sin(A B)=8sin(B 2)

(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^A*(sin(A+B)-sin(A

cos(π/6)=1-2[sin(π/12)]^2所以sin(π/12)=√[(1-cos(π/6)/2]=√[(3-√3)/6]

用正弦得b^2+c^2=a^2+bcb^2+c^2-a^2=bc再用余弦得cosA=（b^2+c^2-a^2）/2bc=1/2向量AC*向量AB=|AC||AB|cosA=4|AC||AB|=8sin

∵（a2+b2）sin（A-B）=（a2-b2）sinC，∴（a2+b2）（sinAcosB-cosAsinB）=（a2-b2）（sinAcosB+cosAsinB），可得sinAcosB（a2+b2

sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]

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

证明：(a²+b²)(sinAcosB-cosAsinB)=(a²-b²)(sinAcosB+cosAsinB)a²sinAcosB-a²c

(a^2+b^2)sin(A-B)=(a^2-b^2)sinC,(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^A*(sin(A+B)-sin(A-B))

用正弦得b^2+c^2=a^2+bcb^2+c^2-a^2=bc再用余弦得cosA=（b^2+c^2-a^2）/2bc=1/2向量AC*向量AB=|AC||AB|cosA=4|AC||AB|=8sin

1.B(直角三角形C为直角)2.B（一个是-1,一个是0）3.B4.B（运用和差化积公式）5.额我必须用导数来做这个题了是不是式子抄错了导数的话很明显就超纲了.6.此题有错由tanA=1/2,cosA

sin(arcsinx)=xcos(arccosx)=xcos(2arcsinx)=1-2[sin(arcsinx)]^2=1-2x^2

(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),化简得sinAsinB(sin2A-sin2B)=0,（因为A、B为三角形内角,则其正弦不为0）sin2A=sin2B2A=2B

求导原计算是y'=dy/dx那么sin'wt=dsinwt/dt=(dsinwt/dwt)*(dwt/dt)(分子分母同时乘以dwt,再分开写成两个式子）=(wt)'*coswt你题目中的结果不正确.

sina+sinb=sin[(a+b)/2+(a-b)/2]+sin[(a+b)/2-(a-b)/2]=sin(a+b)/2cos(a-b)/2+cos(a+b)/2sin(a-b)/2+sin(a+

sin(45°)=COS(315°)=45°=315°

他这是合并同类项(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^Asin(A-B)+sin^Bsin(A-B)=sin^Asin(A+B)-sin^Bsi

1.证明:角BAC为直角,即,证明:向量AB*向量AC=0,即可,向量AB*向量AC=(1+tanx)*sin(x-π/4)+(1-tanx)*sin(x+π/4)=[sin(x-π/4)+sin(x

证明：∵（a2-b2）2=[（a+b）（a-b）]2=[（tanθ+sinθ+tanθ-sinθ）（tanθ+sinθ-tanθ+sinθ）]2=16tan2θsin2θ．又16ab=16（tan2θ

证明：原式化为a2[sin（A-B）-sin（A+B）=-b2[sin（A-B）+sin（A+B）],即a2[sin（A+B）-sin（A-B）=b2[sin（A-B）+sin（A+B）],故2a2c

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(