Đáp án:
$\dfrac{\sqrt{2}}{10}$
Giải thích các bước giải:
$\cos^2a+\sin^2a=1\\
\Rightarrow \sin^2a=1-\cos^2a=1-\left (\dfrac{4}{5} \right )^2=\dfrac{9}{25}\\
\Rightarrow \sin a=\pm \dfrac{3}{5}$
Do $270<a<360\Rightarrow \sin a<0$
$\Rightarrow \sin a=-\dfrac{3}{5}\\
\sin \left ( a+\dfrac{\pi}{4} \right )\\
=\sin a\cos\dfrac{\pi}{4}+\cos a\cos\dfrac{\pi}{4}\\
=-\dfrac{3}{5}.\dfrac{\sqrt{2}}{2}+\dfrac{4}{5}.\dfrac{\sqrt{2}}{2}\\
=\dfrac{\sqrt{2}}{10}$
