enrotlavaec

2022-06-26

$f\left(a\right)=r\mathrm{cos}\left(a\right)-m\mathrm{sin}\left(a\right)$
has the maximum
$\sqrt{{r}^{2}+{m}^{2}}$
(where a, r and m are real numbers).
But I'm not sure how to prove it or even if it's true. Setting $\frac{d}{da}f\left(a\right)=0$ gives $-r\mathrm{sin}\left(a\right)-m\mathrm{cos}\left(a\right)=0$ or $\mathrm{tan}\left(a\right)=-m/r$, so $a=\mathrm{arctan}\left(-m/r\right)$. Plugging that in into wolfram does not give the claimed result...

Anika Stevenson

Expert

$f\left(a\right)=r\mathrm{cos}\left(a\right)-m\mathrm{sin}\left(a\right)=\sqrt{{r}^{2}+{m}^{2}}\left(\frac{r}{\sqrt{{r}^{2}+{m}^{2}}}\mathrm{cos}a-\frac{m}{\sqrt{{r}^{2}+{m}^{2}}}\mathrm{cos}a\right)$
Now suppose $\mathrm{cos}\theta =\frac{r}{\sqrt{{r}^{2}+{m}^{2}}}$, $\mathrm{sin}\theta =\frac{m}{\sqrt{{r}^{2}+{m}^{2}}}$, then you have $f\left(a\right)=\sqrt{{r}^{2}+{m}^{2}}\mathrm{cos}\left(a+\theta \right)$, from which the maximum and minimum values are obvious.

Semaj Christian

Expert

$\frac{{m}^{2}}{r\sqrt{\frac{{m}^{2}}{{r}^{2}}+1}}+\frac{r}{\sqrt{\frac{{m}^{2}}{{r}^{2}}+1}}$
$=\frac{{m}^{2}}{\sqrt{{m}^{2}+{r}^{2}}}+\frac{{r}^{2}}{\sqrt{{m}^{2}+{r}^{2}}}$
$=\frac{{m}^{2}+{r}^{2}}{\sqrt{{m}^{2}+{r}^{2}}}=\sqrt{{m}^{2}+{r}^{2}}$

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