Given $\cos 40 = m$ and $\sin 10 = n$ , can you please express $\sin 50$ in terms of $m$ and $n$ ?

Hint: First, we shall analyze the given data so that we are able to solve the given problem. Here we are given some values and we need to express $\sin 50$ in terms of $m$ and $n$. Before getting into a solution, we need to find the values of $\sin 40$ and $\cos 10$. Then we shall apply these values in $\sin 50$ to obtain the required answer.

Formula to be used:
The required formula to be applied to this problem is as follows.
a) ${\sin ^2}\theta + {\cos ^2}\theta = 1$
b) $\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$

Complete step-by-step answer:
We are given $\cos 40 = m$and $\sin 10 = n$. Then we are asked to express $\sin 50$ in terms of $m$ and $n$.
Before getting into the solution, we need to calculate the values of $\sin 40$ and $\cos 10$.
We all know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Now, we can apply the angle $40$ in the above formula.
Thus, we have ${\sin ^2}40 + {\cos ^2}40 = 1$
\[ \Rightarrow {\sin ^2}40 = 1 - {\cos ^2}40\]
\[ \Rightarrow {\sin ^2}40 = 1 - {m^2}\] (We are given $\cos 40 = m$)
\[ \Rightarrow \sqrt {{{\sin }^2}40} = \sqrt {1 - {m^2}} \] (Here we have taken square roots on both sides of the equation)
\[ \Rightarrow \sin 40 = \sqrt {1 - {m^2}} \]
Thus, we get \[\sin 40 = \sqrt {1 - {m^2}} \]
Similarly, we shall apply the angle $10$ in the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Thus, we have ${\sin ^2}10 + {\cos ^2}10 = 1$
\[ \Rightarrow {\cos ^2}10 = 1 - {\sin ^2}10\]
\[ \Rightarrow {\cos ^2}10 = 1 - {n^2}\] (We are given $\sin 10 = m$)
\[ \Rightarrow \sqrt {{{\cos }^2}10} = \sqrt {1 - {n^2}} \] (Here we have taken square roots on both sides of the equation)
\[ \Rightarrow \cos 10 = \sqrt {1 - {n^2}} \]
Thus, we get \[\cos 10 = \sqrt {1 - {n^2}} \]
Now, we shall get into our solution.
We need to calculate the value of $\sin 50$ in terms of $m$ and $n$.
$\sin 50 = \sin \left( {40 + 10} \right)$ (We have separated the angle for our convenience.)
              $ = \sin 40\cos 10 + \cos 40\sin 10$ (Here we have applied $\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b$)
We have found the required values \[\sin 40 = \sqrt {1 - {m^2}} \]and \[\cos 10 = \sqrt {1 - {n^2}} \]. Also, we are given $\cos 40 = m$ and $\sin 10 = n$. We shall substitute these values in the above equation.
Thus, we have $\sin 50$$ = \sqrt {1 - {m^2}} \sqrt {1 - {n^2}} + mn$ and we have found the required answer.
Therefore, we can express $\sin 50$$ = \sqrt {1 - {m^2}} \sqrt {1 - {n^2}} + mn$ in terms of $m$ and $n$.

Note: Here we have found the values of $\sin 40$ and $\cos 10$by using the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$. We can also find it by using two separate formulae. We can find $\sin 40$by using the formula \[\sin \theta = \sqrt {1 - {{\cos }^2}\theta } \] and $\cos 10$by the formula \[\cos \theta = \sqrt {1 - {{\sin }^2}\theta } \] . Thus, we get the same values \[\sin 40 = \sqrt {1 - {m^2}} \]and \[\cos 10 = \sqrt {1 - {n^2}} \].