Daniell Phillips
2022-01-17
Answered

Prove trigonometric an identity

$\frac{2\mathrm{cos}\alpha -1}{\sqrt{3}-2\mathrm{sin}\alpha}=\mathrm{tan}(\frac{\alpha}{2}+\frac{\pi}{6})$

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Joseph Fair

Answered 2022-01-18
Author has **34** answers

Using Prosthaphaeresis Formulas

$\frac{12}{=}\mathrm{sin}\frac{\pi}{6}=\mathrm{cos}\frac{\pi}{3}$

$\frac{\sqrt{3}}{2}=\mathrm{cos}\frac{\pi}{6}=\mathrm{sin}\frac{\pi}{3}$

$\mathrm{cos}\alpha -\mathrm{cos}\frac{\pi}{3}=2\mathrm{sin}(\frac{\pi}{6}-\frac{\alpha}{2})\mathrm{sin}(\frac{\pi}{6}+\frac{\alpha}{2})$

$\mathrm{sin}\frac{\pi}{3}-\mathrm{sin}\alpha =2\mathrm{sin}(\frac{\pi}{6}-\frac{\alpha}{2})\mathrm{cos}(\frac{\pi}{6}+\frac{\alpha}{2})$

Remember we need$\mathrm{sin}(\frac{\pi}{6}-\frac{\alpha}{2})\ne 0$ to reach at the identity required

Alternatively, we can use Weierstrass Substitution in the left hand side and use$\mathrm{tan}(A+B)$ formula

Remember we need

Alternatively, we can use Weierstrass Substitution in the left hand side and use

RizerMix

Answered 2022-01-19
Author has **438** answers

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