Question

Cum calculez: arccos 1/3 + arccos (-1/3)

Answer 1

Salut,

$x=arccos\dfrac{1}{3}+arccos\left(-\dfrac{1}{3}\right)\Rightarrow cosx=cos\left[arccos\dfrac{1}{3}+arccos\left(-\dfrac{1}{3}\right)\right]=\\\\=\dfrac{1}{3}\cdot\left(-\dfrac{1}3\right)-sin\left(arccos\dfrac{1}3\right)\cdot sin\left[arccos\left(-\dfrac{1}{3}\right)\right]=\\\\=-\dfrac{1}9-\sqrt{1-\left(\dfrac{1}3\right)^2}\cdot\sqrt{1-\left(-\dfrac{1}3\right)^2}=-\dfrac{1}9-\left(1-\dfrac{1}9\right)=-1\Rightarrow\\\\\Rightarrow cosx=-1\Rightarrow\;x=\pi.$

Green eyes.