化简:2sin^4x+3\4sin^2(2x)+5cos^4x-1\2(cos4x+cos2x)
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/24 00:02:40
化简:2sin^4x+3\4sin^2(2x)+5cos^4x-1\2(cos4x+cos2x)化简:2sin^4x+3\4sin^2(2x)+5cos^4x-1\2(cos4x+cos2x)化简:2
化简:2sin^4x+3\4sin^2(2x)+5cos^4x-1\2(cos4x+cos2x)
化简:2sin^4x+3\4sin^2(2x)+5cos^4x-1\2(cos4x+cos2x)
化简:2sin^4x+3\4sin^2(2x)+5cos^4x-1\2(cos4x+cos2x)
2sin^4x+3\4sin^2(2x)+5cos^4x-1\2 (cos4x+cos2x)
=2sin^4x+3\4sin^2(2x)+5cos^4x-1\2【cos^4(x)+sin^4(x)-6sin^2(x)cos^2(x)+cos^2(x)-sin^2(x)】
=【3/2sin^4(x)+6sin^2(x)cos^2(x)+9/2cos^4(x)】 - 1/2 cos^2(x) +1/2 sin^2(x)
=【[(3/2sin^2(x)+9/2cos^2(x)][sin^2(x)+cos^2(x)]】 - 1/2 cos^2(x)+1/2 sin^2(x)
=【[3/2+3cos^2(x)] * 1】- 1/2 cos^2(x)+1/2 sin^2(x)
=3/2+【5/2 cos^2(x)+1/2 sin^2(x)】
=3/2+【1+3/2cos^2(x)】
=5/2+3/2cos^2(x)
黑体中括号为主要变换部分
化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
傅里叶级数作图f(x)=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si
化简sin^4x-sin^2x+cos^2x
化简:sin^4 x-sin^2 x+cos^2 x
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)用matlab画图,求教啊
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
化简sin^4x+cos^2x
化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x)
f(x)=cos(2x-x/3)+2sin(x-π/4)sin(x+π/4)sin(x+π/4)化简.
sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简)
化简【完整步骤!】1.(sin^4x-sin^2x)/secx2.sin^4x-cos^4x【这道题的正确答案是sin^2x-cos^2x】
三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x
2sin(x+π/4)sin(x-π/4)=?
sin^2(π/4 +x)
泰勒公式的为什么㏑( 1 + sin X ) = sin X - ( sin X )²/2 +(sin X )³ /3 -(sin X )∧4 + o (sin ∧4 X ),完全不理解,
化简 sin²x+sin²(x+2π/3)+sin²(x-2π/3)