Đáp án:
$\cos\dfrac{y}{2}=\dfrac{\sqrt{10+\sqrt{10}}}{2\sqrt{5}}$
Giải thích các bước giải:
$1+\tan^2y=\dfrac{1}{\cos^2y}\\
\Rightarrow \cos^2y=\dfrac{1}{1+\tan^2y}\\
=\dfrac{1}{1+3^2}=\dfrac{1}{10}\\
\Rightarrow \cos y=\pm \dfrac{\sqrt{10}}{10}$
Vì $\dfrac{-\pi}{2}<y<0\Rightarrow \cos y=\dfrac{\sqrt{10}}{10}$
$\cos y=2\cos^2\dfrac{y}{2}-1\\
\Rightarrow \cos^2\dfrac{y}{2}=\dfrac{\cos y+1}{2}\\
=\dfrac{\dfrac{\sqrt{10}}{10}+1}{2}=\dfrac{10+\sqrt{10}}{20}\\
\Rightarrow \cos\dfrac{y}{2}=\pm \dfrac{\sqrt{10+\sqrt{10}}}{2\sqrt{5}}$
Vì $\dfrac{-\pi}{2}<y<0\Rightarrow \dfrac{-pi}{4}<\dfrac{y}{2}<0$
$\Rightarrow \cos\dfrac{y}{2}=\dfrac{\sqrt{10+\sqrt{10}}}{2\sqrt{5}}$
