若sin*2A sin*2B sin*2C

1.(sina)^2+(sinb)^2-(sinasinb)^2+(cosacosb)^2=(sina)^2-(sinasinb)^2+1-(cosb)^2+(cosacosb)^2=(sina)^2

由sin^2A+sin^2B-sinAsinB=sin^2C由正弦定理sinA=a/2R,sinB=b/2R,sinC=c/2R则(a/2R)^2+(b/2R)^2-(a/2R)(b/2R)=(c/2

诱导公式f(x)=(1+2cos²x-1)/(4cosx)+asin(x/2)cos(x/2)=(cosx)/2+a/2*sinx=(a/2)sinx+(1/2)cosx=√[(a/2)&s

已经是y=Asin(wx+φ)的形式了A=1w=2π/3φ=π/4

y=2√3sin²(x+π/4)+2cos²x-√3=√3*[2sin²(x+π/4)-1]+2cos²x-1+1由公式2cos²x-1=cos2x和

先化简得出f(x)=1/2cosx+a/2sinx=√[(1/2)^2+(a/2)^2]sin(x+∮)其中(tan∮=1/a)由于f(x)的最大值为2,所以√[(1/2)^2+(a/2)^2]=2所

f(x)=sin²x+2sin2x+3cos²x=1+2sin2x+2cos²x=1+2sin2x+cos2x+1=2sin2x+cos2x+2=√5sin(2x+fai

y=2sin²B+cos((2π/3)-2B)=(1-cos2B)-1/2cos2B+√3/2sin2B=(-3/2cos2B+√3/2sin2B)+1=√3(1/2sin2B-√3/2co

把左式的平方项化成二倍角：sin^2a=1/2(1-cos2a)sin^2p=1/2(1-cos2p);cos^2a=1/2(1+cos2a)cos^2p=1/2(1+cos2p)左式=1/4[(1-

sin(pai/2+x)=cosxcos(pai-x/2)=-cos(x/2)1+cos2x=1+[2(cosx)^2-1]=2(cosx)^2所以f(x)=2(cosx)^2/4cosx+asin(

(asinθ-bcosθ)²=a²+b²,两边同除以a²b²,(sinθ/b-cosθ/a)²=1/a²+1/b²,co

解y=sin(π/6-2X)+cos2x=1/2cos2x-√3/2sin2x+cos2x=3/2cos2x-√3/2sin2x=-√3（1/2sin2x-√3/2cos2x）=-√3sin（2x-π

原式=1/2COSX+asin(x/2)cos(x/2)=1/2COSX+a/2sinx=1/2(cosx+asinx)因为最大值是2所以(√1+a^2)/2=2a=+-√15

由和差化积公式：cosa+cosb=2cos{(a+b)／2}cos{(a-b)／2}得：cos2a+cos2b=2cos(a+b)cos(a-b)又由已知条件4sinasinb=根号2,4cosac

sin(x/3)cos(x/3）+√3cos^2(x/3)=(1/2)sin(2x/3)+(√3/2)[1+cos(2x/3)]=(1/2)sin(2x/3)+(√3/2)cos(2x/3)+√3/2

正弦定理知等价于证sinacosa+sinbcosb+sinccosc=2sinasinbsin（a+b）=2sin^2asinbcosb+2sin^2bsinacosa移项用二倍角公式等价于cos2

f(x)=-sin^2x+2asin(x-π/2)=-(1-cos^2x)+2a(-cosx)=cos^2x-2acosx-1=(cosx-a)^2-a^2-1由于-1

原式=sin^2a+sin^2β-（1-cos^2a)sin^2β+cos^2acos^2β=sin^2a+cos^2asin^2β+cos^2acos^2β=sin^2a+cos^2a(sin^2β

f(x)=sin^2x+asin^2(x/2)=sin^2x+a(1-cosx)=1-cos^2x+a-acosx1=-(cos^2x+acosx)+a+1=-(cos^2x+acosx+a^2/4)

(1)2cos^2wxsinφ=(2cos^2wx-1)sinφ+sinφ=cos2wxsinφ+sinφf(x)=A(sin2wxcosφ+cos2wxsinφ+sinφ)-Asinφ=Asin(2