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In \(\triangle A B C\), if \(a, b\) and \(c\) are in arithmetic progression, then \(\cos A+2 \cos B+\cos C=\)
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The correct answer is:
2
It is given, that in \(\triangle A B C, a, b, c\) are in \(\mathrm{AP}\)
So, \(2 b=a+c\)...(i)
Now, \(\cos A+2 \cos B+2 \cos C\)
\(\begin{aligned}
= & \frac{b^2+c^2-a^2}{2 b c}+2 \frac{a^2+c^2-b^2}{2 a c}+\frac{a^2+b^2-c^2}{2 a b} \\
= & \frac{a b^2+a c^2-a^3+2 a^2 b+2 c^2 b-2 b^3+a^2 c+b^2 c-c^3}{2 a b c} \\
= & \frac{a b^2+a c^2-a^3+2 b\left(a^2+c^2\right)-2 b^3+a^2 c+b^2 c-c^3}{2 a b c} \\
= & \frac{-2 b^2+a c^2-a^3+(a+c)\left(a^2+c^2\right)}{2 a b c} \\
= & \frac{-2 b^2 c-c^3+a^2 c+b^2 c-c^3}{2 a b c} \\
= & \frac{a b^2+2 a c^2+2 c a^2+b^2 c-2 b^3}{2 a b c} \\
= & \frac{b^2(a+c)+2 a c(a+c)-2 b^3}{2 a b c} \\
= & \frac{b^2(2 b)+2 a c(2 b)-2 b^3}{2 a b c}=2
\end{aligned}\)
So, \(2 b=a+c\)...(i)
Now, \(\cos A+2 \cos B+2 \cos C\)
\(\begin{aligned}
= & \frac{b^2+c^2-a^2}{2 b c}+2 \frac{a^2+c^2-b^2}{2 a c}+\frac{a^2+b^2-c^2}{2 a b} \\
= & \frac{a b^2+a c^2-a^3+2 a^2 b+2 c^2 b-2 b^3+a^2 c+b^2 c-c^3}{2 a b c} \\
= & \frac{a b^2+a c^2-a^3+2 b\left(a^2+c^2\right)-2 b^3+a^2 c+b^2 c-c^3}{2 a b c} \\
= & \frac{-2 b^2+a c^2-a^3+(a+c)\left(a^2+c^2\right)}{2 a b c} \\
= & \frac{-2 b^2 c-c^3+a^2 c+b^2 c-c^3}{2 a b c} \\
= & \frac{a b^2+2 a c^2+2 c a^2+b^2 c-2 b^3}{2 a b c} \\
= & \frac{b^2(a+c)+2 a c(a+c)-2 b^3}{2 a b c} \\
= & \frac{b^2(2 b)+2 a c(2 b)-2 b^3}{2 a b c}=2
\end{aligned}\)
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