In a $\triangle A B C, a=6, b=5$ and $c=4$, then $\cos 2 A=$
In a $\triangle A B C, a=6, b=5$ and $c=4$, then $\cos 2 A=$
$-\frac{31}{32}$
$-\frac{15}{16}$
$\frac{31}{32}$
$\frac{15}{16}$
Solution
In a $\triangle A B C, a=6$ units, $b=5$ units and $c=4$ units
As we know that,
$\cos A=\frac{b^2+c^2-a^2}{2 b c}$
$\Rightarrow \cos A=\frac{b^2+c^2-a^2}{2 b c}=\frac{5^2+4^2-6^2}{2 \times 5 \times 4}=\frac{1}{8}$
As we know that,
$\cos 2 A=\cos ^2 A-\sin ^2 A$
$\begin{aligned} & =2 \cos ^2 A-1=2 \times\left(\frac{1}{8}\right)^2-1 \\ & =\frac{1}{32}-1=\frac{-31}{32}\end{aligned}$