求证在三角形ABC中,(1)sinA=sin(B+C) (2)cosa=-cos(B+C)
人气:315 ℃ 时间:2019-10-19 16:28:20
解答
证明:在△ABC中,有:A+B+C=180°
即:A=180°-B-C
所以:
sinA=sin(180°-B-C)=sin[180°- (B+C)]=sin(B+C)
而
cosA=cos(180°-B-C)=cos[180°- (B+C)]=-cos(B+C)
等式得证!
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