If a = b cos

Question:

If $a=b \cos \frac{2 \pi}{3}=c \cos \frac{4 \pi}{3}$, then write the value of $a b+b c+c a$.

Solution:

$a=b \cos 120^{\circ}=c \cos 240^{\circ}$

$\Rightarrow a=-\frac{1}{2} b=-\frac{1}{2} c$

Therefore,

$a b+b c+c a=\frac{-1}{2} b \times b+b \times b+b \times \frac{-1}{2} b$

$=-b^{2}+b^{2}$

$=0$