Betsy Rhone

2021-12-31

Prove that $\frac{\mathrm{sin}\left(x\right)}{2}+{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)\mathrm{tan}\left(\frac{x}{2}\right)\to \mathrm{tan}\left(\frac{x}{2}\right)$

Daniel Cormack

Expert

Use $\mathrm{sin}\left(x\right)=2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)$ then we have
$\frac{\mathrm{sin}x}{2}+{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)\mathrm{tan}\left(\frac{x}{2}\right)=\frac{\mathrm{sin}\frac{x}{2}}{\mathrm{cos}\frac{x}{2}}\underset{=1}{\underset{⏟}{{\mathrm{cos}}^{2}\frac{x}{2}+{\mathrm{sin}}^{2}\frac{x}{2}}}$
$=\mathrm{tan}\frac{x}{2}$

chumants6g

Expert

Fairly obvious:
${\mathrm{sin}}^{2}\frac{x}{2}\mathrm{tan}\frac{x}{2}=\left(1-{\mathrm{cos}}^{2}\frac{x}{2}\right)\mathrm{tan}\frac{x}{2}=\mathrm{tan}\frac{x}{2}-\mathrm{cos}\frac{x}{2}\mathrm{sin}\frac{x}{2}=\mathrm{tan}\frac{x}{2}-\frac{\mathrm{sin}x}{2}$

Vasquez

Expert

With and
$\frac{1}{2}2sc+{s}^{2}\frac{s}{c}=\frac{s{c}^{2}+{s}^{3}}{c}=\frac{s}{c}$

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