Resposta:
a) [tex]\cos(2x)=2\cos^2 x-1.[/tex]
b)
[tex]\mathrm{tg}(a+b)=\left\{\begin{array}{ll} \dfrac{\mathrm{tg\,}a+\mathrm{tg\,}b}{1-\mathrm{tg\,}a\cdot \mathrm{tg\,}b}&\quad\mathrm{se~}\cos a\ne 0\mathrm{~e~}\cos b\ne 0\\\\ -\,\dfrac{1}{\mathrm{tg\,}a}&\quad\mathrm{se~}\cos a\ne 0\mathrm{~e~}\cos b=0\\\\ -\,\dfrac{1}{\mathrm{tg\,}b}&\quad\mathrm{se~}\cos a=0\mathrm{~e~}\cos b\ne 0\\\\ 0&\quad\mathrm{se~}\cos a=\cos b=0\end{array}\right.[/tex]
Explicação passo a passo:
a)
[tex]\begin{array}{l} \cos(2x)=\cos(x+x)\\\\ \cos(2x)=\cos x\cdot \cos x-\mathrm{sen\,}x\cdot \mathrm{sen\,}x\\\\ \cos(2x)=\cos^2 x-\mathrm{sen^2\,}x\end{array}[/tex]
Pela relação trigonométrica fundamental, podemos substituir [tex]\mathrm{sen^2\,}x=1-\cos^2 x,[/tex] e a identidade acima fica
[tex]\begin{array}{l} \cos(2x)=\cos^2 x-(1-\cos^2 x)\\\\ \cos(2x)=\cos^2 x-1+\cos^2 x\\\\ \cos(2x)=2\cos^2 x-1\qquad\checkmark\end{array}[/tex]
b)
Aplicando a definição de tangente, temos
[tex]\begin{array}{l} \mathrm{tg}(a+b)=\dfrac{\mathrm{sen}(a+b)}{\cos(a+b)}\\\\ \mathrm{tg}(a+b)=\dfrac{\mathrm{sen\,}a\cdot \cos b+\mathrm{sen\,}b\cdot \cos a}{\cos a\cdot \cos b-\mathrm{sen\,}a\cdot \mathrm{sen\,}b}\qquad\mathrm{(i)}\end{array}[/tex]
Devemos estudar os casos abaixo:
- Se [tex]\cos a=0[/tex] e [tex]\cos b=0[/tex]
a identidade fica
[tex]\begin{array}{l}\mathrm{tg}(a+b)=\dfrac{\mathrm{sen\,}a\cdot 0+\mathrm{sen\,}b\cdot 0}{0\cdot 0-\mathrm{sen\,}a\cdot \mathrm{sen\,}b}\\\\ \mathrm{tg}(a+b)=\dfrac{0+0}{0-\mathrm{sen\,}a\cdot \mathrm{sen\,}b}\\\\ \mathrm{tg}(a+b)=\dfrac{0}{-\mathrm{sen\,}a\cdot \mathrm{sen\,}b}=0\qquad\checkmark\end{array}[/tex]
- Se [tex]\cos a=0[/tex] e [tex]\cos b\ne 0[/tex]
a identidade fica
[tex]\begin{array}{l} \mathrm{tg}(a+b)=\dfrac{\mathrm{sen\,}a\cdot \cos b+\mathrm{sen\,}b\cdot 0}{0\cdot \cos b-\mathrm{sen\,}a\cdot \mathrm{sen\,}b}\\\\ \mathrm{tg}(a+b)=\dfrac{\mathrm{sen\,}a\cdot \cos b+0}{0-\mathrm{sen\,}a\cdot \mathrm{sen\,}b}\\\\ \mathrm{tg}(a+b)=\dfrac{\mathrm{sen\,}a\cdot \cos b}{-\mathrm{sen\,}a\cdot \mathrm{sen\,}b}\\\\ \mathrm{tg}(a+b)=\dfrac{\cos b}{-\mathrm{sen\,}b}\\\\ \mathrm{tg}(a+b)=\dfrac{1}{-\frac{\mathrm{sen\,}b}{\cos b}}\\\\ \mathrm{tg}(a+b)=-\,\dfrac{1}{\mathrm{tg\,}b}\qquad\checkmark\end{array}[/tex]
- Se [tex]\cos a\ne 0[/tex] e [tex]\cos b=0[/tex]
Procedendo de forma análoga ao caso anterior, obtemos
[tex]\mathrm{tg}(a+b)=-\,\dfrac{1}{\mathrm{tg\,}a}\qquad\checkmark[/tex]
- Se [tex]\cos a\ne 0[/tex] e [tex]\cos b\ne 0[/tex]
Coloque [tex]\cos a\cdot \cos b[/tex] em evidência no numerador e no denominador, e a identidade fica
[tex]\begin{array}{l} \mathrm{tg}(a+b)=\dfrac{\cos a\cdot \cos b\cdot (\frac{\mathrm{sen\,}a}{\cos a}+\frac{\mathrm{sen\,}b}{\cos b})}{\cos a\cdot \cos b\cdot (1-\frac{\mathrm{sen\,}a}{\cos a}\cdot \frac{\mathrm{sen\,}b}{\cos b})}\\\\ \mathrm{tg}(a+b)=\dfrac{\mathrm{tg\,}a+\mathrm{tg\,}b}{1-\mathrm{tg\,}a\cdot \mathrm{tg\,}b}\qquad\checkmark\end{array}[/tex]
Dúvidas? Comente.
Bons estudos!