Betsy Rhone

Answered

2021-12-31

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

Answer & Explanation

Daniel Cormack

Expert

2022-01-01Added 34 answers

Use $\mathrm{sin}\left(x\right)=2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)$ then we have

$\frac{\mathrm{sin}x}{2}+{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)\mathrm{tan}\left(\frac{x}{2}\right)=\frac{\mathrm{sin}\frac{x}{2}}{\mathrm{cos}\frac{x}{2}}\underset{=1}{\underset{\u23df}{{\mathrm{cos}}^{2}\frac{x}{2}+{\mathrm{sin}}^{2}\frac{x}{2}}}$

$=\mathrm{tan}\frac{x}{2}$

chumants6g

Expert

2022-01-02Added 33 answers

Fairly obvious:

$\mathrm{sin}}^{2}\frac{x}{2}\mathrm{tan}\frac{x}{2}=(1-{\mathrm{cos}}^{2}\frac{x}{2})\mathrm{tan}\frac{x}{2}=\mathrm{tan}\frac{x}{2}-\mathrm{cos}\frac{x}{2}\mathrm{sin}\frac{x}{2}=\mathrm{tan}\frac{x}{2}-\frac{\mathrm{sin}x}{2$

Vasquez

Expert

2022-01-09Added 457 answers

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