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Question Number 143958 by bobhans last updated on 20/Jun/21

Answered by bramlex last updated on 20/Jun/21

⇒ { ((sin^2 α+2sin αsin β+sin^2 β=9p)),((cos^2 α+2cos αcos β+cos^2 β=9q)) :} ⇔2+2cos (α−β)=9(p+q) ⇔ 2cos (α−β)=9(p+q)−2 ⇒cos (α−β)=((9(p+q)−2)/2) ⇒sin (α−β)=± (√(1−(((9(p+q)−2)/2))^2 ))

$$\Rightarrow\:\begin{cases}{\mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{2sin}\:\alpha\mathrm{sin}\:\beta+\mathrm{sin}\:^{\mathrm{2}} \beta=\mathrm{9}{p}}\\{\mathrm{cos}\:^{\mathrm{2}} \alpha+\mathrm{2cos}\:\alpha\mathrm{cos}\:\beta+\mathrm{cos}\:^{\mathrm{2}} \beta=\mathrm{9}{q}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{2}+\mathrm{2cos}\:\left(\alpha−\beta\right)=\mathrm{9}\left({p}+{q}\right)\: \\ $$$$\Leftrightarrow\:\mathrm{2cos}\:\left(\alpha−\beta\right)=\mathrm{9}\left({p}+{q}\right)−\mathrm{2} \\ $$$$\Rightarrow\mathrm{cos}\:\left(\alpha−\beta\right)=\frac{\mathrm{9}\left({p}+{q}\right)−\mathrm{2}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha−\beta\right)=\pm\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{9}\left({p}+{q}\right)−\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$

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