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Cos(x h)%Cosx化为关于sin的式子

因为cos(x+h)-cosx=-2sin(x+h/2)sin(h/2) 所以:lim [cos(x+h)-cosx]/h =lim[-2sin(x+h/2)sin(h/2)/h] =lim[-sin(x+h/2)sin(h/2)/(h/2)] =lim[-sin(x+h/2)] =-sinx

和差化积公式: 本题中,a=x+h,b=x

sin(270°-x)=-cosx

sin(x+h)-sinx=sinxcosh+sinhcosx-sinx=sinx(1-2sin²h/2)+2sin(h/2)cos(h/2)cosx-sinx=-2sinxsin²(h/2)+2sin(h/2)cos(h/2)cosx=2sin(h/2)[cos(h/2)cosx-sin(h/2)sinx)=2cos(x+h/2)sin(h/2)

利用积化和差公式: sinXsin2Xsin3X = -(1/2)(cos3X-cosX)sin3X 满意请采纳,谢谢!

12(3-3X)=-X+4 36-36X=-X+4 36X-X=36-4 35X=32 X=32÷35 X=32/35 2(X-3)=-4(X+5) 2X-6=-4X-20 6X=-14 X=-8/3 (X-7)/4-(5X+8)/3=1 两边乘12 3X-21-20X-32=12 17X=-65 X=-65/17 (0.1X-2)/0.6-(0.5-0.3X)/0.02=1 两边乘0.6 0.1X-2-15+9X=0.6 9....

cos(x+y)-cosx=0 2SIN(x/2)SIN(x+y/2)=0 所以SIN(x/2)=0或SIN(x+y/2)=0 x/2=kπ或x+y/2=kπ x=2kπ或y=2kπ-2x 其中k为任意整数

f(x)=2sinxcosx+cos2x =sin2x+cos2x =√2sin(2x+π/4) T=2π/2=π sin函数的单增区间为 -π/2+2kπ,π/2+2kπ k为整数 即 -π/2+2kπ ≤2x+π/4 ≤π/2+2kπ 即 -3π/8+kπ≤x≤π/8+kπ ∵x∈[0,π/3] ∴f(x)max时 即sin=1 即f(x)max=√2 f(x)min=f(π/3)=√2sin(11π/12)

可以用欧拉公式推导 e^[i*(x+h)]=cos(x+h)+i*sin(x+h) e^[i*(x+h)]=e^(ix)*e^(ih)=[cos(x)+i*sin(x)]*[cos(h)+i*sin(h)] =cos(x)cos(h)-sin(x)sin(h)+i*[sin(x)cos(h)+cos(x)sin(h)] 所以cos(x+h)=cos(x)cos(h)-sin(x)sin(h) sin(x+h)=sin(x)cos...

可以,cos(x-3/4π)=sin(x-1/4π) cosx*cos(-3/4π)+sinx*sin(-3/4π)=sinx*cos(-1/4π)-cosx*sin(-1/4π) cosx+sinx=-sinx-cosx(约去了所有的-1/(根号2)) 2cosx=-2sinx tanx=-1 x=Kπ-1/4π(K∈Z(Z即包括0的整数)) 有必要发两遍吗

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