Đáp án:
$\dfrac{1}{2}\cos^2a$
Giải thích các bước giải:
$\cos\left ( \dfrac{\pi}{4}+a \right )\cos\left ( \dfrac{\pi}{4}-a \right )+\dfrac{1}{2}\sin^2a\\
=\left (\cos\dfrac{\pi}{4}\cos a-\sin\dfrac{\pi}{4}\sin a \right )\left ( \cos\dfrac{\pi}{4}\cos a+\sin\dfrac{\pi}{4}\sin a \right )+\dfrac{1}{2}\sin^2a\\
=\left (\dfrac{\sqrt{2}}{2}\cos a-\dfrac{\sqrt{2}}{2}\sin a \right )\left ( \dfrac{\sqrt{2}}{2}\cos a+\dfrac{\sqrt{2}}{2}\sin a \right )+\dfrac{1}{2}\sin^2a\\
=\left ( \dfrac{\sqrt{2}}{2}\cos a\right )^2-\left ( \dfrac{\sqrt{2}}{2}\sin a \right )^2+\dfrac{1}{2}\sin^2a\\
=\dfrac{1}{2}\cos^2a-\dfrac{1}{2}\sin^2a+\dfrac{1}{2}\sin^2a\\
=\dfrac{1}{2}\cos^2a$
