Question

Using complex numbers, prove that the, angles A, B and C of a planar triangle satisfy the relations

(i) cos^2 A + cos^2 B + cos^2 C = 1 − 2 cos A cos B cos C

(ii) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

Answer 1

(i)

$1 − 2 cos A cos B cos C=1-2(\frac{e^{iA}+e^{−iA}}{2})(\frac{e^{iИ}+e^{−iИ}}{2})(\frac{e^{iС}+e^{−iС}}{2})=$

$=1-\frac{1}{4} (e^{i(A+B+C)}+e^{-i(A+B+C)}+e^{i(A+B-C)}+e^{-i(A+B-C)}+e^{i(A-B+C)}+e^{-i(A-B+C)}+e^{i(-A+B+C)}+e^{-i(-A+B+C)})=$

$=1-\frac{1}{4}(e^{i(\pi-2C)}+e^{-i(\pi-2C)}+e^{i(\pi-2B)}+e^{-i(\pi-2B)}+e^{i(\pi-2A)}+e^{-i(\pi-2A)})=$

$=1-\frac{1}{2}(cos(\pi-2C)+cos(\pi-2B)+cos(\pi-2A))=$

$=1+\frac{1}{2}(cos2C+cos2B+cos2A)=cos^2A+cos^2B+cos^2C.$

(ii)

$4sinAsinBsinC=4(\frac{e^{iA}−e^{−iA}}{2i})(\frac{e^{iB}−e^{−iB}}{2i})(\frac{e^{iC}−e^{−iC}}{2i})=$

$=\frac{-e^{i(A+B-C)}+e^{i(-A-B+C)}}{-2i}+\frac{-e^{i(B+C-A)}+e^{i(-B-C+A)}}{-2i}+\frac{-e^{i(C+A-B)}+e^{i(-C-A+B)}}{-2i}+\frac{e^{i(A+B+C)}}{-2i}-\frac{e^{i(-A-B-C)}}{-2i}=$

$=sin(A+B−C)+sin(B+C−A)+sin(C+A−B)+\frac{e^{i\pi}}{-2i}-\frac{e^{-i\pi}}{-2i}=$

$=sin(π−2C)+sin(π−2A)+sin(π−2B)=sin2C+sin2A+sin2B$