x}{2})+sin^2(frac{x}{2})}{cos^2(frac{x}{2})}=1-frac{sin^2(frac{x}{2})}{cos^2(frac{x}{2})}$

$frac{cos(x)(cos^2(frac{x}{2})+sin^2(frac{x}{2}))}{cos^2(frac{x}{2})}=frac{cos^2(frac{x}{2})-sin^2(frac{x}{2})}{cos^2(frac{x}{2})}$

$cos(x)(cos^2(frac{x}{2})+sin^2(frac{x}{2}))=cos^2(frac{x}{2})-sin^2(frac{x}{2})$

$cos(x)(frac{1+cos(x)}{2}+frac{1-cos(x)}{2})=cos^2(frac{x}{2})-sin^2(frac{x}{2})$

$cos(x)(frac{1}{2}+frac{cos(x)}{2}+frac{1}{2}-frac{cos(x)}{2})=cos^2(frac{x}{2})-sin^2(frac{x}{2})$

$cos(x)=cos^2(frac{x}{2})-sin^2(frac{x}{2})$

$cos(x)=frac{1}{2}+frac{cos(x)}{2}+frac{cos(x)}{2}-frac{1}{2}$

$cos(x)=cos(x)$