2 câu trả lời
$\sin10^o.\sin50^o.\sin70^o$
$=\dfrac{-1}{2}(\cos80^o-\cos60^o).\sin50^o$
$=\dfrac{-1}{2}\sin50^o\cos80^o+\dfrac{1}{4}\sin50^o$
$=\dfrac{-1}{4}(\sin130^o-\sin30^o)+\dfrac{1}{4}\sin50^o$
$=\dfrac{-1}{4}\sin50^+\dfrac{1}{4}\sin30^o+\dfrac{1}{4}\sin50^o$
$=\dfrac{1}{8}$
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