1) sina+sin2a+sin3a+sin4a=?2) cos2a-cos4a-cos6a+cos8a=?3) cosa=1/ cos2a-cos6a=?4) sina=2/...

Решите задачу:

$(sina+sin4a)+(sin2a+sin3a) = 2sin\frac{5a}{2}*cos\frac{3a}{2}+2sin\frac{5a}{2}*cos\frac{a}{2}\\ 2sin\frac{5a}{2}(cos\frac{3a}{2}+cos\frac{a}{2})$

$(cos2a+cos8a)-(cos4a+cos6a)=\\ 2cos5a*cos3a-(2cos5a*cosa)=2cos5a(cos3a-cosa)$

$cosa=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\\ sina=\frac{\sqrt{6}}{{3}}\\ cos2a-cos6a=16sina*cosa*(2cos^2a-1)*sina*cosa=\\ 16*\frac{\sqrt{18}}{3}*(2*\frac{1}{3}-1)*\frac{\sqrt{18}}{3}= -\frac{32}{3}$

$sina=\frac{2}{\sqrt{5}}\\ sin5a-sin3a=\\ 2sina*(2(1-2sin^2a)^2-1)=\frac{4}{\sqrt{5}}(2* \frac{9}{25} -1) = \frac{4}{\sqrt{5}}*-\frac{7}{25}=\\ -\frac{28}{25\sqrt{5}}$