sinx-cosx 三次根号sinx coax dx
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令sinx=Tcosx=V所以有T^2+V^2=1所以T^3-v^3=(T-V)(T^2+TV+V^2)=(T-V)(1+TV)因为T-V=根号/2又T^2+V^2=1所以有(T-V)^2=1/2=1
/>tanx=√3则x=kπ+π/3,k∈Z(1)k是偶数,则sinx=√3/2,cosx=1/2(2)k是奇数则sinx=√3/2,cosx=1/2(sinx+cosx)/(sinx-cosx)的值
用sinx2+cosx2=1再问:可以帮我把顺序写出吗。再问:谢谢。再答:就是3cos2x+cos2x=1再答:然后cosx=正负0.5就行了再问:就只有后面这两个步骤吧再问:还是前面那个也要再问:s
sinx+cosx=√2两边同时平方得sin²x+2sinxcosx+cos²x=21+2sinxcosx=22sinxcosx=1sinxcosx=1/2答案:1/2
设t=³√(sinx-cosx)sinx-cosx=t³(sinx+cosx)dx=3t²dt代入易得结果为3/2t²+c回代即可得解
∫(sinx)^3/(cosx+sinx)dx=1/√2*∫(sinx)^3/(sin45*cosx+cos45*sinx)dx=1/√2*∫(sinx)^3/sin(45+x)dx设45+x=t∴d
原式=2√2(cosx*1/2-sinx*√3/2)=2√2(cosxcosπ/3-sinxsinπ/3)=2√2cos(x+π/3)
不等于.如果等于,则两边平方,有sinx3-cosx3=cosx2sinx3移项有cosx3=sinx3(1-cosx2)=sinx5这个肯定不一定满足.
=2*(根号3sinx/2+cosx/2)=2(sinx*cos30°+cosxsin30°)=2sin(x+30°)
√3sinx-cosx=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)
=(sinx)^2+√3sinxcosx=(1-cos2x)/2+√3/2(2sinxcosx)=1/2-1/2cosx+√3/2sin2x=1/2+sin(-π/6)cos2x+cos(-π/6)s
10(根号2/10cosx-根号6/10sinx)=10sin(α+x)其中sinα=根号2/10cosα=根号6/10
(1)a*b=0sin2x-cos2x=0sqr(2)sin(2x-π/4)=0x=π/8+kπ/2,k∈Z(2)f(x)=sqr(2)sin(2x-π/4)x∈(3π/8+kπ,7π/8+kπ),k
sinx+cosx=√2[sinx*(√2/2)+(√2/2)cosx]=√2[sinxcos(π/4)+sin(π/4)cosx]=√2sin(x+π/4)再问:为什么sinx+cosx=√2[si
sinx+cosx=m平方一下得sin^2x+2sinxcosx+cos^2x=m^2sinxcosx=(m^2-1)/2sin^3x+cos^3x=(sinx+cosx)(sin^2-sinxcos
第一个答案:第二个答案:
没错,f(x)=2sin(2x+π/6)周期T=2π/2=π因为-1≤sin(2x+π/6)≤1f(x)max=2f(x)min=-2
原式=2(sinx*√2/2-cosx*√2/2)=2(sinxcosπ/4-cosxsinπ/4)=2sin(x-π/4)
√(1-cosX)=√[(1-cosX)/1]【根号内分子分母同乘以(1+cosx)】:=√[(1-cosX)(1+cosx)/(1+cosx)]=√[(1-cos^2x)/(1+cosx)]=√[s
2√2sin(π/6-x)