1-2sin²3x

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若sin x/2-cos x/2=1/3,则sin x=

两边平方得sin²x/2-2sinx/2cosx/2+cos²x/2=1/91-sinx=1/9sinx=8/9

已知Sin(2x+pi/3)=1/3,求Sin(5Pi/6-4x),

原式=sin[π-(5π/6-4x)]=sin(4x+π/6)=-sin(-4x-π/6)=-cos[π/2-(-4x-π/6)]=-cos(4x+2π/3)=-cos[2(2x+π/3)]=-[1-

在mathematica里输入Plot[Sin[x] Sin[x + 2] - Sin[x + 1]Sin[x + 1]

楼上都错了,图像没问题这个表达式实际是个常数,你可以运行TrigReduce[Sin[x]Sin[x+2]-Sin[x+1]^2]看看,结果为1/2(-1+Cos[2])只不过Plot的自动选择坐标系

化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1

2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[

化简(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1)

tan²-1=sin²x/cos²x-1=(sin²x-cos²x)/cos²x=(sinx+cosx)(sinx-cosx)/cos&su

x*(1+sin^2 x )/sin^2x 不定积分

原式=∫x*(csc^2x+1)=∫x*csc^2x+x(分开积分)前面=-x*cotx+∫cotx=-x*cotx+ln|sinx|后面=1/2x^2记得加C

三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x

用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos

已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x)

sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π

求极限 ((sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1

由和差化积公式分子=2sin[(x^3+x^2)/2]cos[(x^3+x^2-2x)/2]x→0,则(x^3+x^2)/2→0,sin则(x^3+x^2)/2和(x^3+x^2)/2是等价无穷小而c

s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)

x=0:0.1:2*pi;s=2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x);plot(x,s)

证明成立:[cos(3x)-sin(3x)]/(cosx+sinx)=1-2sin(2x).

cos(3x)-sin(3x)=cos(2x+x)-sin(2x+x)=cos2xcosx-sin2xsinx-sin2xcosx-cos2xsinx=(cos^2x-sin^2x)cosx-2sin

sin^2x+cos^2y=1/2 求3sin^2x+sin^2y的最值

sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3(1/2-cos^2y)+sin^2y=1.5-3cos^2y)+sin^2y又有sin^2y+c

已知fx=2/√3sin 2x-2/1[cos^x-sin^x]-1

f(x)=(√3/2)sin2x-(1/2)[(cosx)^2-(sinx)^2]-1=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1f(x)的最大值是0,最小值是-2,

化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x

化简(1)√3sin x+cos x (2)√2(sin x-cos x) (3)√2cos x-√6sin x

2(√3/2sinx+1/2cosx)=2sin(x+π/6)√2*√2(√2/2sinx-√2/2cosx)=2sin(x-π/4)(3)解;2√2(1/2cosx-√3/2sinx)=2√2cos

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x

(x^3-x+1)sin^2x的不定积分

3次分部积分法解用!代表积分号=!(x^3-x+1)(1-cos2x)/2dx=(x^3-x+1)(x/2-sin2x/4)-!(3x^2-1)(x/2-sin2x/4)dx+c=-!(3x^2-1)