Đáp án:
$B=\dfrac{4+\sqrt{2}}{6}$
Giải thích các bước giải:
$B=\sin\left ( 2a+\dfrac{\pi}{4} \right )\\
=\sin2a\cos\dfrac{\pi}{4}+\cos2a\sin\dfrac{\pi}{4}\\
=\dfrac{\sqrt{2}}{2}.2\sin a\cos a+\dfrac{\sqrt{2}}{2}.(1-2\sin^2a)\\
=\sqrt{2}.\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{2}}{2}\left [1-2.\left (\dfrac{1}{\sqrt{3}} \right )^2 \right ]\\
=\dfrac{2}{3}+\dfrac{\sqrt{2}}{6}\\
=\dfrac{4+\sqrt{2}}{6}$
