Question

The maximum afternoon temperature in Granderson is modeled by t=60-30 cos (xexFormula1" title=" \pi " alt=" \pi " align="absmiddle" class="latex-formula">/6) where t represents the maximum afternoon temperature in degrees Fahrenheit in month x, with x=0 representing january, x=1 representing february, and so on.
Find the maximum afternoon temperature in August.

Answer 1

The equation is $t=60-30 cos (x \frac{\pi}{6})$

the -30 expression, is subtracting, if negative, from the 60
if the cosine returned is negative, you'd end up with a +30,
negative * negative = positive

and then the 30 expression will ADD to the 60 amount
so the temperature "t" is highest, when the 30 expression, is
positive and and it's highest

when does that happen, when cosine is negative and at its highest,
well, cosine range is $-1\le cos(\theta ) \le 1$
so the lowest value cosine can provide is -1, when is cosine -1?
well, at $\pi$

so...  let's find a value that makes that expression to $cos(\pi)$

$\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=6 \\\\ thus \\\\ t=60-30 cos (6 \frac{\pi}{6})\implies t=60-30 cos (\pi ) \\\\ t=60-30[-1]\implies t=60+30\implies t=90$

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so...for August, that'll mean


$\bf t=60-30 cos (x \frac{\pi}{6}) \qquad x=7 \\\\ t=60-30 cos (\frac{7\pi}{6})\implies t=60-30\left( -\cfrac{\sqrt{3}}{2} \right) \\\\ t=60+(15\cdot \sqrt{3})\implies t\approx85.98^o$