If $\alpha$ and $\beta$ are the solutions of the equations $$\sin^2x+a\sin x+b=0$$ and $$\cos^2x+c \cos x+d=0$$ then $\sin(\alpha+\beta)=$? I cannot figure out how to start the sum and proceed with it.
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3Do you mean that $\alpha$ and $\beta$ satisfy both equations or is $\alpha$ a solution of the former and $\beta$ of the latter? – egreg Jul 27 '18 at 11:56
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Hint :
$$\begin{align} &\sin^2 \alpha + a\sin \alpha + b = \sin^2 \beta + a\sin \beta + b = 0 \\&\to \sin^2 \alpha - \sin^2 \beta +a\sin \alpha - a\sin \beta = 0 \\&\to (\sin \alpha - \sin \beta)(\sin \alpha + \sin \beta + a) = 0\end{align}$$
In the same manner :
$(\cos \alpha - \cos \beta)(\cos \alpha + \cos \beta + c) = 0$
Davide Morgante
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S.H.W
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$$\implies\sin\alpha+\sin\beta=-\dfrac a1$$
and $$\cos\alpha+\cos\beta=-\dfrac c1$$
On division and using Prosthaphaeresis Formulas, $$\dfrac ac=\tan\dfrac{\alpha+\beta}2$$ assuming $\cos\dfrac{\alpha-\beta}2\ne0$
Now $\sin2A=\dfrac{2\tan A}{1+\tan^2A}$
lab bhattacharjee
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See also : https://math.stackexchange.com/questions/2021356/if-sin-alpha-sin-beta-a-and-cos-alpha-cos-beta-b-then and https://math.stackexchange.com/questions/1490652/prove-that-tan-alpha-beta-2ac-a2-c2 – lab bhattacharjee Jul 27 '18 at 12:37
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@labbhattacharjee the accepted answer clearly shows that alpha and beta are the roots as they have been replaced with x in the solution. – Alapan Bagchi Jul 27 '18 at 13:03
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