sin^2A sin^C-sinAsinC=sin^2B

等式两边乘以4R^2用正弦定理得到a^2-c^2+b^2=ab根据余弦定理c^2=a^2+b^2-2abcosC代入第一个式子得到cosC=1/2因为C是三角形内角所以C=60度

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

sin²A+sin²B=2sin²C由正弦定理a^2+b^2=2c^2代入余弦定理：cosC=(a^2+b^2-c^2)/(2ab)=c^2/(2ab)>0所以：cosC

（1）由已知：2asinA=（2b+c）sinB+（2c+b）sinC,根据正弦定理得：2a2=(2b+c)b+(2c+b)c,即：a2=b2+c2+bc由余弦定理得：a2=b2+c2-2bccosA

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

为直角三角形,利用正弦定理即可得到：因为：sinA/a=sinB/b=sinC/c设上式等于t,则有：sinA=at；sinB=bt;sinC=ct;代入后得到：a^2t^2+b^2t^2=c^2t^

1.sin^2A+sin^2B>sin^2C,不一定是钝角三角形,可以是锐角,也可以是直角.2.sinb=cosA.即b+a=90°.所以为直角三角形

sin^2A+sin^2B=sin^2C=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+sin^2Bcos^2A+2sinAcosAsinBcosB左边减

根据正弦定理：a/sinA=b/sinB=c/sinC=2R,R为该三角形外接圆半径,则：a/2R=sinAb/2R=sinBc/2R=sinC因此：sinA:sinB:sinC=a:b:c=3:2:

sinc=2(sqr(3)/2*cosc-1/2*sinc)1=2(sqr(3)/2*cotc-1/2)cotc=2sqr3/3tanc=sqr3/2sinC=(sqr3/2)/sqr(1+(tanc

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

sin^2A+sin^2B+sin^2C=(1-cosA)/2+(1-cosB)/2+(1-cos^2C)=2-cos(A+B)cos(A-B)-cos^2C=2+cosCsoc(A-B)-cos^2

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

如果是老师布置的作业,我希望你能参考教材独立完成.如果只是个人学习.以下供你参考.现写的,还是热的.为了方便你理解,我注释上.在matlabR2012a下通过.满足条件的n有84个.i=1;%存放结果

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

sin^2A+sin^2B+sin^2C∴1∴2∴2∴2∴0∴0=cos(A+B)·[cos(A+B)+cos(A-B)]=cos(A+B)·2·cosA·cosB而cosC=-cos(A+B)∴co