# Computing $\cos(A+B)$ in a triangle

I came across the following problem.

In $\triangle ABC, \sin A:\sin B : \sin C = 5 : 7 : 9$. Compute $\cos(A + B)$.

The first thing I did was rewrite $\cos(A+B)$ as $\cos A\cos B-\sin A\sin B$.

Using the Law of Sines, I let $a=5,b=7,c=9$. Using the Law of Cosines on angles $A$ and $B$ to obtain

$$\begin{cases}5^2=7^2+9^2-2(7)(9)\cos A\\7^2=5^2+9^2-2(5)(9)\cos B\end{cases}$$

Multiplying the two equations together and simplifying produces

$$\cos A \cos B = \frac{19}{36}$$

However, I can't find $\sin A \sin B$. Is there a way to find it?

Also, would there be another way of solving the problem? Thank you!

Notice $A + B + C = \pi$. By cosine rule, we have $$\cos(A+B) = \cos(\pi - C) = -\cos C = \frac{c^2-a^2-b^2}{2ab}$$ By sine rule, we have $$a : b : c = \sin A : \sin B : \sin C = 5 : 7 :9$$ This means $$\cos(A+B) = \frac{9^2 - 5^2 - 7^2}{2\cdot 5 \cdot 7} = \frac{1}{10}$$
We have $\cos(A+B) = \cos(\pi - C) = -\cos(C)$. By the law of sines, a:b:c = 5:7:9 too. Since similar triangles have the same angles, to compute $\cos(C)$, we can assume $a=5$, $b=7$, and $c=9$. Then using the law of cosines we have $9^2 = 5^2 + 7^2 - 2(5)(7)cos(C).$ This gives $cos(C) = -1/10$. So the answer to your question is $-\cos(C) = 1/10.$
we have $$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$ from $$\frac{\sin(A)}{\sin(B)}=\frac{5}{7}$$ we get $$\cos(A)\cos(B)-\frac{5}{7}\sin^2(B)=\cos(A)\cos(B)-\frac{5}{7}(1-\cos^2(B))$$ here you can plug in your values, and we get $$\frac{1}{10}$$