[petras]

Calcular el valor (R:\( \sqrt{{18+12\sqrt2}}+3+2\sqrt2 \))
\( y = cot[\frac{1}{4}arcsin(\frac{1}{3})] \)

[ani_pascual]

Hola:

Calcular el valor (R:\( \sqrt{{18+12\sqrt2}}+3+2\sqrt2 \))
\( y = cot[\frac{1}{4}arcsin(\frac{1}{3})] \)

Spoiler

Sea \( \sen\alpha=\dfrac{1}{3}\Longrightarrow \cos\alpha=\dfrac{2\sqrt{2}}{3}\Longrightarrow \cos\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1+\cos\alpha}{2}}=\sqrt{\dfrac{1+\dfrac{2\sqrt{2}}{3}}{2}} \). Entonces
\( y=\cot\left(\dfrac{1}{4}\arcsen\left(\dfrac{1}{3}\right)\right)=\dfrac{1}{\tan\left(\dfrac{\alpha}{4}\right)}=\dfrac{1}{\tan\left(\dfrac{\alpha /2}{2}\right)}=\dfrac{\sqrt{1+\cos\left(\dfrac{\alpha}{2}\right)}}{\sqrt{1-\cos\left(\dfrac{\alpha}{2}\right)}}=\sqrt{\dfrac{1+\sqrt{\dfrac{1+\dfrac{2\sqrt{2}}{3}}{2}}}{1-\sqrt{\dfrac{1+\dfrac{2\sqrt{2}}{3}}{2}}}}=\sqrt{\dfrac{\sqrt{6}+\sqrt{3+2\sqrt{2}}}{\sqrt{6}-\sqrt{3+2\sqrt{2}}}}=\\\dfrac{\sqrt{6}+\sqrt{3+2\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}=\left(\sqrt{6}+\sqrt{3+2\sqrt{2}}\right)\sqrt{3-2\sqrt{2}}(3+2\sqrt{2})=(\sqrt{6}\sqrt{3-2\sqrt{2}}+1)(3+2\sqrt{2})=\sqrt{6}\sqrt{3-2\sqrt{2}}(3+2\sqrt{2})+3+2\sqrt{2}=\sqrt{6}\sqrt{3+2\sqrt{2}}+3+2\sqrt{2}=\\\boxed{\sqrt{18+12\sqrt{2}}+3+2\sqrt{2}} \)

[cerrar]

Saludos

[petras]

Hola:

Calcular el valor (R:\( \sqrt{{18+12\sqrt2}}+3+2\sqrt2 \))
\( y = cot[\frac{1}{4}arcsin(\frac{1}{3})] \)

Spoiler

Sea \( \sen\alpha=\dfrac{1}{3}\Longrightarrow \cos\alpha=\dfrac{2\sqrt{2}}{3}\Longrightarrow \cos\left(\dfrac{\alpha}{2}\right)=\sqrt{\dfrac{1+\cos\alpha}{2}}=\sqrt{\dfrac{1+\dfrac{2\sqrt{2}}{3}}{2}} \). Entonces
\( y=\cot\left(\dfrac{1}{4}\arcsen\left(\dfrac{1}{3}\right)\right)=\dfrac{1}{\tan\left(\dfrac{\alpha}{4}\right)}=\dfrac{1}{\tan\left(\dfrac{\alpha /2}{2}\right)}=\dfrac{\sqrt{1+\cos\left(\dfrac{\alpha}{2}\right)}}{\sqrt{1-\cos\left(\dfrac{\alpha}{2}\right)}}=\sqrt{\dfrac{1+\sqrt{\dfrac{1+\dfrac{2\sqrt{2}}{3}}{2}}}{1-\sqrt{\dfrac{1+\dfrac{2\sqrt{2}}{3}}{2}}}}=\sqrt{\dfrac{\sqrt{6}+\sqrt{3+2\sqrt{2}}}{\sqrt{6}-\sqrt{3+2\sqrt{2}}}}=\\\dfrac{\sqrt{6}+\sqrt{3+2\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}=\left(\sqrt{6}+\sqrt{3+2\sqrt{2}}\right)\sqrt{3-2\sqrt{2}}(3+2\sqrt{2})=(\sqrt{6}\sqrt{3-2\sqrt{2}}+1)(3+2\sqrt{2})=\sqrt{6}\sqrt{3-2\sqrt{2}}(3+2\sqrt{2})+3+2\sqrt{2}=\sqrt{6}\sqrt{3+2\sqrt{2}}+3+2\sqrt{2}=\\\boxed{\sqrt{18+12\sqrt{2}}+3+2\sqrt{2}} \)

[cerrar]

Saludos

Excelente..agradecido

Saludos