Salut,
[tex]cos(arccos)=x;\\\\sin(arccosx)=?\ Fie\ p=arccosx,\ deci\ cosp=x;\\\\sin(arccosx)=sinp=\sqrt{1-cos^2p}=\sqrt{1-x^2}.\\\\E=sin\left(arccos\dfrac{3}5+arccos\dfrac{4}5\right)=sin\left(arccos\dfrac{3}5\right)\cdot cos\left(arccos\dfrac{4}5\right)+\\\\+cos\left(arccos\dfrac{3}5\right)\cdot sin\left(arccos\dfrac{4}5\right)=\sqrt{1-\left(\dfrac{3}5\right)^2}\cdot\dfrac{4}5+\dfrac{3}5\cdot\sqrt{1-\left(\dfrac{4}5\right)^2}=\\\\=\dfrac{4}{5}\cdot\dfrac{4}{5}+\dfrac{3}5\cdot\dfrac{3}5=1.\ Simplu,\ nu\ ?\ :-))).[/tex]
Green eyes.
