Question

Calcule sen 105°, cos 105°, tg 105°.

Answer 1

Resposta:

Explicação passo-a-passo:

105° pertence ao 2° quadrante, onde:

sen105° > 0, cos105° < 0 e tg105º < 0

Reduzindo ao primeiro quadrante, fica:

180° - 105° = 75°

$sen105\°=sen(30\°+45\°)=sen30\°cos45\°+sen45\°cos30\°=\frac{1}{2}*\frac{\sqrt{2} }{2} +\frac{\sqrt{2} }{2}*\frac{\sqrt{3} }{2} =\frac{\sqrt{2} }{4} +\frac{\sqrt{6} }{4} =\frac{\sqrt{2}+\sqrt{6} }{4}$            

$cos105\° = cos(30\°+45\°)=cos30\°cos45\°-sen30\°sen45\°=\frac{\sqrt{3} }{2}*\frac{\sqrt{2} }{2} -\frac{1}{2} *\frac{\sqrt{2} }{2}=\frac{\sqrt{6} }{4 }-\frac{\sqrt{2} }{4}=\frac{\sqrt{6}-\sqrt{2} }{4}$

$tg105\°=\frac{sen105\°}{cos105\°} \\\\tg105\°=\frac{\sqrt{2}+\sqrt{6} }{4} :\frac{\sqrt{6} -\sqrt{2} }{\diagup\!\!\!\!4}*\frac{\diagup\!\!\!\!4}{\sqrt{6}-\sqrt{2} } =\frac{(\sqrt{6}+\sqrt{2})(\sqrt{6}+\sqrt{2}) }{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2}) } =\frac{(\sqrt{6}+\sqrt{2})^2 }{\sqrt{6^2}-\sqrt{2^2} }=\frac{6+2\sqrt{12}+2 }{6-2}=\frac{8+4\sqrt{3} }{4} =\frac{\diagup\!\!\!4(2+\sqrt{3}) }{\diagup\!\!\!\!4}= 2+\sqrt{3}$