## University Calculus: Early Transcendentals (3rd Edition)

$$\sin(2\pi-x)=-\sin x$$
$$\sin(2\pi-x)$$ Here we can use the identity that we have proved in Exercise 36: $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ Therefore: $$\sin(2\pi-x)=\sin2\pi\cos x-\cos2\pi\sin x$$ $$\sin(2\pi-x)=0\times\cos x-1\times\sin x$$ $$\sin(2\pi-x)=-\sin x$$