Question

POR FAVOR ME AJUDEM, ESTOU ESTUDANDO CÁLCULO I, E ESSES EXERCÍCIOS SÃO DO LIVRO DO GUIDORIZZI, VOLUME 1

Mostre que, para todo x, com $\frac{x}{2}$ ≠ $0$, tem-se:


a)

$senx = \frac{2*tg(\frac{x}{2}) }{1+ tg^2(\frac{x}{2}) }$


b)

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

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle senx=\frac{2tg(\frac{x}{2})}{1+tg^2(\frac{x}{2})} =\frac{2tg(\frac{x}{2})}{1+\frac{sen^2(\frac{x}{2})}{cos^2(\frac{x}{2})}} =\frac{2tg(\frac{x}{2})}{\frac{cos^2(\frac{x}{2})+sen^2(\frac{x}{2})}{cos^2(\frac{x}{2})}} =\frac{2tg(\frac{x}{2})}{\frac{1}{cos^2(\frac{x}{2})} } =2tg(\frac{x}{2})cos^2(\frac{x}{2})=$

$\displaystyle 2.\frac{sen(\frac{x}{2})}{cos(\frac{x}{2})}.cos^2(\frac{x}{2})=2sen(\frac{x}{2})cos(\frac{x}{2})$

Propriedades:

sen²x+cos²x=1

senA+senB=2.sen[1/2(A+B)]cos[1/2(A-B)]

comparando

1/2(A+B)=x/2 => A+B=x (I)

1/2(A-b)=x/2 => A-B=x (II)

(I)+(II)

2A=2x => A=x

Substituindo A=x em (I)

B=0

Substituindo esses valores A=0 e B=0 nas propriedades:

$\displaystyle senx=2sen(\frac{x}{2})cos(\frac{x}{2})=2.\frac{senx+~sen0}{2} =senx\\\\\\senx=senx~~ (verdadeiro)$

b)

$\displaystyle cosx=\frac{1-tg^2(\frac{x}{2})}{1+tg^2(\frac{x}{2})} =\frac{1-\frac{sen^2(\frac{x}{2})}{cos^2(\frac{x}{2})}}{1+\frac{sen^2(\frac{x}{2})}{cos^2(\frac{x}{2})}} =\frac{\frac{cos^2(\frac{x}{2})-sen^2(\frac{x}{2})}{cos^2(\frac{x}{2})} }{\frac{cos^2(\frac{x}{2})+sen^2(\frac{x}{2})}{cos^2(\frac{x}{2})}} =\frac{\frac{cos^2(\frac{x}{2})-sen^2(\frac{x}{2})}{cos^2(\frac{x}{2})} }{\frac{1}{cos^2(\frac{x}{2})}} =cos^2(\frac{x}{2})-sen^2(\frac{x}{2})=cos2(\frac{x}{2} )\\\\\\cosx= cosx ~~~(verdadeiro)$

Propriedades:

sen²x+cos²x=1

cos2A=cos²A-sen²A