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(2Cos²x%1)/2tAn(π/4%x)sin²(π/4+x)

tan(π/4-x)= (tanπ/4-tanx)/(1+tanπ/4tanx) = (1-tanx)(1+tanx) = (cosx-sinx)/(cosx+sinx) sin^2(π/4+x) = (√2/2sinx+√2/2cosx)^2= 1/2(sinx+cosx)^2 f(x) = [2cos^4x-2cos^2x+1/2]/[tan(π/4-x)sin^2(π/4+x)] =[2cos^4x-2cos^2x+1/2]/[ (...

你的这个式子写得太乱了,请写正规些,这样的题应该不难! 我这里给你一些提示吧: [cos²(x)]²-cos²(x)=cos²(x)[cos²(x)-1]=﹣sin²(x)cos²(x) 其它的式子看不清。

因为(π/4-x)+(π/4+x)=π/2,所以tan(π/4-x)=cot(π/4+x) (2cos²x-1)/2tan(π/4-x)sin²(π/4+x) =sin2x/[2cot(π/4+x)sin²(π/4+x)] =sin2x/{[2cos(π/4+x)/sin(π/4+x)]sin²(π/4+x)} =sin2x/[2sin(π/4+x)cos(π/4+x)] =sin2x/sin(π/2...

[2cos^4(x)-2cos²x+1/2]/[2tan(π/4-x)·sin²(π/4+x)] =1/2*[4cos^4(x)-4cos²x+1]/{2tan[(π/2-π/4)-x]·sin²(π/4+x)} =1/2*[2cos²x-1]²]/{2tan[π/2-(π/4+x)]·sin²(π/4+x)} =1/2*[2cos²x-1]²]/{2co...

f(x)=4cos4x-2cos2x-1tan(π4+x)sin2(π4-x)=4(1+cos2x2)2-2cos2x-1tan(π4+x)cos2(π4+x)=cos22xtan(π4+x)sin(π4+x)=2cos2x.(1)f(-17π12)=2cos17π6=2cos5π6=-3.(2)g(x)=12f(x)+sin2x=cos2x+sin2x=2sin(2x+π4),因为x∈[0,π2],所以π4≤2x+...

tan2x=-2 π/2

证明:2f(x)=(4cos^4x—4cos²x+1)/(2tan(π/4—x)cos²(π/4-x)) =[2cos^2(2x)-1]^2/(2sin(π/4—x)cos(π/4-x)) =(cos2x)^2/sin(π/2-2x)=cos2x (2) f(x)=2/5,所以cos2x=4/5 又x∈(0,π/2),所以2x∈(0,π/2), 所以sin2x=3/...

f(x)=tan(3x+π/4) 1。f(π/9)=tan(π/3+π/4) =(tanπ/3+tanπ/4)/(1-tanπ/3tanπ/4) =(√3+1)/(1-√3) =-√3-2 2. ∵f(a/3+π/4)=2 ∴tan(a+3π/4+π/4) = tan(π+a)=tana=2 sina/cosa=2 sina=2cosa代入sin²a+cos²a=1 cos²a=1/5 ∴cos2a=2cos...

4(cosx)^4-2cos2x-1=[2(cosx)^2]^2-1-2cos2x=[(2(cosx)^2-1]*[2(cosx)^2+1]-2cos2x =cos2x*[2(cosx)^2+1]-2cos2x=cos2x*[2(cos)^2-1]=(cos2x)^2 tan(π/4+x)·[sin(π/4-x)]^2=(1+tanx)/(1-tanx)*[√2/2*cosx- √2/2*sinx]^2 =(cosx+sinx)/(cosx-sinx...

tanx = 1/2 2sinx= cosx 2sinxcosx = (cosx)^2 sin2x = 1/(secx)^2 = 1/[( tanx)^2 +1] =1/(1+ 1/4) = 4/5 [sin(π/4+x)]^2 =[(√2/2)(sinx+cosx) ]^2 =(1/2)(1+sin2x) =(1/2)(1+ 4/5) = 9/10

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