公式
(1)(sinα)^2+(cosα)^2=1
(2)1+(tanα)^2=(secα)^2
(3)1+(cotα)^2=(cscα)^2
证明下面两式,只需将一式,左右同除(sinα)^2,第二个除(cosα)^2即可
(4)对于任意非直角三角形,总有
tanA+tanB+tanC=tanAtanBtanC
由余弦定理:a^2+b^2-c^2-2abcosC=0
正弦定理:a/sinA=b/sinB=c/sinC=2R
得 (sinA)^2+(sinB)^2-(sinC)^2-2sinAsinBcosC=0
转化 1-(cosA)^2+1-(cosB)^2-[1-(cosC)^2]-2sinAsinBcosC=0
即 (cosA)^2+(cosB)^2-(cosC)^2+2sinAsinBcosC-1=0
又 cos(C)=-cos(A+B)=sinAsinB-cosAcosB
得 (cosA)^2+(cosB)^2-(cosC)^2+2cosC[cos(C)+cosAcosB]-1=0
(cosA)^2+(cosB)^2+(cosC)^2=1-2cosAcosBcosC
得证
(8)(sinA)^2+(sinB)^2+(sinC)^2=2+2cosAcosBcosC