a=3 sin^2b sinb sinc=sin^2a-sin^2c

sin²(a+阝)+cos²(a+阝)=1cos²(a+阝)=1-sin²(a+阝)=1-1=0cos(a+阝)=0∴sin2(a+阝)=2sin(a+阝)co

因为sin(a+π/2)=cosa,所以sin(2π/3+a)=sin[π/2+(π/6+a)]=cos(π/6+a)由sin(π/6+a)=（根号3）/3>0,得cos(π/6+a)=(±根号6)/

3(sinA)^2+2(sinB)^2=5sinA(sinA)^2+(sinB)^2=5sinA/2-(sinA)^2/25sinA/2-(sinA)^2/2=-(1/2)(sinA-5/2)^2+2

左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b=sin²a(1-sin

明显的用正弦定理嘛.

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

这种题目比较常见首先对已知条件两边平方得到sinacosa=3/8下面这个公式应该很熟悉的x^3-y^3=(x-y)(x^2+xy+y^2)照着式子代入所要求的式子等于1/2(1+3/8)=11/16

：[sin(a-60)]^2+[sina]^2+[sin(a+60)]^2=[1/2sina-根号3/2cosa]^2+[1/2sina+根号3/2cosa]^2+sin^2a=1/2sin^2a+3

解.2sin²a-sinacosa-3cos²a=(2sina-3cosa)(sina+cosa)=0∵a∈(0,π/2）∴2sina=3cosa即sina=3/√13,cosa=

sin^2(a)+2sin(a)cos(a)+3cos^2(a)/sin(a)cos(a)+cos^2(a)分子分母同除以cos²a,得原式=(tan²a+2tana+3)/(ta

2cos²a+3cosasina-3sin²a=1=sin²a+cos²acos²a+3cosasina-4sin²a=0(cosa+4si

2SINA-SINACOSA-3COSA=0两边同时除以cosA*cosA2tan^2(A)-tanA-3=0tanA=-1或tanA=3/2.A=-π/4或3π/4或{sinA=3/√13和cosA

/>利用积化和差公式,达到裂项的效果.2sinka*sin(a/2)=-cos[(k+1/2)a]+[cos(k-1/2)a]∴2sin(a/2)*(sina+sin2a+sin3a+...+sinn

sin(3π+a)=2sin[(3π/2)+a]3π+a在第三象限,sina

(1)(sina-3cosa)/(sina+cosa)=(tana-3)/(tana+1)=-2/3(2)(2sin^2a-3cos^2a)/(4sin^2a-9cos^2a)=[2(tana)^2-

应该是sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]A=(A+B)/2+(A-B)/2.B=(A+B)/2-(A-B)/2所以sin(A+B)/2cos(A-B)/2+cos(

sinA+sin(A+2π/3)+sin(A-2π/3)=sinA+sinAcos2π/3+cosAsin2π/3+sinAcos2π/3-cosAsin2π/3=sinA+2sinAcos2π/3=

1

cos^4a-sin^4a=3/5(cos²a-sin²a)(cos²a+sin²a)=3/5因为cos²a+sin²a=1所以cos

①f(a)=[(-sina)cosa(-sina)]/[(-cosa)sinacosa]=-tana②tan(-91π/3)=tan(-30π-π/3)=-tan(π/3)=-√3