Usando as fórmulas trigonométricas, calcule: sen^2 ( 3π/8).... Pergunta de ideia deMMchar

January 2020 1 35 Report

Usando as fórmulas trigonométricas, calcule:

sen^2 ( 3π/8)

Lista de comentários

alevini $\mathsf{sen^2\dfrac{3\pi}{8}}$

$\mathsf{sen^267,5^o}$

$\mathsf{sen^2(60^o+7,5^o)}$

Calculando o seno de 7,5 a partir da fórmula de arco metade:

$\mathsf{cos^2\dfrac{\alpha}{2}=\dfrac{1+cos\mbox{ }\alpha}{2}}$

$\mathsf{sen^2\dfrac{\alpha}{2}=\dfrac{1-cos\mbox{ }\alpha}{2}}$

$\mathsf{cos^215^o=\dfrac{1+cos\mbox{ }30^o}{2}}$

$\mathsf{cos^215^o=\dfrac{1+\dfrac{\sqrt3}{2}}{2}}$

$\mathsf{cos^215^o=\dfrac{2+\sqrt3}{4}}$

$\boxed{\mathsf{cos\mbox{ }15^o=\dfrac{\sqrt{2+\sqrt3}}{2}}}$

$\mathsf{cos^27,5^o=\dfrac{1+cos\mbox{ }15^o}{2}}$

$\mathsf{cos^27,5^o=\dfrac{1+\dfrac{\sqrt{2+\sqrt3}}{2}}{2}}$

$\mathsf{cos^27,5^o=\dfrac{2+\sqrt{2+\sqrt3}}{4}}$

$\mathsf{cos\mbox{ }7,5^o=\dfrac{\sqrt{2+\sqrt{2+\sqrt3}}}{2}}$

Calculando o seno de 7,5:

$\boxed{\mathsf{sen^2x+cos^2x=1}}$

$\mathsf{sen^27,5^o+\dfrac{2+\sqrt{2+\sqrt3}}{4}=1}$

$\mathsf{sen^27,5^o=1-\dfrac{2+\sqrt{2+\sqrt3}}{4}}$

$\mathsf{sen^27,5^o=\dfrac{2-\sqrt{2+\sqrt3}}{4}}$

$\boxed{\mathsf{sen\mbox{ }7,5^o=\dfrac{\sqrt{2-\sqrt{2+\sqrt3}}}{2}}}$

Agora, fazendo adição de arcos:

$\boxed{\mathsf{cos\mbox{ }(\alpha+\beta)=cos\mbox{ }\alpha\cdot cos\mbox{ }\beta-sen\mbox{ }\alpha\cdot sen\mbox{ }\beta}}$

$\mathsf{cos\mbox{ }67,5^o=\dfrac{1}{2}\cdot\dfrac{\sqrt{2+\sqrt{2+\sqrt3}}}{2}-\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt{2-\sqrt{2+\sqrt3}}}{2}}$

$\mathsf{cos\mbox{ }67,5^o=\dfrac{\sqrt{2+\sqrt{2+\sqrt3}}}{4}-\dfrac{\sqrt{6-3\sqrt{2+\sqrt3}}}{4}}$

$\mathsf{cos\mbox{ }67,5^o=\dfrac{\sqrt{2+\sqrt{2+\sqrt3}}-\sqrt{6-3\sqrt{2+\sqrt3}}}{4}}$

Elevando ao quadrado:

$\mathsf{cos^267,5^o=\dfrac{(\sqrt{2+\sqrt{2+\sqrt3}})^2-2\cdot\sqrt{2+\sqrt{2+\sqrt3}}\cdot\sqrt{6-3\sqrt{2+\sqrt3}}+(\sqrt{6-3\sqrt{2+\sqrt3}})^2}{4^2}}$

$\mathsf{cos^267,5^o=\dfrac{2+\sqrt{2+\sqrt3}-2\cdot\sqrt{2+\sqrt{2+\sqrt3}}\cdot3\cdot\sqrt{2-\sqrt{2+\sqrt3}}+6-3\sqrt{3+\sqrt3}}{16}}$

$\mathsf{cos^267,5^o=\dfrac{8-2\sqrt{2+\sqrt3}-6\cdot\sqrt{2+\sqrt{2+\sqrt3}}\cdot\sqrt{2-\sqrt{2+\sqrt3}}}{16}}$

$\mathsf{cos^267,5^o=\dfrac{8-2\sqrt{2+\sqrt3}-6\sqrt{4-2+\sqrt3}}{16}}$

$\mathsf{cos^267,5^o=\dfrac{8-2\sqrt{2+\sqrt3}-6\sqrt{2+\sqrt3}}{16}}$

$\mathsf{cos^267,5^o=\dfrac{8-8\sqrt{2+\sqrt3}}{16}}$

$\mathsf{cos\mbox{ }67,5^o=\sqrt{\dfrac{4\cdot(2-2\sqrt{2+\sqrt3}}{16}}}$

$\boxed{\mathsf{cos\mbox{ }67,5^o=\dfrac{\sqrt{2-2\sqrt{2+\sqrt3}}}{2}}}$

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