Question

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1, 2 (:

Answer 1

$sina+cosa= \frac{1}{2}\\\\sina*cosa= \frac{2sina*cosa}{2}= \frac{(sina+cosa)^2-(sin^2a+cos^2a)}{2} =\\\\= \frac{ (\frac{1}{2})^2-1 }{2}= \frac{ \frac{1}{4}-1 }{2}=\frac{- \frac{3}{4} }{2}=- \frac{3}{8}\\\\\\sin^3a+cos^3a=(sina+cosa)(sin^2a-sina*cosa+cos^2a)=\\\\=(sina+cosa)(1-sina*cosa)= \frac{1}{2}*(1-(- \frac{3}{8}))=\\\\= \frac{1}{2}*(1+ \frac{3}{8})= \frac{1}{2}* \frac{11}{8}= \frac{11}{16 }$