Question

Se a tg x/2 = a calcule cos(X)

Answer 1

como não especificaram o quadrante no exercício, vou admitir que o arco pertence ao primeiro quadrante.

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

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

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

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

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

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

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

Answer 2

Partimos da igualdade dada:

    $\mathsf{tg\!\left(\dfrac{x}{2}\right)=a}$

Aplicando a definição de tangente,

    $\mathsf{\Longleftrightarrow\quad \dfrac{sen(\frac{x}{2})}{cos(\frac{x}{2})}=a}\\\\\\ \mathsf{\Longrightarrow\quad sen\!\left(\dfrac{x}{2}\right)=a\cdot cos\!\left(\dfrac{x}{2}\right)}$

Eleve os dois lados ao quadrado:

    $\mathsf{\Longrightarrow\quad sen^2\!\left(\dfrac{x}{2}\right)=a^2\cdot cos^2\!\left(\dfrac{x}{2}\right)}\\\\\\ \mathsf{\Longleftrightarrow\quad 1-cos^2\!\left(\dfrac{x}{2}\right)=a^2\cdot cos^2\!\left(\dfrac{x}{2}\right)}\\\\\\ \mathsf{\Longleftrightarrow\quad 1=a^2\cdot cos^2\!\left(\dfrac{x}{2}\right)+cos^2\!\left(\dfrac{x}{2}\right)}\\\\\\ \mathsf{\Longleftrightarrow\quad 1=(a^2+1)\cdot cos^2\!\left(\dfrac{x}{2}\right)}$

Aplique uma das identidades para o cosseno do arco metade:

    $\mathsf{cos^2(\theta)=\dfrac{1+cos(2\theta)}{2}}$

para $\mathsf{\theta=\dfrac{x}{2},}$  e obtemos

    $\mathsf{\Longleftrightarrow\quad 1=(a^2+1)\cdot \dfrac{1+cos(x)}{2}}\\\\\\ \mathsf{\Longleftrightarrow\quad 2=(a^2+1)\cdot (1+cos(x))}\\\\ \mathsf{\Longleftrightarrow\quad 2=(a^2+1)\cdot 1+(a^2+1)\cdot cos(x)}\\\\ \mathsf{\Longleftrightarrow\quad 2=a^2+1+(a^2+1)\cdot cos(x)}\\\\ \mathsf{\Longleftrightarrow\quad 2-(a^2+1)=(a^2+1)\cdot cos(x)}\\\\ \mathsf{\Longleftrightarrow\quad 2-a^2-1=(a^2+1)\cdot cos(x)}\\\\ \mathsf{\Longleftrightarrow\quad 1-a^2=(a^2+1)\cdot cos(x)}\\\\ \mathsf{\Longleftrightarrow\quad cos(x)=\dfrac{1-a^2}{a^2+1}\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)