Question Number 143958 by bobhans last updated on 20/Jun/21
Answered by bramlex last updated on 20/Jun/21
$$\Rightarrow\:\begin{cases}{\mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{2sin}\:\alpha\mathrm{sin}\:\beta+\mathrm{sin}\:^{\mathrm{2}} \beta=\mathrm{9}{p}}\\{\mathrm{cos}\:^{\mathrm{2}} \alpha+\mathrm{2cos}\:\alpha\mathrm{cos}\:\beta+\mathrm{cos}\:^{\mathrm{2}} \beta=\mathrm{9}{q}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{2}+\mathrm{2cos}\:\left(\alpha−\beta\right)=\mathrm{9}\left({p}+{q}\right)\: \\ $$$$\Leftrightarrow\:\mathrm{2cos}\:\left(\alpha−\beta\right)=\mathrm{9}\left({p}+{q}\right)−\mathrm{2} \\ $$$$\Rightarrow\mathrm{cos}\:\left(\alpha−\beta\right)=\frac{\mathrm{9}\left({p}+{q}\right)−\mathrm{2}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha−\beta\right)=\pm\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{9}\left({p}+{q}\right)−\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$