## Trigonometry (11th Edition) Clone

Express $\cos(180^\circ-\theta)$ as a function of $\theta$ alone. $$\cos(180^\circ-\theta)=-\cos\theta$$
$$\cos(180^\circ-\theta)$$ To be expressed as a function of $\theta$ alone, we need here the identity for difference of cosines: $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ Apply it to $\cos(180^\circ-\theta)$, we have $$\cos(180^\circ-\theta)=\cos180^\circ\cos\theta+\sin180^\circ\sin\theta$$ $$\cos(180^\circ-\theta)=(-1)\times\cos\theta+0\times\sin\theta$$ $$\cos(180^\circ-\theta)=-\cos\theta$$ This is the ultimate function we need to find.