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化简(2Cosx四次方-2Cos²x+½)/2tAn(π/4–...

解:方程式两边同时平方得:(1+COSX)/(1-COSX)+(1-COSX)/(1+COSX)-2=4/(tanx)^2 移项,通分,化简:(SINX)^2+(COSX)^2=1 此为恒等式, 所以,X为任意值。

=lim(2(x²+1)arctanx-πx²)/2x =lim(4xarctanx+2-2πx)/2 =lim(2arctanx-π)/(1/x)+1 =lim(2/(x²+1))/(-1/x²)+1 =-2+1 =-1

»¯¼ò£º 2cos 2;a-1 - 2tan(¦Ð/4-a)sin 2;(¦Ð/4+a) (2cos a-1)/[2tan(¦Ð/4-a)*sin (¦Ð/4+a)] =cos2a/[2tan(¦Ð/4-

解: x∈[0,π/4] 0≤2x≤π/2 π/4≤x+π/4≤π/2 π/2≤2(x+π/4)≤π cos[2(x+π/4)]=1-2sin²[(x+π/4)]=1-2·(4/5)²=-7/25 cos[2(x+π/4)]=cos(2x+π/2)=sin[π/2-(2x+π/2)]=sin(-2x)=-sin(2x) sin(2x)=-cos[2(x+π/4)]=-(-7/25)=7/25 0≤2x≤π/2,cos(2x...

a∈(0,π),b∈(0,π),a+b∈(0,2π),a/2∈(0,π/2), sin(a+b)=5/13>0,所以 a+b∈(0,π) sina=2tan(a/2)/(1+tan²(a/2))=2*(1/2)/(1+1/4)=4/5 cosa=(1-tan²(a/2))/(1+tan²(a/2))=(1-1/4)/(1+1/4)=-3/5 所以a∈(π/2,π),a+b∈(π/2,π) sin²x+c...

sinα^2=[1-cos(2α)]/2 三角函数 降幂公式 sinα^2=[1-cos(2α)]/2 cosα^2=[1+cos(2α)]/2 tanα^2=[1-cos(2α)]/[1+cos(2α)] 倍角公式 sin(2α)=2sinα·cosα=2/(tanα+cotα) cos(2α)=cos²α-sin²α=2cos²α-1=1-2sin²α tan(2α)=2tanα/[。

=lim(-2/(1+x²))/((-1/x²)/(1+1/x)) =lim2(x²+x)/(1+x²) =lim2(1+1/x)/(1/x²+1) =2

F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))] 设x=tant,则dx=sec²tdt, ∴F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))] =∫sec²tdt/[tan²t*(1+tan²t)^(1/2)] =∫sec²tdt/(tan²t*sect) =∫sectdt/tan²t =∫(cos²t/sin²t)*(1...

令t=tan(x/2) ,则sinx=2sin(x/2)cos(x/2)=2tan(x/2)cos²(x/2)=2t/(1+t²) 按此先解出t=tan(x/2) sin²(x/2)=tan²(x/2)cos²(x/2)=tan²(x/2)/[1+tan²(x/2)] 解出sin(x/2)

tan3x=tan(2x+x) =(tan2x+tanx)/(1-tan2x·tanx) =[2tanx/(1-tan²x)+tanx]/{1-[2tan²x/(1-tan²x)]} =(3tanx-tan³x)/(1-3tan²x)

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