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(2Cos²x%1)/2tAn(π/4%x)sin²(π/4+x)

分母=2tan(π/4-x)*cos^2[π/2-(π/4+x)] =2[sin(π/4-x)/cos(π/4-x)]*cos^2(π/4-x) =2sin(π/4-x)cos(π/4-x) =sin[2(π/4-x)] =sin(π/2-2x) =cos2x 分子=1/2[4cos^4(x)-4cos^2(x)+1] =1/2(2cos^2x-1)^2 =1/2(cos2x)^2 所以原式=1/2(cos2x)^2/cos2x=1/...

[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...

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...

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]/[ (...

先从题干给的关系入手,求出tanx的值 分子分母同除以cosx tanx+1/2tanx-3=3 然后解出tanx的值为2 1.根据诱导公式(务必熟记)tan(π-x)=-tanx=-2(我口算的,你自己再算一遍吧) 2.首先,分子乘以sin²x+cos²x(即1) 变为5sin²x+5...

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...

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。

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

原式=1/4tanx+1/4 由tan[(π/4)-x]=-1/3 再解tanx,得结果

(1)原式=22[12sin(π4-x)+32cos(π4-x)]=22[sinπ6sin(π4-x)+cosπ6cos(π4-x)]=22cos(π6-π4+x)=22cos(x-π12)(2)原式=cos2α1?tanα1+tanα[1?cos(π2+2α)]

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