Answer
$$\sin40^\circ\cos50^\circ+\cos40^\circ\sin 50^\circ=1$$
Work Step by Step
$$X=\sin40^\circ\cos50^\circ+\cos40^\circ\sin 50^\circ$$
From the identity of the sum of sines:
$$\sin A\cos B+\cos A\sin B=\sin(A+B)$$
So here $X$ actually follows the above identity with $A=40^\circ$ and $B=50^\circ$.
Therefore, $X$ can also be rewritten as
$$X=\sin(40^\circ+50^\circ)$$
$$X=\sin90^\circ$$
$$X=1$$
In conclusion, $$\sin40^\circ\cos50^\circ+\cos40^\circ\sin 50^\circ=1$$
