Помогите номер 350.алгебра

Question 1

Question 2

$1)sin(\alpha+\beta)=sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)\\0\ \textless \ \alpha\ \textless \ 90\Rightarrow sin(\alpha)\ \textgreater \ 0\Rightarrow sin(\alpha)=\sqrt{1-cos^2(\alpha)}=\\=\sqrt{1-0.36}=0.8\\sin(\alpha+30^\circ)=0.8*{\sqrt3\over2}+0.6*{1\over2}=0.4\sqrt3+0.3\\\\ 2)cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\\0\ \textless \ \alpha\ \textless \ 90\Rightarrow cos(\alpha)\ \textgreater \ 0\Rightarrow cos(\alpha)=\sqrt{1-sin^2(\alpha)}=\\=\sqrt{1-{1\over2}}={sqrt2\over2}\\cos(\alpha+60^\circ)={\sqrt2\over2}*{1\over2}-{\sqrt2\over2}*{\sqrt3\over2}={\sqrt2\over4}(1-\sqrt3)\\\\ 3)sin(\alpha-\beta)=sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)\\270\ \textless \ \alpha\ \textless \ 360\Rightarrow sin(\alpha)\ \textless \ 0\Rightarrow sin(\alpha)=-\sqrt{1-sin^2(\alpha)}=\\=-\sqrt{1-{1\over4}}=-{\sqrt3\over2}\\180\ \textless \ \beta\ \textless \ 270\Rightarrow cos(\beta)\ \textless \ 0\Rightarrow cos(\beta)=-\sqrt{1-sin^2(\beta)}=\\=-\sqrt{1-0.16}=-{0.2\sqrt{21}}\\sin(\alpha-\beta)=0.3\sqrt7+0.2\\cos(\alpha+\beta)=-0.1\sqrt{21}-0.2\sqrt3\\\\ 4)cos(\alpha-\beta)=cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)\\cos(\alpha)\ \textless \ 0\Rightarrow cos(\alpha)=-\sqrt{1-sin^2(\alpha)}=-{\sqrt5\over3}\\sin(\beta)\ \textless \ 0\Rightarrow sin(\beta)=-\sqrt{1-cos^2(\alpha)}=-{\sqrt7\over4}\\sin(\alpha+\beta)=-{1\over2}+{\sqrt{35}\over12}\\cos(\alpha-\beta)={\sqrt5\over4}-{\sqrt7\over6}$ )