landdenaw

2022-06-24

If $\frac{\mathrm{sin}x}{\mathrm{sin}y}=3$ and $\frac{\mathrm{cos}x}{\mathrm{cos}y}=\frac{1}{2}$, then find $\frac{\mathrm{sin}2x}{\mathrm{sin}2y}+\frac{\mathrm{cos}2x}{\mathrm{cos}2y}$

Eli Shaffer

$\mathrm{sin}y=3\mathrm{sin}x$ and $\mathrm{cos}y=2\mathrm{cos}x$
Thus,
$1={\mathrm{sin}}^{2}y+{\mathrm{cos}}^{2}y=9{\mathrm{sin}}^{2}x+4{\mathrm{cos}}^{2}x=4+5{\mathrm{sin}}^{2}x,$
which is impossible.If $\mathrm{sin}y=\frac{1}{3}\mathrm{sin}x$ we obtain:
$1={\mathrm{sin}}^{2}y+{\mathrm{cos}}^{2}y=\frac{1}{9}{\mathrm{sin}}^{2}x+4{\mathrm{cos}}^{2}x=\frac{1}{9}+\frac{35}{9}{\mathrm{sin}}^{2}x.$
Now, you can get ${\mathrm{sin}}^{2}x$,${\mathrm{sin}}^{2}y$ and the rest for you.

aligass2004yi

hint
$\mathrm{sin}\left(y\right)=\frac{1}{3}\mathrm{sin}\left(x\right)$
$\mathrm{cos}\left(y\right)=2\mathrm{cos}\left(x\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\mathrm{sin}}^{2}\left(x\right)+36{\mathrm{cos}}^{2}\left(x\right)=9$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}35{\mathrm{cos}}^{2}\left(x\right)=8$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\mathrm{cos}}^{2}\left(x\right)=\frac{8}{35}$
the second expression is
$\frac{16-35}{64-35}=-19/29$
the final result is
$3/2-19/29=49/58$

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