Prove that the given trigonometric expression $\sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{1-\sqrt{2}\sin {{35}^{\circ }}}{2\sqrt{2}}$ is valid .

Hint: Now consider the LHS of the equation. Now to the expression, we will use the formula
$2\cos A \sin B = \sin (A + B) – \sin (A - B)$ and form a new equation. Now again we know that $\sin (90 + x) = \cos x$. Again we will substitute this in the obtained equation and simplify. Now we will substitute the value of $\cos {{45}^{\circ }}$ which is nothing but $\dfrac{1}{\sqrt{2}}$ in the equation. Now we will take LCM in the numerator and divide the numerator and the denominator by 2 to arrive at the required equation.

Complete step-by-step solution:
Now consider the LHS of the given equation which is $\sin {{50}^{\circ }}\cos {{85}^{\circ }}$ .
We know that $2\cos A \sin B = \sin (A + B) – \sin (A - B)$
Hence we can say that $\cos A\sin B=\dfrac{\sin \left( A+B \right)-\sin \left( A-B \right)}{2}$ .
Now comparing the LHS of above equation with $\sin {{50}^{\circ }}\cos {{85}^{\circ }}$ we get $A={{85}^{\circ }}$ and $B={{50}^{\circ }}$
Hence using the formula by substituting the values of A and B we get,
$\begin{align}
  & \sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{\sin \left( {{85}^{\circ }}+{{50}^{\circ }} \right)-\sin \left( {{85}^{\circ }}-{{50}^{\circ }} \right)}{2} \\
 & \Rightarrow \sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{\sin \left( {{135}^{\circ }} \right)-\sin \left( {{35}^{\circ }} \right)}{2} \\
 & \Rightarrow \sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{\sin \left( {{90}^{\circ }}+{{45}^{\circ }} \right)-\sin \left( {{35}^{\circ }} \right)}{2} \\
\end{align}$
Now we know that the value of $\sin \left( 90+x \right)=\cos x$ .
Hence using this result in the above equation we get,
$\sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{\cos \left( {{45}^{\circ }} \right)-\sin \left( {{35}^{\circ }} \right)}{2}$
Now again we know that the value of $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ . Substituting the value in the above equation we get.
$\sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{\dfrac{1}{\sqrt{2}}-\sin \left( {{35}^{\circ }} \right)}{2}$
Now taking LCM in the numerator we get,
$\Rightarrow \sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{\dfrac{1-\sqrt{2}\sin \left( {{35}^{\circ }} \right)}{\sqrt{2}}}{2}$
Now dividing the numerator and denominator in the above equation by 2 we get,
$\Rightarrow \sin {{50}^{\circ }}\cos {{85}^{\circ }}=\dfrac{1-\sqrt{2}\sin \left( {{35}^{\circ }} \right)}{2\sqrt{2}}$
This is nothing but the RHS of the given equation.
Hence the result is proved.

Note: Now note that the formula form $2\cos A \sin B = \sin (A + B) – \sin (A - B)$,
$2\cos A \cos B = \cos (A + B) + \cos (A − B)$ and $2\sin A \sin B = \cos (A − B) − \cos (A + B).$ Since all these formulas are similar not to be confused among them.