已知a-b=8π/3,且a≠kπ,求(1-cos(π-a))/(csc(a/2)-sin(a/2))的商减去4sin(π/4 -b/4)的差的最大值及如题.打错了,是求函数f(x)=(1-cos(π-a))/(csc(a/2)-sin(a/2)) -4sin(π/4 -b/4)的最大值及最大值条件
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![已知a-b=8π/3,且a≠kπ,求(1-cos(π-a))/(csc(a/2)-sin(a/2))的商减去4sin(π/4 -b/4)的差的最大值及如题.打错了,是求函数f(x)=(1-cos(π-a))/(csc(a/2)-sin(a/2)) -4sin(π/4 -b/4)的最大值及最大值条件](/uploads/image/z/14443355-11-5.jpg?t=%E5%B7%B2%E7%9F%A5a-b%3D8%CF%80%2F3%2C%E4%B8%94a%E2%89%A0k%CF%80%2C%E6%B1%82%EF%BC%881-cos%28%CF%80-a%29%EF%BC%89%2F%28csc%28a%2F2%29-sin%28a%2F2%29%29%E7%9A%84%E5%95%86%E5%87%8F%E5%8E%BB4sin%28%CF%80%2F4+-b%2F4%29%E7%9A%84%E5%B7%AE%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%8F%8A%E5%A6%82%E9%A2%98.%E6%89%93%E9%94%99%E4%BA%86%2C%E6%98%AF%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%3D%EF%BC%881-cos%28%CF%80-a%29%EF%BC%89%2F%28csc%28a%2F2%29-sin%28a%2F2%29%29++-4sin%28%CF%80%2F4+-b%2F4%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%8F%8A%E6%9C%80%E5%A4%A7%E5%80%BC%E6%9D%A1%E4%BB%B6)
已知a-b=8π/3,且a≠kπ,求(1-cos(π-a))/(csc(a/2)-sin(a/2))的商减去4sin(π/4 -b/4)的差的最大值及如题.打错了,是求函数f(x)=(1-cos(π-a))/(csc(a/2)-sin(a/2)) -4sin(π/4 -b/4)的最大值及最大值条件
已知a-b=8π/3,且a≠kπ,求(1-cos(π-a))/(csc(a/2)-sin(a/2))的商减去4sin(π/4 -b/4)的差的最大值及
如题.
打错了,是求函数f(x)=(1-cos(π-a))/(csc(a/2)-sin(a/2)) -4sin(π/4 -b/4)的最大值及最大值条件
已知a-b=8π/3,且a≠kπ,求(1-cos(π-a))/(csc(a/2)-sin(a/2))的商减去4sin(π/4 -b/4)的差的最大值及如题.打错了,是求函数f(x)=(1-cos(π-a))/(csc(a/2)-sin(a/2)) -4sin(π/4 -b/4)的最大值及最大值条件
f(x)=(1-cos(π-a))/(csc(a/2)-sin(a/2)) -4sin(π/4 -b/4)
=(1-cos(π-a))/(csc(a/2)-sin(a/2)) -4sin(π/12+a/4)
= ( 2-2sin(a/2)^2)sin(a/2)/(1- sin(a/2)^2) -4sin(π/12+a/4)
= 2sin(a/2) -4sin(π/12+a/4)
然后求导f‘(x)= cos(a/2)-cos(π/12+a/4)=0得最大值
即:cos(a/2)=cos(π/12+a/4)
a/2=π/12+a/4
a=π/3
此时f(x)=1-2=-1
其实你已经算的差不多了,临门一脚就ok!