Answer
$cos~A = 2~cos^2~B-1$
$A = 2B$
Work Step by Step
We can use the law of cosines to find $A$:
$a^2 = b^2+c^2-2bc~cos~A$
$2bc~cos~A = b^2+c^2-a^2$
$cos~A = \frac{b^2+c^2-a^2}{2bc}$
$A = arccos(\frac{b^2+c^2-a^2}{2bc})$
$A = arccos(\frac{4^2+5^2-6^2}{(2)(4)(5)})$
$A = arccos(0.125)$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{6^2+5^2-4^2}{(2)(6)(5)})$
$B = arccos(0.75)$
Note the following identity:
$cos~2x = 2~cos^2~x-1$
We can verify that $cos~A = 2~cos^2~B-1$:
$2~cos^2~B-1$
$= 2~(0.75)^2-1$
$= 2~(\frac{3}{4})^2-1$
$= 2~(\frac{9}{16})-1$
$= \frac{18}{16}-1$
$= \frac{2}{16}$
$= \frac{1}{8}$
$= 0.125$
$= cos~A$
Therefore:
$A = 2B$
