if-tan-r-s-and-sin-s-r-show-that-cos-r-2-s-

Question Number 33466 by mondodotto@gmail.com last updated on 17/Apr/18

if \tan β = \frac{r}{\sqrt{s}} and \sin β = \frac{\sqrt{s}}{r}

show that \cos β = \sqrt{r^{2} + s}

Commented by MJS last updated on 17/Apr/18

\cos β = \sqrt{1 - \sin^{2} β} = \frac{\sqrt{r^{2} - s}}{∣ r ∣}

\sqrt{r^{2} + s} = \frac{\sqrt{r^{2} - s}}{∣ r ∣} \Rightarrow s = r^{2} \frac{1 - r^{2}}{1 + r^{2}}

\cos β = \frac{\sin β}{\tan β}

\sqrt{r^{2} + s} = \frac{s}{r^{2}} \Rightarrow s = \frac{r^{3}}{2} (r \pm \sqrt{4 + r^{2}})

r^{2} \frac{1 - r^{2}}{1 + r^{2}} = \frac{r^{3}}{2} (r \pm \sqrt{4 + r^{2}}) \Rightarrow

\Rightarrow r = \sqrt{- 2 + \sqrt{5}}; s = \frac{7}{2} - \frac{3}{2} \sqrt{5}

β \approx 51.83 °

{it}^{'} s only true in this singular case

Commented by mondodotto@gmail.com last updated on 17/Apr/18

thanx for your help sir

Answered by Rasheed.Sindhi last updated on 17/Apr/18

\tan β = \frac{\sin β}{\cos β}

\frac{r}{\sqrt{s}} = \frac{\frac{\sqrt{s}}{r}}{\cos β}

\cos β = \frac{\frac{\sqrt{s}}{r}}{\frac{r}{\sqrt{s}}} = \frac{\sqrt{s}}{r} \times \frac{\sqrt{s}}{r} = \frac{s}{r^{2}} \neq \sqrt{r^{2} + s} in general .