How do you solve \[\sin \left( \alpha +\beta \right)\] when we are given \[\sin \alpha =\dfrac{12}{13}\] and \[\cos \beta =\dfrac{-4}{5}?\]

Hint: We are given the value of \[\sin \alpha \] and \[\cos \beta \] and we are asked to find the value of \[\sin \left( \alpha +\beta \right).\] To find the value of \[\sin \left( \alpha +\beta \right)\] we will first calculate the value of \[\cos \alpha \] and \[\sin \beta .\] To do so we will need to knowledge of quadrant and the sign of sin and cos in these quadrants. After that, we will find the value of \[\sin \left( \alpha +\beta \right).\] We will use \[\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \] to find the required value.

Complete step-by-step solution:
We are given \[\sin \alpha =\dfrac{12}{13}\] and we have \[\cos \beta =\dfrac{-4}{5}.\] We are asked to find the value of \[\sin \left( \alpha +\beta \right).\] We know that \[\sin \left( \alpha +\beta \right)\] is given as \[\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta .\] So to find the value of \[\sin \left( \alpha +\beta \right)\] we will first need to find the value of \[\sin \beta \] and \[\cos \alpha .\] Now, to calculate these value, we will use the identity given as \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.\] So, using this, we will find our value.
We will start with \[\sin \alpha =\dfrac{12}{13}.\] We have \[\sin \alpha =\dfrac{12}{13}.\] So using the identity \[{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\] and putting \[\sin \alpha =\dfrac{12}{13}\] in the above equation, we get,
\[{{\left( \dfrac{12}{13} \right)}^{2}}+{{\cos }^{2}}\alpha =1\]
So, we get,
\[\Rightarrow \dfrac{144}{169}+{{\cos }^{2}}\alpha =1\]
On simplifying, we get,
\[\Rightarrow \cos \alpha =1-\dfrac{144}{169}=\dfrac{25}{169}\]
So, we get,
\[\Rightarrow \cos \alpha =\pm \sqrt{\dfrac{25}{169}}\]
As, \[\sqrt{25}=5\] and \[\sqrt{169}=13,\] so, we get,
\[\Rightarrow \cos \alpha =\pm \dfrac{5}{13}\]
So, we have that for \[\sin \alpha =\dfrac{12}{13},\] we have \[\cos \alpha =\dfrac{5}{13}\] and also \[\cos \alpha =\dfrac{-5}{13}.....\left( i \right)\]
Now, we will use \[\cos \beta =\dfrac{-4}{5}\] and we know that \[{{\cos }^{2}}\beta +{{\sin }^{2}}\beta =1.\] So, using \[\cos \beta =\dfrac{-4}{5},\] we get,
\[\Rightarrow {{\left( \dfrac{-4}{5} \right)}^{2}}+{{\sin }^{2}}\beta =1\]
Simplifying, we will get,
\[\Rightarrow {{\sin }^{2}}\beta =1-\dfrac{16}{25}=\dfrac{9}{25}\]
So, we get,
\[\Rightarrow \sin \beta =\pm \dfrac{3}{5}\]
Now, \[\cos \beta =\dfrac{-4}{5}\] is possible in the second quadrant and third quadrant and sin is positive as well as negative in these two quadrants. So, \[\sin \beta =\dfrac{+3}{5}\] and \[\sin \beta =\dfrac{-3}{5}\] and both are possible.
So, for \[\cos \beta =\dfrac{-4}{5}\] we get, \[\sin \beta =\dfrac{3}{5}\] and \[\sin \beta =\dfrac{-3}{5}.....\left( ii \right)\]
So, we get, for the set of values and we will list them as
\[\text{A}\text{. }\sin \alpha =\dfrac{12}{13},\cos \alpha =\dfrac{5}{13},\sin \beta =\dfrac{3}{5},\cos \beta =\dfrac{-4}{5}\]
\[\text{B}\text{. }\sin \alpha =\dfrac{12}{13},\cos \alpha =\dfrac{-5}{13},\sin \beta =\dfrac{3}{5},\cos \beta =\dfrac{-4}{5}\]
\[\text{C}\text{. }\sin \alpha =\dfrac{12}{13},\cos \alpha =\dfrac{5}{13},\sin \beta =\dfrac{-3}{5},\cos \beta =\dfrac{-4}{5}\]
\[\text{D}\text{. }\sin \alpha =\dfrac{12}{13},\cos \alpha =\dfrac{-5}{13},\sin \beta =\dfrac{-3}{5},\cos \beta =\dfrac{-4}{5}\]
Now we will use these values on the formula \[\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \] and find our answer. We will use the value one by one.
a. For part A,
\[\sin \left( \alpha +\beta \right)=\dfrac{12}{13}\times \left( \dfrac{-4}{5} \right)+\dfrac{5}{13}\times \dfrac{3}{5}\]
So, we get,
\[\Rightarrow \sin \left( \alpha +\beta \right)=\dfrac{-33}{65}\]
b. For part B,
\[\sin \left( \alpha +\beta \right)=\dfrac{12}{13}\times \left( \dfrac{-4}{5} \right)+\left( \dfrac{-5}{13} \right)\times \dfrac{3}{5}\]
\[\Rightarrow \sin \left( \alpha +\beta \right)=\dfrac{-63}{65}\]
c. For part C,
\[\sin \left( \alpha +\beta \right)=\dfrac{12}{13}\times \left( \dfrac{-4}{5} \right)+\left( \dfrac{5}{13} \right)\times \left( \dfrac{-3}{5} \right)\]
\[\Rightarrow \sin \left( \alpha +\beta \right)=\dfrac{-63}{65}\]
d. For part D,
\[\sin \left( \alpha +\beta \right)=\dfrac{12}{13}\times \left( \dfrac{-4}{5} \right)+\left( \dfrac{-5}{13} \right)\times \left( \dfrac{-3}{5} \right)\]
\[\Rightarrow \sin \left( \alpha +\beta \right)=\dfrac{-33}{65}\]
Hence, we get two solutions. We have the value of \[\sin \left( \alpha +\beta \right)\] as \[\dfrac{-63}{65}\] and \[\sin \left( \alpha +\beta \right)\] as \[\dfrac{-33}{65}.\]

Note: While solving the square root we need to be careful with the sign when we do root, it will give positive and negative values. We need to compare which out of the positive or negative is possible by checking the quadrant in which sin and cos are taking positive or negative values. We have a table showing sign as