## Trigonometry (11th Edition) Clone

$$\cos210^\circ=\cos150^\circ\cos60^\circ-\sin150^\circ\sin60^\circ$$ 17 is matched with I.
$$\cos210^\circ$$ We can write $210^\circ$ as the sum of $150^\circ$ and $60^\circ$. Thus, $$\cos210^\circ=\cos(150^\circ+60^\circ)$$ Here we can apply the sum identity for cosine for $\cos(150^\circ+60^\circ)$. $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ So, if we replace $A=150^\circ$ and $B=60^\circ$, we would have $$\cos210^\circ=\cos150^\circ\cos60^\circ-\sin150^\circ\sin60^\circ$$ This fits with choice I. Therefore, we should match 17 with I.