lượng giác khó

N

newstarinsky

Rút gọn không còn dấu căn:
[TEX]A = \sqrt{2+\sqrt{2+2.cosa}} (0\leq a<2\pi )[/TEX]

Ta có
$A=\sqrt{2+\sqrt{2(1+cosa)}}
=\sqrt{2+2|cos(\frac{a}{2})|}$
TH1 $0\leq a \leq \pi\Rightarrow cos(\frac{a}{2})\geq 0$

$A=\sqrt{2+2cos(\frac{a}{2})}=2cos(\frac{a}{4})$
TH2 $\pi\leq a < 2\pi\Rightarrow cos(\frac{a}{2})\leq 0$
$A=\sqrt{2-2cos(\frac{a}{2})}=2sin(\frac{a}{4})$
 
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