Question

Докажите тождество cosx=1-tg^2*x/2 дробь 1+tg^2*x/2 заранее ♥

Answer 1

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

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

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

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

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

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

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

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

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

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

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

$cos(x)=cos(x)$