## Trigonometry (11th Edition) Clone

$$\sin40^\circ\cos50^\circ+\cos40^\circ\sin 50^\circ=1$$
$$X=\sin40^\circ\cos50^\circ+\cos40^\circ\sin 50^\circ$$ 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, $$\sin40^\circ\cos50^\circ+\cos40^\circ\sin 50^\circ=1$$