sin^2A sin^2B sin^2C=2√3sinAsinBsinC

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sin^2A sin^2B sin^2C=2√3sinAsinBsinC
在三角形abc中,若bsin(90度+A)=asin(90度-B)求三角形形状

cosA=acosB,由正弦定理sinBcosA=sinAcosB,得sin(A-B)=0,得A=B,故为等腰三角形.

证明恒等式 三角比1. sin^2a+sin^2b-sin^2asin^2b+cos^2acos^2b=12. 2(1-

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

在△ABC中,已知sin^2+Asin^2B-sinAsinB=sin^2C.且ab=4.则三角形面积为?

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

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

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

函数y=sin(2π/3 x+π/4)化成y=Asin(wx+φ)的形式

已经是y=Asin(wx+φ)的形式了A=1w=2π/3φ=π/4

函数f(x)=(1+cos2x)/4sin(π/2+x)-asin(x/2)*cos(π-x/2)的最大值为2,试确定常

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

已知asin(θ+α)=bsin(θ+β),求证

asin(θ+α)=bsin(θ+β)a(sinθcosα+cosθsinα)=b(sinθcosβ+cosθsinβ)asinθcosα+acosθsinα=bsinθcosβ+bcosθsinβ移

y=2sin²B+cos((2π/3)-2B)化简成y=Asin(ωx+φ)

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

二倍角的三角函数sin²asin²p+cos²acos²p=(1/2)(1+co

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-

asinθ-bcosθ=根号a^2+b^2,(sin^2θ)/m^2+(cos^2θ)/n^2=1/(a^2+b^2)

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

在三角形ABC中,sinC=根号2/2,(c-b)sin^2A+bsin^2B=cSin^2C,求三个角的度数.

先有已知和正弦定理得:(sinC-sinB)sin^2A+sinBsin^2B=sinCSin^2C∴sinC=sinB或sin^A=sin^B+Sin^C+sinBsinC(1)sinC=sinB,

sin(x/3)cos(x/3)+√3cos^2(x/3) 化成Asin(Ωx+&)的形式

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

在三角形ABC中求证 aCOS A+bCOS B+cCOS C=2aSIN B SIN C

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

化简:sin^2a+sin^2β-sin^2asin^2β+cos^2acos^2β

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

△ABC中角A,B,C所对的边分别为a,b,c.已知asin²B/2+bsin²A/2=c/2

(1)∵asin²B/2+bsin²A/2=c/2∴a(1-cosB)+b(1-cosA)=ca-(a²+c²-b²)/2c+b-(b²+c

设f(x)=sin^2 x+asin^2 (x/2),求f(x)最大值

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)

函数f(x)=A(sin2wxcosφ +2cos^2wx*sinφ )-Asinφ 明天交,

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

已知x/acosθ+y/bsinθ=1,x/asinθ-y/bcosθ=1,则x^2/a^2+y^2/b^2=

x/acosθ+y/bsinθ=1x^2/a^2cosθ^2+y^2/b^2sinθ^2+2xy/absinθcosθ=1x/asinθ-y/bcosθ=1x^2/a^2sinθ^2+y^2/b^2c