Question

Calcule cos(2 arc sen 3/5) 

Answer 1

Seja α o arco em questão

$\displaystyle \arcsin{\frac{3}{5}}=\alpha~\Leftrightarrow~\sin{\alpha}=\frac{3}{5}~~\wedge~~0\ \textless \ \alpha\ \textless \ \frac{\pi}{4}$

Logo


$\displaystyle \cos{\left(2\arcsin{\frac{3}{5}}\right)}=\cos{2\alpha}=1-2\sin^2{\alpha}=1-2\left(\frac{3}{5}\right)^2=\frac{7}{25}$

Answer 2

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$\sf cos\bigg(2~arc~sen\bigg(\dfrac{3}{5}\bigg)\bigg)\\\underline{\rm fac_{\!\!,}a}~arc~sen\bigg(\dfrac{3}{5}\bigg)=\theta\implies sen(\theta)= \dfrac{3}{5}\\\sf cos\bigg(2~arc~sen\bigg(\dfrac{3}{5}\bigg)\bigg)=cos(2\theta)\\\sf cos(2\theta)=1-2sen^2(\theta)\\\sf cos(2\theta)=1-2\cdot\bigg(\dfrac{3}{5}\bigg)^2\\\sf cos(2\theta)=1-2\cdot\dfrac{9}{25}=\dfrac{25-18}{25}\\\sf cos(2\theta)=\dfrac{7}{25}\\\underline{\rm portanto:}$

$\Large\boxed{\boxed{\boxed{\boxed{\sf cos\bigg(2~arc~sen\bigg(\dfrac{3}{5}\bigg)\bigg)=\dfrac{7}{25}}}}}\blue{\checkmark}$