Bài 3 trang 154 sgk đại số 10


Nội dung bài giảng

Bài 3. Rút gọn các biểu thức

a) \(\sin(a + b) + \sin(\frac{\pi}{2}- a)\sin(-b)\).

b) \(cos(\frac{\pi }{4} + a)\cos( \frac{\pi}{4} - a) +  \frac{1 }{2} \sin^2a\)

c) \(\cos( \frac{\pi}{2} - a)\sin( \frac{\pi}{2} - b) - \sin(a - b)\)

Giải

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

b) \(\cos( \frac{\pi }{4} + a)\cos(\frac{\pi }{4}- a) + \frac{1 }{2}\sin^2a\)

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

\( =\frac{1}{2}\cos 2a +  \frac{1}{4}(1 - \cos 2a) = \frac{1+\cos 2a}{4 }= \frac{1 }{2}\cos^2 a\)

c) \(\cos( \frac{\pi}{2} - a)\sin( \frac{\pi}{2} - b) - \sin(a - b) = \sin a\cos b - \sin a\cos b + \sin b\cos a\)

                                                               \(= \sin b\cos a\)