Giải Khám phá 4 trang 22 Toán 11 tập 1 Chân trời

$cosa.cosb = \frac{1}{2}[cos(a-b)+cos(a+b)]$

$cos\frac{\alpha +\beta }{2} .cos\frac{\alpha -\beta }{2} = \frac{1}{2}[cos(\frac{\alpha +\beta }{2} -\frac{\alpha -\beta }{2})+cos(\frac{\alpha +\beta }{2} +\frac{\alpha -\beta }{2})]$

$cos\frac{\alpha +\beta }{2} .cos\frac{\alpha -\beta }{2} = \frac{1}{2}.(cos\alpha +cos\beta )$

$sina.sinb = \frac{1}{2}[cos(a+b)-cos(a-b)]$

$sin\frac{\alpha +\beta }{2} .sin\frac{\alpha -\beta }{2} = \frac{1}{2}[cos(\frac{\alpha +\beta }{2} +\frac{\alpha -\beta }{2})-cos(\frac{\alpha +\beta }{2} -\frac{\alpha -\beta }{2})]$

$sin\frac{\alpha +\beta }{2} .sin\frac{\alpha -\beta }{2} = \frac{1}{2}.(cos\alpha-cos\beta )$

$sina.cosb = \frac{1}{2}[sin(a-b)+sin(a+b)]$

$sin\frac{\alpha +\beta }{2} .cos\frac{\alpha -\beta }{2} = \frac{1}{2}[sin(\frac{\alpha +\beta }{2} -\frac{\alpha -\beta }{2})+sin(\frac{\alpha +\beta }{2} +\frac{\alpha -\beta }{2})]$

$sin\frac{\alpha +\beta }{2} .cos\frac{\alpha -\beta }{2} = \frac{1}{2}(sin\beta +sin\alpha )$

$sinb.cosa = \frac{1}{2}[sin(b-a)+sin(a+b)]$

$sin\frac{\alpha -\beta }{2} .cos\frac{\alpha +\beta }{2} = \frac{1}{2}[sin(\frac{\alpha -\beta }{2}-\frac{\alpha +\beta }{2} )+sin(\frac{\alpha +\beta }{2} +\frac{\alpha -\beta }{2})]$

$sin\frac{\alpha -\beta }{2} .cos\frac{\alpha +\beta }{2} = \frac{1}{2}(sin\alpha -sin\beta )$