Leland Morrow

## Answered question

2022-07-01

$\text{If}\phantom{\rule{thickmathspace}{0ex}}\sqrt{9-8\mathrm{sin}{50}^{\circ }}=a+b\mathrm{csc}{50}^{\circ }\text{, then}\phantom{\rule{thickmathspace}{0ex}}ab=\text{?}$

### Answer & Explanation

Hadley Cunningham

Beginner2022-07-02Added 20 answers

$\sqrt{9-8\mathrm{sin}{50}^{\circ }}$
$=\mathrm{csc}{50}^{\circ }\sqrt{9{\mathrm{sin}}^{2}{50}^{\circ }-8{\mathrm{sin}}^{3}{50}^{\circ }}$

$=\mathrm{csc}{50}^{\circ }\sqrt{9{\mathrm{sin}}^{2}{50}^{\circ }-8\mathrm{sin}{50}^{\circ }\left(1-{\mathrm{cos}}^{2}{50}^{\circ }\right)}$
$=\mathrm{csc}{50}^{\circ }\sqrt{9{\mathrm{sin}}^{2}{50}^{\circ }-8\mathrm{sin}{50}^{\circ }+8\mathrm{sin}{50}^{\circ }{\mathrm{cos}}^{2}{50}^{\circ }}$

$=\mathrm{csc}{50}^{\circ }\sqrt{9{\mathrm{sin}}^{2}{50}^{\circ }-8\mathrm{sin}{50}^{\circ }+4\mathrm{sin}{100}^{\circ }\mathrm{cos}{50}^{\circ }}$

$=\mathrm{csc}{50}^{\circ }\sqrt{9{\mathrm{sin}}^{2}{50}^{\circ }-8\mathrm{sin}{50}^{\circ }+2\left(\mathrm{sin}{150}^{\circ }+\mathrm{sin}{50}^{\circ }\right)}$
$=\mathrm{csc}{50}^{\circ }\sqrt{9{\mathrm{sin}}^{2}{50}^{\circ }-6\mathrm{sin}{50}^{\circ }+2\mathrm{sin}{150}^{\circ }}$

$=\mathrm{csc}{50}^{\circ }\sqrt{9{\mathrm{sin}}^{2}{50}^{\circ }-6\mathrm{sin}{50}^{\circ }+1}$
$=\mathrm{csc}{50}^{\circ }\sqrt{\left(3\mathrm{sin}{50}^{\circ }-1{\right)}^{2}}$
$\text{(Taking the positive root.)}$
$=\mathrm{csc}{50}^{\circ }\left(3\mathrm{sin}{50}^{\circ }-1\right)$
$=3-\mathrm{csc}{50}^{\circ }$
So a=3 and b=−1

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