Question

Rút gọn biểu thức:

\(B=cos\left(\frac{\pi}{4}+a\right).cos\left(\frac{\pi}{4}-a\right)+\frac{1}{2}sin^2\)

\(C=cos\left(\frac{\pi}{2}-a\right).sin\left(\frac{\pi}{2}-b\right)-sin\left(a-b\right)\)

Answer 1

áp dụng công thức biến đổi tích thành tổng:

\(B=cos\left(\frac{\pi}{4}+a\right).cos\left(\frac{\pi}{4}-a\right)+\frac{1}{2}sin^2a\)

\(B=\frac{1}{2}\left(cos2a+cos\frac{\pi}{2}\right)+\frac{1}{2}sin^2a\)

\(B=\frac{1}{2}cos2a+\frac{1}{2}sin^2a=\frac{1}{2}\left(cos^2a-sin^2a+sin^2a\right)\)

\(B=\frac{1}{2}cos^2a\)

Answer 2

\(C=cos\left(\frac{\pi}{2}-a\right).sin\left(\frac{\pi}{2}-b\right)-sin\left(a-b\right)\)

\(C=sina.cosb-\left(sina.cosb+sinb.cosa\right)\)

\(C=sinb.cosa\)