1)Вычислить [latex]cos \alpha [/latex], [latex]sin2 \alpha[/latex] Если [latex]sin\alpha= \frac{9}{13} [/latex] и [latex] \frac{ \pi }{2} \ \textless \ \alpha \ \textless \ \pi [/latex]

sin(4α+π6)==sin4α×cosπ6+cos4α×sinπ6==sin4α×3√2+cos4α×12==2sin2α×cos2α×3√2+(cos22α−sin22α)×12==2sinα×cosα×cos2α×3√+cos22α−sin22α2==2sinα×cosα×(cos2α−sin2α)×3√+(cos2α−sin2α)(cos2α−sin2α)−sin22α2==23√sinαcosα(cos2α−sin2α)+(cos2α−sin2α)(cos2α−sin2α)−(2sinα×cosα)(2sinα×cosα)2==23√sinαcosα(cos2α−sin2α)+(cos2α−sin2α)(cos2α−sin2α)2−2sin2α×cos2α==2×3√sinαcosα(cos2α−sin2α)+cos4α−(sin2α×cos2α)−(sin2α×cos2α)+sin4α2−2sin2α×cos2α==23√sinαcosα(cos2α−sin2α)+(cos2α−sin2α)22−2sin2α×cos2α==23√(sinα×cos3α−sin3α×cosα)+(1−2sin2α)22−2sin2α×cos2α==23√(sinα×cos3α−sin3α×cosα)+12−2sin2α+2sin4α−2sin2α×cos2α==2sinα(3√×cos3α−3√sin2α×cosα+14sinα−sinα+sin3α−sinα×cos2α)=Right side:sin(2α+π5)=sin2x×cosπ5+cos2α×sinπ5==2sinαcosα×cosπ5+(cos2α−sin2α)sinπ5==2sinαcosα×cosπ5+(1−2sin2α)sinπ5==2sinαcosα×cosπ5+sinπ5−2sin2α×sinπ5==2sinα(cosα×cosπ5+sinπ52sinα−sinα×sinπ5)I can't get any further on either. Bringing them together I get:2sinα(3√×cos3α−3√sin2α×cosα+14sinα−sinα+sin3α−sinα×cos2α)=2sinα(cosα×cosπ5+sinπ52sinα−sinα×sinπ5)