Câu i)$sin^2A+sin^2B+sin^2C=\frac{1-cos2A+1-cos2B+1}2+1-cos^2C$$=1+\frac{1}2(cos2A+cos2B)+1-cos^2C$$=2+cos(A+B)cos(A-B)-cos^2C$$=2+cos(A+B).cos(A-B)-cos^2(A+B)$$=2+cos(A+B)[cos(A-B)-cos(A+B)]$$=2-cos(A+B).2cosAcosB$$=2+2cosAcosBcosC$
Câu i)$
\sin^2A+
\sin^2B+
\sin^2C=\
dfrac{1-
\cos2A+1-
\cos2B+1}2+1-
\cos^2C
\\=1+\
dfrac{1}2
\left(
\cos2A+
\cos2B
\right)+1-
\cos^2C
\\=2+
\cos
\left(A+B
\right)
\cos(A-B)-
\cos^2C
\\=2+
\cos(A+B)
\cos(A-B)-
\cos^2(A+B)
\\=2+
\cos
\left(A+B
\right)
\left[
\cos(A-B)-
\cos(A+B)
\right]
\\=2-
\cos(A+B)
\times2
\cos
A
\cos
B
\\=2+2
\cos
A
\cos
B
\cos
C$