Tiffany Russell

2021-12-30

how to show that $\mathrm{csc}x-\mathrm{csc}\left(\frac{\pi }{3}+x\right)+\mathrm{csc}\left(\frac{\pi }{3}-x\right)=3\mathrm{csc}3x$?
My attempt:
$LHS=\mathrm{csc}x-\mathrm{csc}\left(\frac{\pi }{3}+x\right)+\mathrm{csc}\left(\frac{\pi }{3}-x\right)$
$=\frac{1}{\mathrm{sin}x}-\frac{1}{\mathrm{sin}\left(\frac{\pi }{3}+x\right)}+\frac{1}{\mathrm{sin}\left(\frac{\pi }{3}-x\right)}$
$=\frac{\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}+x\right)+\mathrm{sin}\left(\frac{\pi }{3}+x\right)\mathrm{sin}\left(\frac{\pi }{3}-x\right)-\mathrm{sin}\left(\frac{\pi }{3}-x\right)\mathrm{sin}x}{\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}+x\right)\mathrm{sin}\left(\frac{\pi }{3}-x\right)}$
$=\frac{4}{\mathrm{sin}3x}\left(\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}+x\right)+\mathrm{sin}\left(\frac{\pi }{3}+x\right)\mathrm{sin}\left(\frac{\pi }{3}-x\right)-\mathrm{sin}\left(\frac{\pi }{3}\right)\mathrm{sin}x\right)$
$=\frac{4}{\mathrm{sin}3x}\left(\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}+x\right)-\mathrm{sin}\left(\frac{\pi }{3}-x\right)\mathrm{sin}\left(\mathrm{sin}x-\mathrm{sin}\left(\frac{\pi }{3}+x\right)\right)\right)$
$=\frac{4}{\mathrm{sin}3x}\left(\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}+x\right)-\mathrm{sin}\left(\frac{\pi }{3}-x\right)\left(2\mathrm{sin}\frac{-\pi }{6}\mathrm{cos}\left(x+\frac{\pi }{6}\right)\right)\right)$

Jordan Mitchell

We have
$\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}+x\right)+\mathrm{sin}\left(\frac{\pi }{3}-x\right)\mathrm{cos}\left(x+\frac{\pi }{6}\right)$
$=\mathrm{sin}\left(x\right)\left[\mathrm{sin}\left(\frac{\pi }{3}\right)\mathrm{cos}\left(x\right)+\mathrm{cos}\left(\frac{\pi }{3}\right)\mathrm{sin}\left(x\right)\right]+\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\mathrm{cos}\left(x\right)-\mathrm{cos}\left(\frac{\pi }{3}\right)\mathrm{sin}\left(x\right)\right)\left(\mathrm{cos}\left(\frac{\pi }{6}\right)\mathrm{cos}x-\mathrm{sin}\left(\frac{\pi }{6}\right)\mathrm{sin}\left(x\right)\right)$
$=\mathrm{sin}\left(x\right)\left[\frac{\sqrt{3}}{2}\mathrm{cos}\left(x\right)+\frac{12}{\mathrm{sin}\left(x\right)}\right]+\left(\frac{\sqrt{3}}{2}\mathrm{cos}\left(x\right)-\frac{12}{\mathrm{sin}\left(x\right)}\right)\left(\frac{\sqrt{3}}{2}\mathrm{cos}\left(x\right)-\frac{12}{\mathrm{sin}\left(x\right)}\right)$
$=\frac{\sqrt{3}}{2}\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)+{\frac{12}{\mathrm{sin}}}^{2}x+{\frac{34}{\mathrm{cos}}}^{2}\left(x\right)-\frac{\sqrt{3}}{2}\mathrm{cos}\left(x\right)\mathrm{sin}\left(x\right)+{\frac{14}{\mathrm{sin}}}^{2}\left(x\right)$
or
$\frac{3}{4}\left({\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)\right)=\frac{34}{}$

Mason Hall

$\frac{1}{\mathrm{sin}x}-\frac{1}{\mathrm{sin}\left(\frac{\pi }{3}+x\right)}+\frac{1}{\mathrm{sin}\left(\frac{\left\{}{\left\{}-x\right)}=$
$=\frac{\mathrm{sin}\left(\frac{\pi }{3}-x\right)\mathrm{sin}\left(\frac{\pi }{3}+x\right)+\mathrm{sin}x\left(\mathrm{sin}\left(\frac{\pi }{3}+x\right)-\mathrm{sin}\left(\frac{\pi }{3}-x\right)\right)}{\mathrm{sin}x\mathrm{sin}\left(\frac{\pi }{3}+x\right)\mathrm{sin}\left(\frac{\pi }{3}-x\right)}$
$=\frac{\frac{1}{2}\left(\mathrm{cos}2x-\mathrm{cos}\frac{2\pi }{3}\right)+\mathrm{sin}x\cdot 2\mathrm{sin}x\mathrm{cos}\frac{\pi }{3}}{\mathrm{sin}x\left(\frac{\sqrt{3}}{2}\mathrm{cos}x+\frac{12}{\mathrm{sin}x}\right)\left(\frac{\sqrt{3}}{2}\mathrm{cos}x-\frac{12}{\mathrm{sin}x}\right)}$
$=\frac{\frac{12}{-}{\mathrm{sin}}^{2}x+\frac{14}{+}{\mathrm{sin}}^{2}x}{\mathrm{sin}x\left({\frac{34}{\mathrm{cos}}}^{2}x-{\frac{14}{\mathrm{sin}}}^{2}x\right)}=\frac{3}{\mathrm{sin}x\left(3\left(1-{\mathrm{sin}}^{2}x\right)-{\mathrm{sin}}^{2}x\right)}=\frac{3}{\mathrm{sin}3x}$

Vasquez