cos^2 x=tan^2 x sinx
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tan(3x/2)-tan(x/2)-2sinx/(cosx+cos2x)=sin(3x/2)/cos(3x/2)-sin(x/2)/cos(x/2)-2sinx/(cosx+cos2x)=[sin(
左边=(1-sin²x/cos²x)/(1+sin²x/cos²x)上下乘cos²x=(cos²x-sin²x)/(cos
是[1-(tanx)^2]/[1+(tanx)^2]=(cosx)^2-(sinx)^2=========证明:[1-(tanx)^2]/[1+(tanx)^2]={[1-(tanx)^2]*(cos
-cosxtanx再问:根据上题求f(-31π/3)的值再答:上面的答案可化简为-sinxf(-31π/3)=f(-π/3)=-1/2
你的题目中有点错误.
因为sin²x+cos²x=1所以:sin²x-sinxcosx+cos²x=(sin²x-sinxcosx+cos²x)/(sin
已知cosx=cosαcosβ,求证tan[(x+α)/2]tan[(x-α)/2]=tan²(β/2)证明:左边={sin[(x+α)/2]sin[(x-α)/2]}/{cos[(x+α)
tan,正切;sin,正弦;cos,余弦tan(x+y)tan(x-y)=sin(x+y)/cos(x+y)*sin(x-y)/cos(x-y)=sin(x+y)sin(x-y)/[cos(x+y)c
...由韦达定理tana+tanb=-4tana*tanb=3tan(a+b)=(tana+tanb)/(1-tana*tanb)=2cos^2(a+b)=1/[1+tan^2(a+b)]=1/5si
左边=(sin²x/cos²x-cos²x/sin²x)/(sin²x-cos²x)=[(sin^4x-cos^4x)/cos²x
我觉得应该是tana+1/tana=(sin²a+cos²a)/(sinacosa)=1/(sinacosa)你那个,举个例子:a=30°左边=√3+√3/3右边=1/4+(3/4
sin^2x+cos^2x=1所以左边=tan^2x-cot^2x=sin^2x/cos^2x-cos^2x/sin^2x=(sin^4x-cos^4x)/sin^2xcos^2x=(sin^2x+c
(1/sin^2x)+(1/cos^2x)-(1/tan^2x)=[(cosx)^2+(sinx)^2]/(sinxcosx)^2-(cosx)^2/(sinx)^2=1/(sinxcosx)^2-(
左边=(1-2sinxcosx)/(cos²x-sin²x)=(sin²x+cos²x-2sinxcosx)/(cos²x-sin²x)=(
原式=sin3x/2/cos3x/2-sinx/2/cosx/2-sinx/cos3x/2cosx/2=(sin3x/2cosx/2-sinx/2cos3x/2)/cos3x/2cosx/2-sinx
题目有没有问题呢?怎么觉得应该是求tan(x+y)tanx呢?∵2cos(2x+y)=cosy∴2cos(x+x+y)=cos(x+y-x)∴2cosxcos(x+y)-2sinxsin(x+y)=c
(1+tanx)/(1-tanx)=3+2根号21+tanx=(3+2根号2)-(3+2根号)tanx2(2+根号2)tanx=2(根号2+1)tanx=(根号2+1)/(2+根号2)=根号2/2(s
tan(π/4+x)=3(1+tanx)/(1-tanx)=3tanx=1/2sin2x-2cos^2x=(2sinxcosx-2cos^2x)/(sin^2x+cos^2x)=(2tanx-2)/(
sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)
cos(2x+y)=3cosycos(x+y+x)=3cos(x+y-x)cos(x+y)cosx-sin(x+y)sinx=3[cos(x+y)cosx+sin(x+y)sinx]2cos(x+y)