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tAn x=1/2,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

∵ tanπ/4=1 ∴ tan(π/4-x)=(1-tanx) / (1 + tanx)=1/2 2-2tanx=1 + tanx 所以 tanx=1/3 sinx/cosx=tanx=1/3 3sinx=cosx 代入三角恒等式: sin²x + cos²x=1 解得:sin²x=1/10 所以,sinxcosx=sinx * (3sinx)=3sin&#...

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

解析: 令α=arcsin[(1-x²)^½ ,β=arctan[x/(1-x²)^½],其中0

[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...

判断间断点的类型还是要从定义出发,肯定不会错的,求解方法是一样的

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...

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/...

cosx/1+sinx =【cos²(x/2)-sin²(x/2)】/【sin²(x/2)+2sin(x/2)cos(x/2)+cos²(x/2)】 =[cos(x/2)+sin(x/2)][cos(x/2)-sin(x/2)]/[cos(x/2)+sin(x/2)]² =[cos(x/2)-sin(x/2)]/[cos(x/2)+sin(x/2)] =√2sin(π/4-x/2)/√2cos...

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