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高二数学,三角函数
化简两题
sin(a+b)cos(y-b)-cos(b+a)sin(b-y)
sin(a-b)sin(b-y)-cos(a-b)cos(y-b)
人气:107 ℃ 时间:2020-06-07 02:41:31
解答
1、原式=sin(a+b)cos(y-b)+cos(a+b)sin(y-b)
=sin[(a+b)+(y-b)]
=sin(a+y)
2、原式=-[cos(a-b)cos(b-y)-sin(a-b)sin(b-y)]
=-cos[(a-b)+(b-y)]
=-cos(a-y)
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