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Hi. One of my math books seems to indicate that
$\displaystyle \sqrt {2 - 2\,sin\,t\,sin\,5t - 2\,cos\,t\,cos\,5t$
is equal to
$\displaystyle \sqrt {2 - 2\,cos\,4t}$
The transformation is I imagine quite obvious to everyone else, but if somebody could bear with me and show the hidden steps, it would help my learning.
Hello, infraRed!
We are expected to know this identity:
. . $\displaystyle \cos(A\pm B) \:=\:\cos A\cos B \mp\sin A\sin B$
Quote:
My math book indicates that: .$\displaystyle \sqrt {2 - 2\sin t\sin 5t - 2\cos t\cos5t} \;=\; \sqrt {2 - 2\cos4t}$
The left side has:
$\displaystyle 2 - 2\sin t\sin5t - 2\cos t\cos5t$
. . $\displaystyle =\;2 - 2(\cos 5t\cos t + \sin5t\sin t) $
. . $\displaystyle =\;2 - 2\cos(5t-t)$
. . $\displaystyle =\;2 - 2\cos4t$
