sinx-sina化成2cos[(x a) 2]·sin[(x-a) 2]

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sinx-sina化成2cos[(x a) 2]·sin[(x-a) 2]
cos(A-π/2)=-sinA

错了cos(A-π/2)=sinA

将根号3sinx-cosx+2cos(x-π/6)化成一个角的三角函数式

√3sinx-cosx+2cos(x-π/6)=2sin(x-π/6)+2cos(x-π/6)=2√2sin[(x-π/6)+π/4]=2√2sin(x+π/12)

已知函数f(x)=cos^2x/2-sinx/2cosx/2-1/2,若f(a)=3*根号2/10,求sina

f(x)=cos^2x/2-sinx/2cosx/2-1/2=1/2(cosx+1-sinx)-1/2=(cosx-sinx)/2=√2/2cos(x+π/4)f(a)=√2/2cos(x+π/4)=

证明 cosa/(1+sina0-sina/(1+cosa)=2(cosa-sina)/(1+sina+cos)

左边通分=(cosa+cos²a-sina-sin²a)/(1+sina)(1+cosa)=[(cosa-sina)+(cosa+sina)(cosa-sina)]/(1+sina

∫(1+sinx) / cos^2 x dx

∫(1+sinx)/(cosx)^2dx=∫[(secx)^2+tanxsecx]dx=tanx+secx+C

∫(0,π/2)cos(sinx)dx

=sin(cos(π/2-0)dx再问:������˼

cos(a-π/2)=sina

cos(a-π/2)=cos-(π/2-a)=cos(π/2-a)=sina

谁帮帮我解决几道数学题将下列函数化成标准形式 y=2sinx9cosx+sinx) y=cos平方x+2根号3sinxc

1.y=2sinx(cosx+sinx)=2sinxcosx+2sin平方x=sin2x+1-cos2x=根号2sin(2x-四分之π)+12.y=cos平方x+2根号3sinxcosx-sin平方x

cos(x)^2+sinx 求范围.

cos²x+sinx=1-sin²x+sinx=-(sin²x-sinx+1/4)+5/4=-(sinx-1/2)²+5/4-1

已知sin a+cos a=1/2 sina*cosa=?cos a/sina+sina/cosa=?

sina+cosa=1/2,那么1+2sinacosa=1/4所以sinacosa=-3/8cosa/sina+sina/cosa=1/(sinacosa)=-8/3

化简(cos^2 a/sina+sina)*tana

(cos^2a/sina+sina)*tana=(1-2sin^2a)/sina+sina)*tana=((1/sina-2sina)+sina)*tana=(1/sina(1-sin^2a))*si

f(x)=cos2x+2sinx如何化成关于sin或cos的最简形式!

f(x)=cos2x+2sinx=[1-2sin^2(x)]+2sinx=-2sin^2(x)+2sinx+1

sinx+sina如何化成2sin(x-a)/2cos(a+x)/2

sin(A+B)=sinAcosB+cosAsinB(1)sin(A-B)=sinAcosB-cosAsinB(2)(1)+(2)2sinAcosB=sin(A+B)+sin(A-B)A=(x-a)/

数学下面的x是被她cosa+cosx=1/2 sina+sinx/=1/3求cos(a~x)

cosa+cosβ=1/2得(cosa+cosβ)^2=1/4即cos^2a+2cosacosβ+cos^2β=1/4(1)sina+sinβ=1/3得(sina+sinβ)^2=1/9即sin^2a

证明(1-cos^2a)/(sina+cosa)-(sina+cosa)/(tan^2a-1)=sina+cosa

(sina+cosa)/(tan^2a-1)=(sina+cosa)/(sin^2a/cos^2a-cos^2a/cos^2a)=(sina+cosa)/((sin^2a-cos^2a)/cos^2a

cos^2x sinx 不定积分

∫[(cosx)∧2]sinxdx=-∫(cosx)∧2d(cosx)=-(cosx)∧3/3+C

证明(1-cos^2a)/(sina-cosa)-(sina+cosa)/(tan^2-1)=sina+cosa

证:(1-cos^2a)/(sina-cosa)-(sina+cosa)/(tan^2a-1)=sin^2a/(sina-cosa)-(sina+cosa)/(sin^2a/cos^2a-cos^2a

证明2(cosx-sinx)/(1+sinx+cosx)=cosx/(1+sinx)-sinx/(1+cos)

证明:右边通分=(cosx+cos²x-sinx-sin²x)/(1+sinx+cosx+sinxcosx)=[cosx-sinx+(cosx+sinx)(cosx-sinx)]/