已知sin(a-π)是方程6x=1-根号x

(x-2)(5x+3)=00

由5x²-7x-6=0得x=2或x=-3/5所以sinα=-3/5,sin²α=9/25,cos²α=16/25所以sin²[（2k+1/2）π-α]+cos&

(5x+3)(x-2)=0∴tana=-0.6或tana=2我没看错题目的表达式的话……化简得,原式=[cosa*(-cosa)*(tana)^2]/[sina*(-sina)*(-cota)]=[-

由根与系数的关系sina+cosa=2/3sinacosa=a/3sina^2+cosa^2=(sina+cosa)^2-2sinacosa=4/9-2a/3=1a=-5/6

tan(A+B)=-12sin^2(A+B)-3sin(A+B)cos(A+B)+cos^2(A+B)=[2sin^2(A+B)-3sin(A+B)cos(A+B)+cos^2(A+B)]/[sin^

tana+tanb=-6tanatanb=7tan(a+b)=tana+tanb/1-tanatanb=-6/1-7=1∴sin(a＋b)=cos(a＋b)

sin(α+3π/2)=-cosαsin(3π/2-α)=-cosαtan(2π-α)=-tanαtan(π-α)=-tanαcos(π/2-α)=sinαcos(π/2+α)=-sinα原式=[(c

5x²-7x-6=0(5x+3)(x-2)=0x=-3/5x=2>1取sinα=-3/5cos(2π-α)cos(π+α)tan²(2π-α)/sin(π-α)sin(2π-α)c

〔sin（-α-3/2π）sin（3/2π-α）*tan（2π-α）〕/〔cos（π/2-α）cos（π/2+α）*cos（π-α）〕=cosαcosαtanα/sinαsinαcosα=sinα/s

5x^2-7x-6=0(5x+3)(x-2)=0x1=-3/5x2=2由于sina是方程的根,故有sina=-3/5化简得到：原式=-cosa*(-cosa)*tan^2a/(-sina)*(-sin

sina是方程5x^2-7x-6=0的根,a是第三象限角∴sina=-3/5∴cosa=-4/5tana=3/4∴sin(-a-3π/2)sin(3π/2-a)tan2^a）/（cos(π/2-a)c

f(x)=sin(x+π/6)+sin(x-π/6)+cosx+af(x)=根3sinx+cosx+af(x)=2sin(x+π/6)+a因为f(X)最大值为1所以2+a=1a=-1

1、f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a=2sinxcosπ/6+cosx+a=√3sinx+cosx+a=√[(√3)&su

sin(A+B)/cos(A+B)=tan(A+B)=(tanA+tanB)/[1-tanA*tanB]=(-6)/(1-7)=1

f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=sinx*cos(π/6)+cosx*sin(π/6)+sinx*cos(π/6)-cosx*sin(π/6)+cosx+a=（√3

tana+tanb=6tana*tanb=7tan(a+b)=(tana+tanb)/(1-tanatanb)=6/(1-7)=-1sin(a+b)/cos(a+b)=tan(a+b)=-1sin(a

x2-2x-3=0x=-1或x=3锐角a,tana>0tana=3(cosa+sina)(cosa-sina)/2sina*cosa=(cos^2a-sin^2a)/(2cosa*sina)分子分母同

tana+tanb=-6tana*tanb=7tan(a+b)=(tana+tanb)/(1-tanatanb)=7/7=1所以sin(a+b)=cos(a+b)

由韦达定理可以得到tan(a)+tan(b)=3,tan(a)*tan(b)=-3,所以tan(a+b)=3/4.也就是说sin(a+b)/cos(a+b)=3/4.因此sin(a+b)=3/5,co

韦达定理sinα+cosα=2/3sinαcosα=a/3因为(sinα)^2+(cosα)^2=1则(sinα+cosα)^2-2sinαcosα=14/9-2a/3=1a=-5/6