## Trigonometry (11th Edition) Clone

$$\sin100^\circ\cos10^\circ-\cos100^\circ\sin10^\circ=1$$
$$X=\sin100^\circ\cos10^\circ-\cos100^\circ\sin10^\circ$$ Recall the sine difference identity that $$\sin A\cos B-\cos A\sin B=\sin(A-B)$$ Looking back at $X$, we find $X$ is indeed the above identity with $A=100^\circ$ and $B=10^\circ$. Therefore, we can rewrite $X$ as $$X=\sin(100^\circ-10^\circ)$$ $$X=\sin90^\circ$$ $$X=1$$ In conclusion, $$\sin100^\circ\cos10^\circ-\cos100^\circ\sin10^\circ=1$$