Ответ:
[tex]tg3\alpha=tg(2\alpha+\alpha)=\frac{tg2\alpha+tg\alpha}{1-tg2\alpha*tg\alpha}=\frac{\frac{2tg\alpha}{1-tg^{2}\alpha}+tg\alpha}{1-\frac{2tg\alpha }{1-tg^{2}\alpha}*tg\alpha}=\frac{2tg\alpha+tg\alpha(1-tg^{2}\alpha)}{1-tg^{2}\alpha-2tg^{2}\alpha}=\frac{3tg\alpha-tg^{3}\alpha}{1-3tg^{2}\alpha}[/tex]
Sin5α = Sin(3α + 2α) = Sin3αCos2α + Sin2αCos3α = (3Sinα — 4Sin³α)(1 — Sin²α) + (4Cos³α — 3Cosα)*2SinαCosα = 3Sinα — 6Sin³α -4Sin³α + 8Sin⁵α + 2SinαCos²α(4Cos²α — 3) = 3Sinα — 10Sin³α + 8Sin⁵α +2Sinα(1 — Sin²α)(4 — 4Sin²α — 3 = 3Sinα — 10Sin³α + 8Sin⁵α + (2Sinα — 2Sin³α)(1 — 4Sin²α) = 3Sinα — 10Sin³α + 8Sin1⁵α +2Sinα — 8Sin³α — 2Sin³α + 8Sin⁵α = 5Sinα — 20Sin³α + 16Sin⁵α