1.求函数y=3sin(x/6-7π/12)的单调递增区间.2.化简{[tan(π-x)]/[sin(π-α)*sin(3π/2-α)]}+{[sin(2π-α)*cos(α-3π/2)]/[sin(3π/2+α)*cos(2π-α)]}.
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![1.求函数y=3sin(x/6-7π/12)的单调递增区间.2.化简{[tan(π-x)]/[sin(π-α)*sin(3π/2-α)]}+{[sin(2π-α)*cos(α-3π/2)]/[sin(3π/2+α)*cos(2π-α)]}.](/uploads/image/z/4029588-36-8.jpg?t=1.%E6%B1%82%E5%87%BD%E6%95%B0y%3D3sin%28x%2F6-7%CF%80%2F12%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4.2.%E5%8C%96%E7%AE%80%7B%5Btan%28%CF%80-x%29%5D%2F%5Bsin%28%CF%80-%CE%B1%29%2Asin%283%CF%80%2F2-%CE%B1%29%5D%7D%2B%7B%5Bsin%282%CF%80-%CE%B1%29%2Acos%28%CE%B1-3%CF%80%2F2%29%5D%2F%5Bsin%283%CF%80%2F2%2B%CE%B1%29%2Acos%282%CF%80-%CE%B1%29%5D%7D.)
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1.求函数y=3sin(x/6-7π/12)的单调递增区间.2.化简{[tan(π-x)]/[sin(π-α)*sin(3π/2-α)]}+{[sin(2π-α)*cos(α-3π/2)]/[sin(3π/2+α)*cos(2π-α)]}.
1.求函数y=3sin(x/6-7π/12)的单调递增区间.
2.化简{[tan(π-x)]/[sin(π-α)*sin(3π/2-α)]}+{[sin(2π-α)*cos(α-3π/2)]/[sin(3π/2+α)*cos(2π-α)]}.
1.求函数y=3sin(x/6-7π/12)的单调递增区间.2.化简{[tan(π-x)]/[sin(π-α)*sin(3π/2-α)]}+{[sin(2π-α)*cos(α-3π/2)]/[sin(3π/2+α)*cos(2π-α)]}.
应该是没问题的~
回答这是对的
被抢先了
你可以放心看
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我认为我的对
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