sin(×−π÷6) cos(×−π÷3),2sin2x÷2

（Ⅰ）f(x)=sin(x+π6)-cos(x+π3)+cosx=32sinx+12cosx-（12cosx-32sinx）+cosx=3sinx+cosx=2sin（x+π6），∵ω=1，∴T=2π

cos(π/6+α)=sin(π/3-α),sin(π/4+α)=cos(π/4-α)所以,原式=sin(π/4-α)cos(π/3-α)-sin(π/3-α)cos(π/4-α)=sin(π/4-α

f(x)=[cos(派/6)]^2+sin(派/6)cos(派/6)=[(根号3)/2]^2+[(1/2)*(根号3)/2]=(3/4)+(根号3)/4=(3+根号3)/4.

因为sin(x+π/2)=sin[π/2-(-x)]=cos(-x)=cosx,cos(x+π/2)=cos[π/2-(-x)]=sin(-x)=-sinx,所以：sinˆ6(x+π/2)+

cos5/12πcosπ/12+cosπ/12sinπ/6=cosπ/12（cos5/12π+sinπ/6）=cosπ/12（cos(π/2-π/12)+sinπ/6）=cosπ/12（sin(π/1

sinx=cos(π/2-x)这没有问题吧sin(π/3)=cos(π/2-π/3)=cos(π/6)

√2sin(x/2+π/4)=sinx/2+cosx/2tan(x/2-π/4)=-cot(x/2+π/4)a*b=√2sin(x+π/4)f(x)的导数=√2cos(x+π/4)f(x)加上f(x)

右边2sin(π/6+A)=2sin(π/6)cosA+2sinAcos(π/6)=cosA+根号3sin(A)=左式.得证#

由题意知，sinα+cosαsinα−cosα=2，分子和分母同除以cosα得，tanα+1tanα−1=2，解得tanα=3，∵sin（α-5π）•sin（3π2-α）=-sinα•（-cosα）=

应该是你打错了吧?!原式应该是cos(π/3+a)+sin(π/6+a)=cosπ/3cosa-sinπ/3sina+sinπ/6cosa+cosπ/6sina=cosa/2-(根号3)/2sina+

原式=(1/2)cosa-(√3/2)sina+(1/2)cosa+(√3/2)sina=cosa

sin(π+π/6)-cos(π+π/4)cos(-π/2)+1=-sin(π/6)+cos(π/4)cos(+π/2)+1=-1/2+cos(π/4)*0+1=1/2

√3；cosπ/6=√3/2,sinπ/6=1/2所以cosπ/6+√3sinπ/6=√3/2+√3/2=√3希望对你有帮助

是不是sin^2a+cos^2(π/6+a)+1/2sin(2a+π/6)=(1-cos2a)/2+(1+cos(π/3+2a))/2+1/2sin(2a+π/6)=1-1/2cos2a+1/4cos

sin^2(a)+cosa*cos(pi/3+a)-sin^2(pi/6-a)=sin^2(a)+cosa*(cospi/3cosa-sinpi/3sina)-[sinpi/6cosa-sinacos

sin(-20/3π)cos(43/6π)+sin(35/6π)cos-(17/3π)=sin(-2/3π)cos(7/6π)+sin(-1/6π)cos(1/3π)=sin(1/3π)cos(1/6

利用积化和差公式和二倍角公式：sin^2α+cosαcos(π/3+a)-sin^2(π/6-α)=(1-cos2a)/2+[cos(2a+π/3)+cos(π/3)]/2-[1-cos(π/3-2a

1sin²α+cosαcos(π/3+α)-sin²(π/6-α)=sin²α+cosα(cosπ/3cosα-sinπ/3sina)-(sinπ/6cosα-cosπ/

sin(-19π/3)*cos(19π/6)=-sin(19π/3)*cos(19π/6)=-sin(6π+π/3)*cos(3π+π/6)=-sin(π/3)*(-cosπ/6)=√3/2*√3/2

[cos^4(π/3+α)]-[cos(π/6-α)]^2=[cos(π/3+α)]^4-[sin(π/3+α)]^4=[cos(π/3+α)]^2-[sin(π/3+α)]^2=cos(2π/3+2