How do I determine the sign of $\sin\theta$ in this question. 
$\alpha$ and $\beta$ are two different values of $\theta$ lying between $0$ and $2\pi$ which satisfy the equation $\begin{equation}6\cos\theta + 8\sin\theta = 9\end{equation}$. Find $\sin(\alpha+\beta).$  

I solved and got $\cos(\alpha+\beta) = \frac{-28}{100}$.
So $\sin(\alpha+\beta) = \pm\frac{24}{25}$.
I think the sign of $\sin(\alpha+\beta)$ could be either of '+' and '-', however my book says it must be '+'. How do I determine the sign?
 A: Your book is correct. We don't need to find either $\alpha$ or $\beta$, i.e. to solve the given linear equation in $\cos\theta$ and $\sin\theta$, which could be transformed into a quadratic equation in $\tan\frac{\theta}{2}$. 
[EDIT. Since you've commented that your course was a beginner's one, I edited this answer to add the derivation of all trigonomatric identities I used, from the addition and subtraction formulas.] 
Equating the equations satisfied by $\alpha$ and $\beta$
\begin{eqnarray*}
6\cos \alpha +8\sin \alpha  &=&9\tag{1} \\
6\cos \beta +8\sin \beta  &=&9,\tag{2}
\end{eqnarray*}
and then simplyfying and rearranging, we get
\begin{eqnarray*}
3\left( \cos \alpha -\cos \beta \right)  &=&4\left( \sin \beta -\sin \alpha
\right)  .\tag{3}
\end{eqnarray*}
Applying the sum-to-product identities
\begin{eqnarray*}
\cos \alpha -\cos \beta  &=&-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha
-\beta }{2} \tag{4}\\
\sin \beta -\sin \alpha  &=&-2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha
+\beta }{2},\tag{5}
\end{eqnarray*}
assuming$^1$ that  $\sin \frac{\alpha-\beta }{2}\ne 0 $ we have that
\begin{eqnarray*}
-3\times 2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}
&=&-4\times 2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}\tag{6} \\
3\sin \frac{\alpha +\beta }{2} &=&4\cos \frac{\alpha +\beta }{2}\tag{7} \\
\tan \frac{\alpha +\beta }{2} &=&\frac{4}{3}.\tag{8}
\end{eqnarray*}
Finally expressing $\sin \left( \alpha +\beta \right) $ in terms of $\tan 
\frac{\alpha +\beta }{2}$, using the double-angle identity we find 
\begin{equation*}
\sin \left( \alpha +\beta \right) =\frac{2\tan \frac{\alpha +\beta }{2}}{
1+\tan ^{2}\frac{\alpha +\beta }{2}}=\frac{2\left( \frac{4}{3}\right) }{
1+\left( \frac{4}{3}\right) ^{2}}=\frac{24}{25},\tag{9}
\end{equation*}
which agrees with the $\cos \left( \alpha +\beta \right) $ value you found:
\begin{equation*}
\cos \left( \alpha +\beta \right) =\frac{1-\tan ^{2}\frac{\alpha +\beta }{2}
}{1+\tan ^{2}\frac{\alpha +\beta }{2}}=\frac{1-\left( \frac{4}{3}\right) ^{2}
}{1+\left( \frac{4}{3}\right) ^{2}}=-\frac{7}{25}.\tag{10}
\end{equation*}
EDIT. Justification of:


*

*Identity $(4)$
\begin{eqnarray*}
\cos \left( a\pm b\right)  &=&\cos a\cos b\mp \sin a\sin b \\
&&\text{If }\left\{ 
\begin{array}{c}
a+b=\alpha  \\ 
a-b=\beta ,
\end{array}
\right. \text{ then }\left\{ 
\begin{array}{c}
a=\frac{\alpha +\beta }{2} \\ 
b=\frac{\alpha -\beta }{2}
\end{array}
\right. \text{ and} \\
\cos \alpha  &=&\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
-\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \\
\cos \beta  &=&\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
+\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2} \\
\cos \alpha -\cos \beta  &=&-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha
-\beta }{2}.
\end{eqnarray*}

*Identity $(5)$
\begin{eqnarray*}
\sin \left( a\pm b\right)  &=&\sin a\cos b\pm \sin b\cos a \\
&&\text{If }\left\{ 
\begin{array}{c}
a+b=\alpha  \\ 
a-b=\beta ,
\end{array}
\right. \text{ then }\left\{ 
\begin{array}{c}
a=\frac{\alpha +\beta }{2} \\ 
b=\frac{\alpha -\beta }{2}
\end{array}
\right. \text{ and} \\
\sin \alpha  &=&\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
+\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\
\sin \beta  &=&\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}
-\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\
\sin \beta -\sin \alpha  &=&-2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha
+\beta }{2}.
\end{eqnarray*}

*identities $(9)$ and $(10)$
\begin{eqnarray*}
\sin \left( \alpha +\beta \right)  &=&2\sin \frac{\alpha +\beta }{2}\cos 
\frac{\alpha +\beta }{2}=\frac{2\sin \frac{\alpha +\beta }{2}\cos \frac{
\alpha +\beta }{2}}{\cos ^{2}\frac{\alpha +\beta }{2}+\sin ^{2}\frac{\alpha
+\beta }{2}} \\
&=&\frac{2\tan \frac{\alpha +\beta }{2}}{1+\tan ^{2}\frac{\alpha +\beta }{2}}
\\
&& \\
\cos \left( \alpha +\beta \right)  &=&\cos ^{2}\frac{\alpha +\beta }{2}-\sin
^{2}\frac{\alpha +\beta }{2}=\frac{\cos ^{2}\frac{\alpha +\beta }{2}-\sin
^{2}\frac{\alpha +\beta }{2}}{\cos ^{2}\frac{\alpha +\beta }{2}+\sin ^{2}
\frac{\alpha +\beta }{2}} \\
&=&\frac{1-\tan ^{2}\frac{\alpha +\beta }{2}}{1+\tan ^{2}\frac{\alpha +\beta 
}{2}}
\end{eqnarray*}


--
$^1$ The assumption $\sin \frac{\alpha -\beta }{2}\neq 0$ is valid because if $
0\neq \frac{\alpha -\beta }{2}=\pi $, then without loss of generality we might
assume that $0=\beta <\alpha =2\pi $, but we would get a contradition:
\begin{eqnarray*}
6\cos \left( 2\pi \right) +8\sin \left( 2\pi \right)  &\neq &9 \\
6\cos \left( 0\right) +8\sin \left( 0\right)  &\neq &9.
\end{eqnarray*}
A: HINT:
You won't face such confusion if you adapt the following method:
Use Weierstrass substitution  to form a Quadratic Equation  in $\tan \frac x2$
Then use Trigonometric Addition Formulas to find $\displaystyle\tan\left(\frac{\alpha+\beta}2\right)$
Finally use this
A: $$\sin \theta=\frac{2\tan \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}}$$
$$\cos \theta=\frac{1-\tan^2 \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}}$$
Substitute the above two equations in the main equation and let $\tan \frac{\theta}{2} = x$. Now you get a quadratic equation $x$ from where you can find $x=\tan \frac{\theta}{2}$.
Then find $\sin (\alpha+\beta)$ from here.
