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

	Answered by bramlex last updated on 20/Jun/21
	$⇒ { ((sin^2 α+2sin αsin β+sin^2 β=9p)),((cos^2 α+2cos αcos β+cos^2 β=9q)) :} ⇔2+2cos (α−β)=9(p+q) ⇔ 2cos (α−β)=9(p+q)−2 ⇒cos (α−β)=((9(p+q)−2)/2) ⇒sin (α−β)=± (√(1−(((9(p+q)−2)/2))^2 ))$
	$\Rightarrow {\begin{cases} \sin^{2} α + 2 \sin α \sin β + \sin^{2} β = 9 p \\ \cos^{2} α + 2 \cos α \cos β + \cos^{2} β = 9 q \end{cases}$ $\Leftrightarrow 2 + 2 \cos (α - β) = 9 (p + q)$ $\Leftrightarrow 2 \cos (α - β) = 9 (p + q) - 2$ $\Rightarrow \cos (α - β) = \frac{9 (p + q) - 2}{2}$ $\Rightarrow \sin (α - β) = \pm \sqrt{1 - {(\frac{9 (p + q) - 2}{2})}^{2}}$
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