Выразить: а) tg 3a через tg a; б) sin 5a через sin a. Created by wwcltrl. algebra-ru

Выразить: а) tg 3a через tg a; б) sin 5a через sin a...

wwcltrl @wwcltrl

July 2022 1 5 Report

Выразить: а) tg 3a через tg a; б) sin 5a через sin a

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$tg3\alpha=tg(2\alpha+\alpha)=\frac{tg2\alpha+tg\alpha}{1-tg2\alpha*tg\alpha}=\frac{\frac{2tg\alpha}{1-tg^{2}\alpha}+tg\alpha}{1-\frac{2tg\alpha }{1-tg^{2}\alpha}*tg\alpha}=\frac{2tg\alpha+tg\alpha(1-tg^{2}\alpha)}{1-tg^{2}\alpha-2tg^{2}\alpha}=\frac{3tg\alpha-tg^{3}\alpha}{1-3tg^{2}\alpha}$

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

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