Bài 7 trang 155 SGK Đại số 10

Biến đổi thành tích các biểu thức sau

a) \(1 - \sin x\);                    b) \(1 + \sin x\);

c) \(1 + 2\cos x\);                  d) \(1 - 2\sin x\)  

Lời giải

a) \(1 - \sin x = \sin \frac{\pi }{2} - \sin x \)

\(= 2\cos \frac{\frac{\pi }{2}+x}{2}\sin \frac{\frac{\pi}{2}-x}{2}\)

\(= 2 \cos \left ( \frac{\pi }{4} +\frac{x}{2}\right )\sin\left ( \frac{\pi }{4} -\frac{x}{2}\right )\)

b) \(1 + \sin x = \sin \frac{\pi }{2} + \sin x \)

\(= 2\sin \left ( \frac{\pi }{4} +\frac{x}{2}\right )\cos \left ( \frac{\pi }{4} -\frac{x}{2}\right )\)

c) \(1 + 2\cos x = 2( \frac{1}{2} + \cos x) \)

\(= 2(\cos \frac{\pi}{3} + \cos x) \)

\(= 4\cos \left ( \frac{\pi }{6} +\frac{x}{2}\right )\cos \left ( \frac{\pi }{6} -\frac{x}{2}\right )\)

d) \(1 - 2\sin x = 2( \frac{1}{2} - \sin x) \)

\(= 2(\sin \frac{\pi}{6} - \sin x)\)

\(= 4\cos \left ( \frac{\pi }{12} +\frac{x}{2}\right )\sin \left ( \frac{\pi }{12} -\frac{x}{2}\right )\)