Автор - bogdaneuzzz

довести тотожність
а) tg8a-ctg8a=-2ctg16a
б) (cos(π/2-5a)-sin(π+3a))(sin(π/2+3a)-cos(π+5a))/1+cos(2π-2a)=sin8a

Ответ

Проверено экспертом

Автор - Universalka

1)tg8alpha-Ctg8alpha=frac{Sin8alpha}{Cos8alpha}-frac{Cos8alpha}{Sin8alpha}=frac{Sin^{2}8alpha-Cos^{2}8alpha}{Sin8alpha Cos8alpha}=frac{-Cos16alpha}{Sin8alpha Cos8alpha}=frac{-2Cos16alpha}{2Sin8alpha Cos8alpha}=frac{-2Cos16alpha}{Sin16alpha}=-2Ctg16alpha\\-2Ctg16alpha=-2Ctg16alpha

2)frac{(Cos(frac{pi}{2}-5alpha)-Sin(pi+3alpha))(Sin(frac{pi }{2}+3alpha)-Cos(pi+5alpha))}{1+Cos(2pi-2alpha)} =frac{(Sin5alpha+Sin3alpha)(Cos3alpha+Cos5alpha)}{1+Cos2alpha}=frac{2Sin4alpha Cosalpha*2Cos4alpha Cosalpha}{1+Cos2alpha}=frac{2Sin4alpha*Cos4alpha*2Cos^{2}alpha   }{2Cos^{2}alpha}=Sin8alpha\\Sin8alpha=Sin8alpha

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