In a $\triangle A B C, \cos \left(\frac{B+2 C+3 A}{2}\right)+\cos \left(\frac{A-B}{2}\right)$ is equal to
In a $\triangle A B C, \cos \left(\frac{B+2 C+3 A}{2}\right)+\cos \left(\frac{A-B}{2}\right)$ is equal to
- $-1$
- $0$
- $1$
- $2$
Solution
In $\triangle A B C$,
$
\begin{gathered}
A+B+C=1 \pi \\
\therefore \quad \cos \left(\frac{B+2 C+3 A}{2}\right)+\cos \left(\frac{A-B}{2}\right) \\
=\cos \left(\frac{2 A+2 B+2 C-B+A}{2}\right)+\cos \left(\frac{A-B}{2}\right) \\
=\cos \left(\frac{2 \pi+(A-B)}{2}\right)+\cos \left(\frac{A-B}{2}\right)
\end{gathered}
$
$
\begin{aligned}
& =\cos \left(\pi+\left(\frac{A-B}{2}\right)\right)+\cos \left(\frac{A-B}{2}\right) \\
& =-\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A-B}{2}\right)=0
\end{aligned}
$
Alternative
As we know in a $\triangle A B C$,
$
\begin{gathered}
A+B+C=\pi \\
\therefore \quad \cos \left(\frac{B+2 C+3 A}{2}\right)+\cos \left(\frac{A-B}{2}\right) \\
=2 \cos \left(\frac{2 C+4 A}{4}\right) \cos \left(\frac{2 A+2 B+2 C}{4}\right) \\
=2 \cos \left(\frac{C+2 A}{2}\right) \cos \left(\frac{\pi}{2}\right)=0
\end{gathered}
$
Asked in: AP EAMCET 2004
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