作业帮a,b,c属于(0,π/2),a=cosa,b=sin(cosb) c=cos(sinc) 试比较a,b,c大小 要详细
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a,b,c属于(0,π/2),a=cosa,b=sin(cosb) c=cos(sinc) 试比较a,b,c大小 要详细的答案.
考虑函数f(x) = x-cos(x), ∵f'(x) = 1+sin(x) ≥ 0, ∴f(x)单调递增.
∵a = cos(a), ∴f(a) = 0.
∵b ∈ (0,π/2), ∴cos(b) > 0, ∴b = sin(cos(b)) < cos(b) (当x > 0, 有sin(x) < x).
即得f(b) < 0 = f(a), 又∵f(x)单调递增, ∴b < a.
∵c ∈ (0,π/2), ∴0 < sin(c) < c < π/2, ∴c = cos(sin(c)) > cos(c) (cos(x)在(0,π/2)单调递减).
即得f(c) > 0 = f(a), 又∵f(x)单调递增, ∴c > a.
于是b < a < c.
∵a = cos(a), ∴f(a) = 0.
∵b ∈ (0,π/2), ∴cos(b) > 0, ∴b = sin(cos(b)) < cos(b) (当x > 0, 有sin(x) < x).
即得f(b) < 0 = f(a), 又∵f(x)单调递增, ∴b < a.
∵c ∈ (0,π/2), ∴0 < sin(c) < c < π/2, ∴c = cos(sin(c)) > cos(c) (cos(x)在(0,π/2)单调递减).
即得f(c) > 0 = f(a), 又∵f(x)单调递增, ∴c > a.
于是b < a < c.
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