Question

Rút gọn BT giúp mik với mik cảm mơn nhiều ạ

\(\text{sin(π/8+a)^2 - sin(π/8 - a)^2 - \dfrac{\sqrt{2}}{2} sin2a}\)

Answer 1

Đề bài là: \(sin^2\left(\dfrac{\pi}{8}+a\right)-sin^2\left(\dfrac{\pi}{8}-a\right)-\dfrac{\sqrt{2}}{2}sin2a\) đúng không nhỉ?

\(=\dfrac{1}{2}-\dfrac{1}{2}cos\left(\dfrac{\pi}{4}+2a\right)-\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{\pi}{4}-2a\right)-\dfrac{\sqrt{2}}{2}sin2a\)

\(=\dfrac{1}{2}\left[cos\left(\dfrac{\pi}{4}-2a\right)-cos\left(\dfrac{\pi}{4}+2a\right)\right]-\dfrac{\sqrt{2}}{2}sin2a\)

\(=sin\left(\dfrac{\pi}{4}\right).sin2a-\dfrac{\sqrt{2}}{2}sin2a=\dfrac{\sqrt{2}}{2}sin2a-\dfrac{\sqrt{2}}{2}sin2a=0\)