enrotlavaec

Answered

2022-06-26

In a book I'm reading, a claim is made that
$f\left(a\right)=r\mathrm{cos}\left(a\right)-m\mathrm{sin}\left(a\right)$
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...

Answer & Explanation

Anika Stevenson

Expert

2022-06-27Added 19 answers

$f\left(a\right)=r\mathrm{cos}\left(a\right)-m\mathrm{sin}\left(a\right)=\sqrt{{r}^{2}+{m}^{2}}\left(\frac{r}{\sqrt{{r}^{2}+{m}^{2}}}\mathrm{cos}a-\frac{m}{\sqrt{{r}^{2}+{m}^{2}}}\mathrm{cos}a\right)$
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

2022-06-28Added 12 answers

From your link,
$\frac{{m}^{2}}{r\sqrt{\frac{{m}^{2}}{{r}^{2}}+1}}+\frac{r}{\sqrt{\frac{{m}^{2}}{{r}^{2}}+1}}$
$=\frac{{m}^{2}}{\sqrt{{m}^{2}+{r}^{2}}}+\frac{{r}^{2}}{\sqrt{{m}^{2}+{r}^{2}}}$
$=\frac{{m}^{2}+{r}^{2}}{\sqrt{{m}^{2}+{r}^{2}}}=\sqrt{{m}^{2}+{r}^{2}}$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?