在△ABC中,AC/AB=cosB/cosC. (Ⅰ)证明B=C: (Ⅱ)若cosA=-1/3,求sin(4B+π3)的值.
在△ABC中,
=
.
(Ⅰ)证明B=C:
(Ⅱ)若cosA=-
,求sin(4B+
)的值.
AC |
AB |
cosB |
cosC |
(Ⅰ)证明B=C:
(Ⅱ)若cosA=-
1 |
3 |
π |
3 |
数学人气:255 ℃时间:2020-01-31 21:58:28
优质解答
(Ⅰ)证明:在△ABC中,由正弦定理及已知得sinBsinC=cosBcosC.于是sinBcosC-cosBsinC=0,即sin(B-C)=0.因为-π<B-C<π,从而B-C=0.所以B=C;(Ⅱ)由A+B+C=π和(Ⅰ)得A=π-2B,故cos2B=-cos(π-2B)=-cos...
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