作业帮sinx-sina化成2cos[(x a) 2]·sin[(x-a) 2]
来源：学生作业帮助网 编辑：作业帮 时间：2024/08/26 19:33:40
错了cos(A-π/2)=sinA
√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)
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)=
左边通分=(cosa+cos²a-sina-sin²a)/(1+sina)(1+cosa)=[(cosa-sina)+(cosa+sina)(cosa-sina)]/(1+sina
∫(1+sinx)/(cosx)^2dx=∫[(secx)^2+tanxsecx]dx=tanx+secx+C
=sin(cos(π/2-0)dx再问：������˼
cos(a-π/2)=cos-(π/2-a)=cos(π/2-a)=sina
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
cos²x+sinx=1-sin²x+sinx=-(sin²x-sinx+1/4)+5/4=-(sinx-1/2)²+5/4-1
sina+cosa=1/2,那么1+2sinacosa=1/4所以sinacosa=-3/8cosa/sina+sina/cosa=1/（sinacosa）=-8/3
（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=[1-2sin^2(x)]+2sinx=-2sin^2(x)+2sinx+1
sin(A+B)=sinAcosB+cosAsinB(1)sin(A-B)=sinAcosB-cosAsinB(2)(1)+(2)2sinAcosB=sin(A+B)+sin(A-B)A=(x-a)/
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
(sina+cosa)/(tan^2a-1)=(sina+cosa)/(sin^2a/cos^2a-cos^2a/cos^2a)=(sina+cosa)/((sin^2a-cos^2a)/cos^2a
∫[(cosx)∧2]sinxdx＝－∫(cosx)∧2d(cosx)=－(cosx)∧3／3＋C
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：右边通分=(cosx+cos²x-sinx-sin²x)/(1+sinx+cosx+sinxcosx)=[cosx-sinx+(cosx+sinx)(cosx-sinx)]/