Question

Sobre transformações trigonométricas;

I. sen75° = sen30° + sen45°

II. sen 75° = sen30° . sen45° + cos45°.cos30°

III. tg75° = ( tg30° + tg45° ) / ( 1 - tg30° . tg45° )

É correto afirmar:

a) Somente III está correta.
b) I e II estão corretas.
c) Todas estão corretas.
d) Todas estão incorretas.
e) A afirmação II está correta.

Answer 1

Boa tarde Roger!

Solução!

Vamos escrever o seno e o cosseno de 75º usando uma identidade.

$Sen(75\º)=(30\º+45\º)\\\\\\\\ Cos(a)\times sen(b)+Sen(b) \times cos(a)\\\\\\\ Sen(30\°)\times sen(45\º)+Sen(45\º) \times cos(30\º)\\\\\\\\ \dfrac{ \sqrt{1} }{2}\times \dfrac{ \sqrt{2} }{2} + \dfrac{ \sqrt{2} }{2} \times \dfrac{ \sqrt{3} }{2} \\\\\\\\ \dfrac{ \sqrt{2} }{4}+ \dfrac{ \sqrt{6} }{4}\\\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}\\\\\\ Sen(75\º)=\dfrac{ \sqrt{2}+ \sqrt{6} }{4}$


$Cos(75\º)=(30\º+45\º)\\\\\\ Cos(a)\times Cos(b)-Sen(a) \times sen(b)\\\\\\\ Cos(30\º)\times Cos(45\º)-Sen(30\º) \times sen(45\º)\\\\\\\ \dfrac{ \sqrt{3} }{2} \times \dfrac{ \sqrt{2} }{2}- \dfrac{1}{2} \times \dfrac{ \sqrt{2} }{2}\\\\\\ \dfrac{ \sqrt{6} }{4} - \dfrac{ \sqrt{2} }{4} \\\\\\ \dfrac{ \sqrt{6}- \sqrt{2} }{4}\\\\\\ Cos(75\º)= \dfrac{ \sqrt{6}- \sqrt{2} }{4}$


$I\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}=Sen(30\º)+Sen(45\º)\\\\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}= \dfrac{1}{2}+ \dfrac{ \sqrt{2} }{2} \\\\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}= \dfrac{1+ \sqrt{2} }{2} \\\\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4} \ne \dfrac{1+ \sqrt{2} }{2}\Rightarrow \boxed{Falso}$


$II\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}= Sen(30\º) \times Sen(45\º)+ Cos(45\º) \times Cos(30\º)\\\\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}= \dfrac{1}{2} \times \dfrac{ \sqrt{2} }{2} + \dfrac{ \sqrt{2} }{2} \times \dfrac{ \sqrt{3} }{2} \\\\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}= \dfrac{ \sqrt{2} }{4} + \dfrac{ \sqrt{6} }{4}\\\\\\ \dfrac{ \sqrt{2}+ \sqrt{6} }{4}= \dfrac{ \sqrt{2}+ \sqrt{6}}{4}\Rightarrow \boxed{Verdadeiro}$

Na tangente vamos usar a propriedade comutativa da soma para provarmos a igualdade.


$III\\\\ Tag(75\º)= \dfrac{Sen(75\º)}{Cos(75\º)}\\\\\\ Tag(75\º)= \dfrac{\sqrt{2}+ \sqrt{6}}{ \dfrac{4}{ \dfrac{\sqrt{6}- \sqrt{2}}{4} } } \\\\\\\ Tag(75\º)= \dfrac{\sqrt{2}+ \sqrt{6} }{4} \times \dfrac{4}{ \sqrt{6}- \sqrt{2} } \\\\\\ Tag(75\º)= \dfrac{\sqrt{2}+ \sqrt{6} }{ \sqrt{6} - \sqrt{2} }$

Feito isso vamos multiplicar pelo conjugado.

$Tag(75\º)= \dfrac{\sqrt{6}+ \sqrt{2} }{ \sqrt{6} - \sqrt{2} }\\\\\\\\ Tag(75\º)= \dfrac{\sqrt{6}+ \sqrt{2} }{ \sqrt{6} - \sqrt{2} }\times \dfrac{\sqrt{6} + \sqrt{2}}{\sqrt{6} +\sqrt{2} }\\\\\\\\\\ Tag(75\º)= \dfrac{( \sqrt{6}+ \sqrt{2})^{2}}{( \sqrt{6} - \sqrt{2})( \sqrt{6} + \sqrt{2}}$

$Tag(75\º)= \dfrac{6+2\sqrt{12}+2 }{6- \sqrt{12} + \sqrt{12} -2} \\\\\\\\ Tag(75\º)= \dfrac{6+2(2\sqrt{3})+2 }{6 -2} \\\\\\\\ Tag(75\º)= \dfrac{8+4\sqrt{3} }{4} \\\\\\\\ Tag(75\º)= 4( \dfrac{2+\sqrt{3} }{4}) \\\\\\\\ Tag(75\º)= 2+\sqrt{3}$

Finalmente!
$2+\sqrt{3}= \dfrac{tag(30\º)+tag(45\º)}{1-tag(30\º).tag(45\º)}\\\\\\ 2+\sqrt{3}= \dfrac{\dfrac{ \sqrt{3} }{3} +1}{1- \dfrac{ \sqrt{3} }{3}. 1}\\\\\\ 2+\sqrt{3}= \dfrac{\dfrac{ \sqrt{3}+3 }{3} }{3- \dfrac{ \sqrt{3} }{3}}\\\\\\ 2+\sqrt{3}= \dfrac{ \sqrt{3}+3}{3} \times \dfrac{3}{3- \sqrt{3}}\\\\\\ 2+\sqrt{3}= \dfrac{ \sqrt{3}+3}{3- \sqrt{3} }\\\\\\\\ 2+\sqrt{3}= \dfrac{ \sqrt{3}+3}{3- \sqrt{3} }\times\dfrac{ 3+\sqrt{3}}{3+ \sqrt{3} }$

$2+\sqrt{3}= \dfrac{ (\sqrt{3}+3)^{2} }{(3- \sqrt{3})\times(3+ \sqrt{3}) }\\\\\\\ 2+\sqrt{3}= \dfrac{( \sqrt{3})^{2}+2.(3).( \sqrt{3})+(3)^{2}}{(3)^{2}-3 \sqrt{3}+3 \sqrt{3}-( \sqrt{3})^{2}}\\\\\\\ 2+\sqrt{3}= \dfrac{3+6 \sqrt{3}+9}{9-3}\\\\\\\ 2+\sqrt{3}= \dfrac{12+6 \sqrt{3}}{6}\\\\\\\ 2+\sqrt{3}= \dfrac{6(2+ \sqrt{3})}{6}\\\\\\\ 2+\sqrt{3}= 2+ \sqrt{3}\Rightarrow \boxed{Verdadeiro}$

Conclusão: A alternativa II e III estão corretas,porem não consta nas opções.

Boa tarde!
Bons estudos!