cos(x+π/3)=1/4 sin(x+4π/3)=?
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 16:52:23
cos(x+π/3)=1/4sin(x+4π/3)=?cos(x+π/3)=1/4sin(x+4π/3)=?cos(x+π/3)=1/4sin(x+4π/3)=?不等.因为sin(x+4π/3)=si
cos(x+π/3)=1/4 sin(x+4π/3)=?
cos(x+π/3)=1/4 sin(x+4π/3)=?
cos(x+π/3)=1/4 sin(x+4π/3)=?
不等.
因为sin(x+4π/3)=sin(π+x+π/3)= -sin (x+π/3).
等于正负4分之根号下15!
已知函数f(x)=sin(x+4分之7π)+cos(x-4分之3π),x∈R(1)求函数最小值和最小正周期
f(x)=sin(x+2π-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-π/4-π/2) \\公式 cos(x-π/2)=sinx\\
=sin(x-π/4)+sin(x-π...
全部展开
已知函数f(x)=sin(x+4分之7π)+cos(x-4分之3π),x∈R(1)求函数最小值和最小正周期
f(x)=sin(x+2π-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-3π/4)
=sin(x-π/4)+cos(x-π/4-π/2) \\公式 cos(x-π/2)=sinx\\
=sin(x-π/4)+sin(x-π/4)
=2sin(β-π/4)
所以最小正周期为2π,函数最小值为 -2
收起
已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值第一题1/2cos x-√3/2sin x第二题√3sin x+cos x第三题√2[sin x-cos x]第四题√2cos x-√6sin x
求证(3-sin^4 x-cos^4 x)/2cos^2 x=1+tan^2 x+sin^2 x
求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2).
求证:1-cos^4x-sin^4x/1-cos^6x-sin^6x=2/3
证明1-sin^6x-cos^2x/1-sin^4x-cos^4x=3/2
证明sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=3/4
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x
求函数y=cos(9/2π+x)+sin^2x的最大值和最小值还有一题 已知sinx+cosx=1/2,求sin^3x+cos^3x和sin^4x+cos^2x
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x
已知sin(x+π/3)=1/4,求sin(2π/3-x)+cos²(π/6-x)
急,matlab解非线性方程组用什么函数呢 方程如下所示:(cos(x(1))*cos(x(3))-sin(x(1))*cos(x(2))*sin(x(3)))*(194-x(4))*x(7)+(cos(x(1))*sin(x(3))+sin(x(1))*cos(x(2))*cos(x(3)))*(145-x(5))*x(7)+sin(x(1))*sin(x(2))*x(6)*x(7)-1=0;(-sin(x(1
求证(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)-(1/4)sin^2(2x)-cos^4x=sin^2x
化简sin(x+7π/4)+cos(x-3π/4)步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2-
sin(5x)-sin(3x)= 根号2cos(4x)1.sin(5x)-sin(3x)= (根号2)cos(4x)解方程2.sin(x)+sin(2x)+sin(3x)=1+cos(x)+cos(2x)
sin(x)+cos(x)=2/3 根号2倍(sin(2x-π/4)+1)/1+tan(x)=?2sin(x)*cos(x)+2sin(x)^2)/(1+tan(x))=2sin(x)*cos(x)怎么来的。
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x