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(2Cos²x%1)/2tAn(π/4%x)sin²(π/4+x)

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

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/...

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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...

令tan(x-π/4)=0 x-π/4=kπ x=π/4+kπ 在(0,2π) x可取,π/4,5π/4, 则,间断点,两个

-π/2

1.f(x)=sin(π+x)cos(π-x)tan(π-x)/tan(π+x)cos(3π/2+x) =(-sinx)(-cosx)(-tanx)/(tanx)(sinx) =-cosx 2.若x是第二象限角,且cos(x-π/3)=1/5>0,sin(x-π/3)=2√6/5 cosx=cos[(x-π/3)+π/3]=cos(x-π/3)cosπ/3-sin(x-π/3)sinπ/3=1/5*1/2-2√6/5*√3/...

sin(π/4+x)sin(π/4-x) =sin[π/2-(π/4-x)]sin(π/4-x) =cos(π/4-x)sin(π/4-x) =1/2sin(π/2-2x) =1/2cos2x=1/6 cos2x=1/3 x在(pai/2,pai),则2x在(pai,2pai) 所以:sin2x=-√[1-(cos2x)^2]=-2√2/3 sin4x=2sin2xcos2x=-2*1/3*2√2/3=-4√2/9 然后再求出co...

sin(x)=2sin(x/2)cos(x/2),cos(x)=2cos(x/2)^2 -1 代入等式右边,前提 x不等于(2k+1)π,k为整数,因为cos(x/2)不能为0。

(1)由2x+π4≠π2+kπ,k∈Z,得:x≠π8+kπ2,k∈Z,所以f(x)的定义域为{x|x≠π8+kπ2,k∈Z},f(x)的最小正周期为π2;(2)由f(α2)=2cos2α,得tan(α+π4)=2cos2α,sin(α+π4)cos(α+π4)=2(cos2α-sin2α),整理得:sinα+cosαcosα-sinα=2(cosα+sin...

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