# Rewrite equation in new form

Write the equation:

$\frac{24}{13}e^{\frac{-t}{2}}[19\sin(\frac{13t}{8})+13\cos(\frac{13t}{8})]$

In the form:

__$\cos[\frac{13t}{8}-\tan^{-1}(\frac{19}{13})]$

I have to fill in the blank spot. Any help?

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Hint: if $\theta = \tan^{-1}(19/13)$, $\cos(\theta) = 13/ R$ and $\sin(\theta) = 19/ R$ for some $R$. What is $R$? Note that $\sin(\theta) \sin(13 t/8) + \cos(\theta) \cos(13t/8) = \cos(13 t/8 - \theta)$.
$$a \sin(\theta) + b \cos(\theta) = \sqrt{a^2 + b^2} \times \cos(\theta - \arctan(a/b))$$