Ensino Médio ⇒ Trigonometria Tópico resolvido

[futuromilitar]

Provar a igualdade:

Código: Selecionar tudo

[tex3]arc sen \frac{1}{\sqrt{5}}+arc \cos\frac{3}{\sqrt{10}}=\frac{\pi }{4}[/tex3]

[futuromilitar]

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[alevini98]

[tex3]\arcsen{\frac{1}{\sqrt5}}=a\\\\\boxed{\sen{a}=\frac{1}{\sqrt5}}\\\\\sen^2{a}+\cos^2{a}=1\\\\\frac{1}{5}+\cos^2{a}=1\\\\\boxed{\cos{a}=\frac{2}{\sqrt5}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\arcsen{\frac{1}{\sqrt5}}=a\\\\\boxed{\sen{a}=\frac{1}{\sqrt5}}\\\\\sen^2{a}+\cos^2{a}=1\\\\\frac{1}{5}+\cos^2{a}=1\\\\\boxed{\cos{a}=\frac{2}{\sqrt5}}[/tex3]

[tex3]\arccos{\frac{3}{\sqrt{10}}}=b\\\\\boxed{\cos{b}=\frac{3}{\sqrt{10}}}\\\\\sen^2{b}+\cos^2{b}=1\\\\\sen^2{b}+\frac{9}{10}=1\\\\\boxed{\sen{b}=\frac{1}{\sqrt{10}}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\arccos{\frac{3}{\sqrt{10}}}=b\\\\\boxed{\cos{b}=\frac{3}{\sqrt{10}}}\\\\\sen^2{b}+\cos^2{b}=1\\\\\sen^2{b}+\frac{9}{10}=1\\\\\boxed{\sen{b}=\frac{1}{\sqrt{10}}}[/tex3]

[tex3]\arcsen{\frac{1}{\sqrt5}}+\arccos{\frac{3}{\sqrt{10}}}=\frac{\pi}{4}\\\\a+b=\frac{\pi}{4}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\arcsen{\frac{1}{\sqrt5}}+\arccos{\frac{3}{\sqrt{10}}}=\frac{\pi}{4}\\\\a+b=\frac{\pi}{4}[/tex3]

Aplicando seno dos dois lados (poderia ser qualquer outra função trigonométrica, mas iremos optar pelo seno):

Iremos rearranjar apenas o primeiro lado da equação.

[tex3]\sen{(a+b)}=\sen{\frac{\pi}{4}}\\\\\sen{a}\cdot\cos{b}+\sen{b}\cdot\cos{a}=\frac{\sqrt2}{2}\\\\\frac{1}{\sqrt5}\cdot\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{10}}\cdot\frac{2}{\sqrt5}=\frac{\sqrt2}{2}\\\\\frac{3}{5\sqrt2}+\frac{2}{5\sqrt2}=\frac{\sqrt2}{2}\\\\\frac{5}{5\sqrt2}=\frac{\sqrt2}{2}\\\\\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\\\\\boxed{\frac{\sqrt2}{2}=\frac{\sqrt2}{2}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sen{(a+b)}=\sen{\frac{\pi}{4}}\\\\\sen{a}\cdot\cos{b}+\sen{b}\cdot\cos{a}=\frac{\sqrt2}{2}\\\\\frac{1}{\sqrt5}\cdot\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{10}}\cdot\frac{2}{\sqrt5}=\frac{\sqrt2}{2}\\\\\frac{3}{5\sqrt2}+\frac{2}{5\sqrt2}=\frac{\sqrt2}{2}\\\\\frac{5}{5\sqrt2}=\frac{\sqrt2}{2}\\\\\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\\\\\boxed{\frac{\sqrt2}{2}=\frac{\sqrt2}{2}}[/tex3]