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.

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/21 17:47:38
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.a,b,c属于(0,π/2),a=cosa,b=sin(cosb)c=cos(

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.
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.

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,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. 两个正弦 、余弦定理的题目1.cos a =1/17,cos (a+b)= -47/51 ,a、b属于(0,π/2),求cos b.2.cos (a+b)=4/5,cos (a-b)= -4/5,a+b属于(7π/4,2π),a-b属于(3π/4,π),求cos 2a.谢八辈祖宗! 求证:a^2(cos^2b-cos^2c)+b^2(cos^c-cos^2a)+c^2(cos^2a-cos^2b)=0 abc属于r,cosa+cosb+cosc=2cos[(a-b+c)/3],sina+sinb+sinc=-2sin[(a-b+c)/3],求cos[(a-b)/3]cos[(b-c)/3]cos[(c-a)/3]=? 是否存在a,b,a属于(-π/2,π/2),b属于(0,π),使等式sin(3π-a)=根2cos(π/2-b),根3cos(-a)=-根2cos(π+b) 已知cos(a-b)=-3/15 cosb=4/5 求cos(a-2b)a属于(π/2,π)b属于(0,π/2)求cos(a-2b) 已知A(a,0)B(0,b)C(cos,α sinα)三点共线,a,b大于0,α属于0,π/2,求1/a^2+1/b^2的最小值.c点是cosα,sinα 已知角a、b属于(0,π),且cos(2a+b)-2cos(a+b)cosa=3/5,求sin2b的值.如题. 设a、b、c属于区间(0,π/2),且满足等式sina=a、sin(cosb)=b、cos(sinc)=c,试比较a、b、c的大小. 已知cosa=1/7,cos(a+b)=-11/14,且a属于(0,π/2),a+b属于(π/2,π),求b 已知sina=2/3,a属于(π/2,π),cosb=-3/4,b属于(π,3π/2),求cos(a+b)和cos(a-b) cos(a-b)=-4/5,cos(a+b)=4/5 a属于(π/2,π)b属于(3π/2,2π),求cos2a 设cos(a-b/2)=-1/9,sin(a/2-b)=2/3,其中a属于(π/2,π),b属于(0,π/2),求cos(a+b) 设cos(a-b/2)=-1/9,sin(a/2-b)=2/3,其中a属于(π/2,π),b属于(0,π/2),求cos(a+b) 已知cosa=1/7,cos(a+b)=-11/14,且a属于(0,π/2),a+b属于(π/2,π),求cosb的值? sin²2a + sin(2a)cos(a) -cos(2a) = 1 a属于(0,π/2) 已知sin(a+2b)=4/5,cos(a+b)=12/13(a+2b,a+b)属于(0~2π)求sinb 非线性方程解析解-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos(b)-y0*sin(a)*co